Redox Reactions and Electrochemistry MCQs with Answers – Part 7 (NEET Chemistry)

Explanation: \(\textbf{Use oxygen consumption to determine the cathode extent:}\) The oxygen-electrode reaction consumes one mole of O2 for each stoichiometric event: \[ \mathrm{O_2+2H_2O+4e^-\rightarrow4OH^-} \] Thus, the number of moles of this reaction occurring is: \[ \xi_{\mathrm{cathode}}=0.0250\ \mathrm{mol} \] \(\textbf{Determine the electron transfer:}\) Four moles of electrons are accepted per mole of oxygen: \[ n(e^-)=4(0.0250) \] \[ n(e^-)=0.100\ \mathrm{mol} \] \(\textbf{Calculate water consumed at the oxygen electrode:}\) The cathode consumes two water molecules per oxygen molecule. Therefore: \[ n(\mathrm{H_2O})_{\mathrm{cathode,consumed}}=2(0.0250) \] \[ n(\mathrm{H_2O})_{\mathrm{cathode,consumed}}=0.0500\ \mathrm{mol} \] \(\textbf{Determine the hydrogen-electrode reaction extent:}\) The anode produces four electrons per occurrence of: \[ \mathrm{2H_2+4OH^-\rightarrow4H_2O+4e^-} \] Consequently: \[ \xi_{\mathrm{anode}}=\frac{0.100}{4} \] \[ \xi_{\mathrm{anode}}=0.0250\ \mathrm{mol} \] \(\textbf{Calculate water formed directly at the hydrogen electrode:}\) Four moles of water form per mole of the displayed anode reaction. Thus: \[ n(\mathrm{H_2O})_{\mathrm{anode,formed}}=4(0.0250) \] \[ n(\mathrm{H_2O})_{\mathrm{anode,formed}}=0.100\ \mathrm{mol} \] \(\textbf{Determine the net water production:}\) Water appears on opposite sides of the two half-reactions. The net amount formed is: \[ n(\mathrm{H_2O})_{\mathrm{net}}=0.100-0.0500 \] \[ n(\mathrm{H_2O})_{\mathrm{net}}=0.0500\ \mathrm{mol} \] \(\textbf{Verify with the overall molecular reaction:}\) Adding the half-reactions gives: \[ \mathrm{2H_2+O_2\rightarrow2H_2O} \] Therefore: \[ n(\mathrm{H_2O})_{\mathrm{net}}=2n(\mathrm{O_2}) \] \[ n(\mathrm{H_2O})_{\mathrm{net}}=2(0.0250) \] \[ n(\mathrm{H_2O})_{\mathrm{net}}=0.0500\ \mathrm{mol} \] \(\textbf{Interpret local versus overall water production:}\) The hydrogen electrode produces more water than the complete cell forms overall. This is not a contradiction because the oxygen electrode simultaneously consumes water. The overall balanced equation contains only the difference between these local contributions. Reading the overall equation alone would not reveal the amount of water generated specifically at the hydrogen electrode. Conversely, using only the anode equation would overstate net water production. The result requires combining the two alkaline half-reactions while preserving their local chemical meanings. \(\textbf{Final answer:}\) \[ \boxed{\left(n(\mathrm{H_2O})_{\mathrm{anode,formed}},n(\mathrm{H_2O})_{\mathrm{net}}\right)=(0.100,\ 0.0500)\ \mathrm{mol}} \]

Explanation: \(\textbf{Identify the complete fuel-cell reaction quotient:}\) For the balanced reaction: \[ \mathrm{2H_2+O_2\rightarrow2H_2O} \] the reaction quotient is: \[ Q_r=\frac{a_{\mathrm{H_2O}}^2}{(p_{\mathrm{H_2}}/p^\circ)^2(p_{\mathrm{O_2}}/p^\circ)} \] The water activity appears squared because two moles of water form per mole of reaction as written. \(\textbf{Evaluate the initial quotient:}\) Initially all the relevant relative activities equal unity: \[ Q_i=\frac{1.00^2}{1.00^2(1.00)} \] \[ Q_i=1 \] \(\textbf{Apply the unchanged-emf requirement:}\) The Nernst equation is: \[ E=E^\circ-\frac{0.0591}{4}\log Q_r \] At fixed temperature and unchanged reaction stoichiometry, equal values of the reversible emf require equal reaction quotients. Therefore: \[ Q_f=Q_i=1 \] \(\textbf{Substitute the final gas conditions:}\) Let the required final water activity be \(a_w\). Then: \[ Q_f=\frac{a_w^2}{(1.00)^2(0.0100)} \] The unchanged-emf condition gives: \[ 1=\frac{a_w^2}{0.0100} \] \(\textbf{Solve for water activity:}\) \[ a_w^2=0.0100 \] \[ a_w=\sqrt{0.0100} \] \[ a_w=0.100 \] The positive root is the physically meaningful activity. \(\textbf{Verify the compensating Nernst effects:}\) Decreasing oxygen relative pressure from \(1.00\) to \(0.0100\) increases the reaction quotient by a factor of one hundred if other activities remain unchanged. Decreasing water activity from \(1.00\) to \(0.100\) decreases its squared quotient contribution by a factor of one hundred. The two changes cancel: \[ Q_f=\frac{(0.100)^2}{0.0100} \] \[ Q_f=1 \] \(\textbf{Interpret the chemical compensation:}\) Reducing oxygen pressure alone would reduce the forward fuel-cell potential because oxygen is a reactant. Reducing the activity of the water product instead favours water formation and increases the forward potential. Because water has stoichiometric coefficient two, the required change in its activity is the square root of the oxygen-pressure factor. The emf can therefore remain unchanged even when individual chemical activities change substantially. This conclusion concerns the stipulated reversible activity model, not an assumption that pure liquid water necessarily has a variable activity under standard conditions. \(\textbf{Final answer:}\) \[ \boxed{a_{\mathrm{H_2O},f}=0.100} \]

Explanation: \(\textbf{Determine the net voltage driving charging:}\) The two dry-cell emfs reinforce each other in series: \[ E_{\mathrm{dry,total}}=2(1.50) \] \[ E_{\mathrm{dry,total}}=3.00\ \mathrm V \] The lead accumulator opposes the charging current with \(2.00\ \mathrm V\). The available net driving voltage is therefore: \[ E_{\mathrm{net}}=3.00-2.00 \] \[ E_{\mathrm{net}}=1.00\ \mathrm V \] \(\textbf{Determine the complete series resistance:}\) Both dry-cell internal resistances and the accumulator's effective resistance contribute: \[ R_{\mathrm{total}}=0.250+0.250+0.500 \] \[ R_{\mathrm{total}}=1.00\ \Omega \] \(\textbf{Calculate the charging current:}\) \[ I=\frac{E_{\mathrm{net}}}{R_{\mathrm{total}}} \] \[ I=\frac{1.00}{1.00} \] \[ I=1.00\ \mathrm A \] \(\textbf{Determine charge transferred during the interval:}\) \[ Q=It \] \[ Q=(1.00)(1930) \] \[ Q=1930\ \mathrm C \] \(\textbf{Convert external charge to electron amount:}\) \[ n(e^-)=\frac QF \] \[ n(e^-)=\frac{1930}{96500} \] \[ n(e^-)=0.0200\ \mathrm{mol} \] \(\textbf{Calculate acid regenerated in the accumulator:}\) The overall charging reaction produces two moles of H2SO4 per two moles of transferred electrons. Thus: \[ n(\mathrm{H_2SO_4})_{\mathrm{regenerated}}=0.0200\ \mathrm{mol} \] \(\textbf{Determine zinc oxidation in each dry cell:}\) The same electrical charge passes through each of the two series-connected dry cells. It is not divided between them. For either zinc anode: \[ n(\mathrm{Zn})_{\mathrm{consumed}}=\frac{n(e^-)}2 \] \[ n(\mathrm{Zn})_{\mathrm{consumed}}=\frac{0.0200}{2} \] \[ n(\mathrm{Zn})_{\mathrm{consumed}}=0.0100\ \mathrm{mol} \] \(\textbf{Calculate zinc mass consumed in each cell:}\) \[ m(\mathrm{Zn})=(0.0100)(65.4) \] \[ m(\mathrm{Zn})=0.654\ \mathrm g \] \(\textbf{Verify the series-device comparison:}\) Each dry cell consumes \(0.0100\ \mathrm{mol}\) of zinc. Together, the two dry cells consume \(0.0200\ \mathrm{mol}\) of zinc, but each transfers the full \(0.0200\ \mathrm{mol}\) of electrons through its own electrode process. The lead accumulator likewise receives that electron amount in the charging direction. \(\textbf{Interpret the different energy-device roles:}\) The dry cells undergo their spontaneous primary-cell discharge chemistry and supply electrical energy. The lead accumulator undergoes the reverse of its spontaneous discharge reaction and stores chemical energy. Series connection adds the dry-cell emfs but not their charge capacities. The resulting current depends on the excess applied emf after accounting for the opposing accumulator voltage and all series resistances. The chemical amounts follow from the common current and each device's own electrode electron stoichiometry. \(\textbf{Final answer:}\) \[ \boxed{\left(n(\mathrm{H_2SO_4})_{\mathrm{regenerated}},m(\mathrm{Zn})_{\mathrm{per\ dry\ cell}}\right)=(0.0200\ \mathrm{mol},\ 0.654\ \mathrm g)} \]

Explanation: \(\textbf{Extract the cell emf from the experimental graph:}\) The terminal-voltage relation is: \[ V=E-Ir \] At zero discharge current, there is no internal ohmic voltage loss in this model. Therefore, the vertical intercept gives: \[ E=2.10\ \mathrm V \] \(\textbf{Determine internal resistance from the graph slope:}\) The slope of the voltage–current line is: \[ \frac{\Delta V}{\Delta I}=-r \] Using the two specified points: \[ -r=\frac{1.90-2.10}{4.00-0} \] \[ -r=-0.0500\ \Omega \] \[ r=0.0500\ \Omega \] \(\textbf{Calculate the current with the specified resistor:}\) The internal resistance and external resistor are in series. Thus: \[ I=\frac{E}{R+r} \] \[ I=\frac{2.10}{0.650+0.0500} \] \[ I=\frac{2.10}{0.700} \] \[ I=3.00\ \mathrm A \] \(\textbf{Determine the loaded terminal voltage:}\) \[ V=IR \] \[ V=(3.00)(0.650) \] \[ V=1.95\ \mathrm V \] The same result follows from: \[ V=2.10-(3.00)(0.0500) \] \[ V=1.95\ \mathrm V \] \(\textbf{Calculate total transferred charge:}\) \[ Q=It \] \[ Q=(3.00)(965) \] \[ Q=2895\ \mathrm C \] \(\textbf{Convert the charge to electron amount:}\) \[ n(e^-)=\frac QF \] \[ n(e^-)=\frac{2895}{96500} \] \[ n(e^-)=0.0300\ \mathrm{mol} \] \(\textbf{Determine the discharge reaction extent:}\) The balanced accumulator reaction transfers two electrons per mole of reaction. Therefore: \[ \xi=\frac{0.0300}{2} \] \[ \xi=0.0150\ \mathrm{mol} \] \(\textbf{Calculate acid consumption:}\) Two moles of H2SO4 are consumed per mole of reaction. Thus: \[ n(\mathrm{H_2SO_4})_{\mathrm{consumed}}=2\xi \] \[ n(\mathrm{H_2SO_4})_{\mathrm{consumed}}=2(0.0150) \] \[ n(\mathrm{H_2SO_4})_{\mathrm{consumed}}=0.0300\ \mathrm{mol} \] \(\textbf{Interpret the combined measurement:}\) The voltage–current graph provides the electrical properties required to determine the operating current. The circuit then fixes the electrical charge delivered over the specified time. The amount of sulphuric acid consumed is determined by the chemical electron stoichiometry of the accumulator. A lower terminal voltage does not directly specify a particular amount of acid consumption; the current and operating time must also be known. The problem therefore connects an electrical measurement with the material change accompanying battery discharge. \(\textbf{Final answer:}\) \[ \boxed{\left(V_{\mathrm{terminal}},n(\mathrm{H_2SO_4})_{\mathrm{consumed}}\right)=(1.95\ \mathrm V,\ 0.0300\ \mathrm{mol})} \]

Explanation: \(\textbf{Statement I — Correct:}\) A hydrogen–oxygen fuel cell requires continuing delivery of its electrochemical reactants. Hydrogen undergoes oxidation, while oxygen undergoes reduction. When a required reactant feed is interrupted, the corresponding electrode process cannot continue indefinitely at its former rate. Restoring the appropriate feeds can permit electrical output to resume, provided the electrodes and other components remain functional. The observation for P is therefore consistent with a fuel cell. \(\textbf{Statement II — Correct:}\) Q contains electrode materials characteristic of a discharged lead accumulator. During discharge, Pb and PbO2 are converted to PbSO4. During suitable charging, the reverse reactions restore Pb at the negative plate and PbO2 at the positive plate. This chemically reversible operating cycle is the defining useful behaviour of the secondary cell described. \(\textbf{Statement III — Incorrect:}\) R has the stated construction of a conventional zinc–carbon dry cell. Its zinc container is an active electrode rather than merely an inert container. During discharge: \[ \mathrm{Zn\rightarrow Zn^{2+}+2e^-} \] Zinc is therefore oxidised and consumed. The primary-cell classification does not mean that its electrode materials remain chemically unchanged; it means that normal electrical regeneration is not the intended operating mode. \(\textbf{Statement IV — Correct:}\) P can resume operation because fresh chemical reactants are supplied to its electrodes. In Q, external electrical energy drives conversion of accumulated discharge products back into the charged electrode materials. The two procedures involve distinct material and energy flows. A fuel cell is not charged in the same way as a lead accumulator merely because both devices can operate again after an interruption. \(\textbf{Interpret the comparative evidence:}\) The recovery observations identify how each device accesses or restores chemical reactants. P depends on ongoing fuel and oxidant delivery, Q on reversible internal active-material conversion, and R on a finite primary-cell reaction inventory. The fact that all three deliver electrical energy during normal operation does not make their chemical maintenance procedures interchangeable. Statements I, II and IV are correct.

Explanation: \(\textbf{Determine the fuel-cell oxygen requirement:}\) The overall fuel-cell reaction consumes one mole of O2 for every two moles of H2. Therefore: \[ n(\mathrm{O_2})_{\mathrm{consumed}}=\frac{n(\mathrm{H_2})_{\mathrm{consumed}}}{2} \] \[ n(\mathrm{O_2})_{\mathrm{consumed}}=\frac{0.0400}{2} \] \[ n(\mathrm{O_2})_{\mathrm{consumed}}=0.0200\ \mathrm{mol} \] \(\textbf{Determine fuel-cell electron transfer:}\) Hydrogen oxidation is a two-electron process per hydrogen molecule: \[ \mathrm{H_2\rightarrow2H^++2e^-} \] Consequently: \[ n(e^-)_{\mathrm{fuel}}=2(0.0400) \] \[ n(e^-)_{\mathrm{fuel}}=0.0800\ \mathrm{mol} \] \(\textbf{Determine the charge passing through both cells:}\) The same external charge passes through the coupled devices. Thus: \[ Q=(0.0800)(96500) \] \[ Q=7720\ \mathrm C \] \(\textbf{Apply the chlorine-evolution current efficiency:}\) Only \(75.0\%\) of the anodic charge produces chlorine. Therefore: \[ n(e^-)_{\mathrm{chlorine}}=0.750(0.0800) \] \[ n(e^-)_{\mathrm{chlorine}}=0.0600\ \mathrm{mol} \] \(\textbf{Convert productive electrons into chlorine molecules:}\) The chlorine-evolution half-reaction is: \[ \mathrm{2Cl^-\rightarrow Cl_2+2e^-} \] Two electrons are released per chlorine molecule. Therefore: \[ n(\mathrm{Cl_2})=\frac{0.0600}{2} \] \[ n(\mathrm{Cl_2})=0.0300\ \mathrm{mol} \] \(\textbf{Check the charge allocated to the side reaction:}\) The remaining anodic electron-equivalent amount is: \[ n(e^-)_{\mathrm{side}}=0.0800-0.0600 \] \[ n(e^-)_{\mathrm{side}}=0.0200\ \mathrm{mol} \] This accounts for the \(25.0\%\) of anodic charge that does not yield chlorine. \(\textbf{Verify the fuel-cell cathode electron balance:}\) The fuel cell consumes \(0.0200\ \mathrm{mol}\) of oxygen. Oxygen reduction accepts four electrons per oxygen molecule: \[ 4(0.0200)=0.0800\ \mathrm{mol\ e^-} \] This equals the electron amount supplied by hydrogen oxidation. \(\textbf{Interpret the coupled chemical processes:}\) The fuel cell converts the chemical energy of hydrogen and oxygen into electrical energy. The electrolyser uses that electrical energy to drive a non-spontaneous electrode process. The common external charge is fixed by the fuel-cell hydrogen consumption under complete current efficiency. However, the amount of chlorine formed is smaller than its ideal Faraday-law value because only part of the electrolyser's anodic current produces chlorine. The two devices therefore have different chemical products and different current-efficiency conditions despite carrying the same electrical charge. \(\textbf{Final answer:}\) \[ \boxed{\left(n(\mathrm{O_2})_{\mathrm{fuel}},n(\mathrm{Cl_2})_{\mathrm{electrolyser}}\right)=(0.0200,\ 0.0300)\ \mathrm{mol}} \]

Explanation: \(\textbf{Determine the useful charging reaction extent:}\) The intended reverse accumulator reaction is: \[ \mathrm{2PbSO_4+2H_2O\rightarrow Pb+PbO_2+2H_2SO_4} \] Two moles of PbSO4 are reconverted per two moles of electrons transferred. Thus, the useful electron amount required to restore all the stated PbSO4 is: \[ n(e^-)_{\mathrm{restoration}}=0.0400\ \mathrm{mol} \] \(\textbf{Calculate sulphuric acid regenerated:}\) The reverse reaction produces two moles of H2SO4 per two moles of PbSO4 reconverted. Therefore: \[ n(\mathrm{H_2SO_4})_{\mathrm{regenerated}}=0.0400\ \mathrm{mol} \] \(\textbf{Determine the charge available after complete restoration:}\) The total electron amount supplied is \(0.0600\ \mathrm{mol}\). After restoration: \[ n(e^-)_{\mathrm{excess}}=0.0600-0.0400 \] \[ n(e^-)_{\mathrm{excess}}=0.0200\ \mathrm{mol} \] \(\textbf{Identify the subsequent oxygen-evolution stoichiometry:}\) During the stipulated water-electrolysis stage, oxygen evolves at the oxidation electrode. In acidic notation: \[ \mathrm{2H_2O\rightarrow O_2+4H^++4e^-} \] Four moles of electrons are released per mole of oxygen produced. \(\textbf{Calculate oxygen evolution after restoration:}\) \[ n(\mathrm{O_2})=\frac{n(e^-)_{\mathrm{excess}}}{4} \] \[ n(\mathrm{O_2})=\frac{0.0200}{4} \] \[ n(\mathrm{O_2})=0.00500\ \mathrm{mol} \] \(\textbf{Calculate accompanying hydrogen evolution as a check:}\) At the cathode, water or hydrogen ions are reduced to hydrogen. Two electrons correspond to one hydrogen molecule. Therefore: \[ n(\mathrm{H_2})_{\mathrm{overcharge}}=\frac{0.0200}{2} \] \[ n(\mathrm{H_2})_{\mathrm{overcharge}}=0.0100\ \mathrm{mol} \] The gas ratio is: \[ n(\mathrm{H_2}):n(\mathrm{O_2})=2:1 \] \(\textbf{Check the allocation of all supplied electrons:}\) The active-material restoration consumes: \[ 0.0400\ \mathrm{mol\ e^-} \] The subsequent water electrolysis consumes: \[ 0.0200\ \mathrm{mol\ e^-} \] Their sum equals the stipulated total: \[ 0.0600\ \mathrm{mol\ e^-} \] \(\textbf{Interpret the change of electrode process:}\) The initial charging stage restores the accumulator's active materials and regenerates sulphuric acid. Once the specified PbSO4 inventory has been fully restored, additional current cannot be assigned to further conversion of that same finite inventory. Under the imposed model, the excess charge instead decomposes water and produces gases. The appearance of oxygen is therefore a consequence of passing charge beyond that required for chemical restoration. A correct calculation must distinguish the finite active-material charging stage from the later side-reaction stage. \(\textbf{Final answer:}\) \[ \boxed{\left(n(\mathrm{H_2SO_4})_{\mathrm{regenerated}},n(\mathrm{O_2})_{\mathrm{overcharge}}\right)=(0.0400,\ 0.00500)\ \mathrm{mol}} \]

Explanation: \(\textbf{Determine the discharge reaction extent in P:}\) The lead-accumulator reaction transfers two electrons per mole of reaction. Therefore: \[ \xi_P=\frac{n(e^-)}2 \] \[ \xi_P=\frac{0.0200}{2} \] \[ \xi_P=0.0100\ \mathrm{mol} \] \(\textbf{Calculate the acid consumed in P:}\) Two moles of H2SO4 are consumed per mole of reaction: \[ n(\mathrm{H_2SO_4})_{\mathrm{consumed}}=2(0.0100) \] \[ n(\mathrm{H_2SO_4})_{\mathrm{consumed}}=0.0200\ \mathrm{mol} \] Thus: \[ n(\mathrm{H_2SO_4})_{P,f}=0.100-0.0200 \] \[ n(\mathrm{H_2SO_4})_{P,f}=0.0800\ \mathrm{mol} \] \(\textbf{Calculate water produced and remaining in P:}\) The overall reaction produces two water molecules per reaction event. Therefore: \[ n(\mathrm{H_2O})_{\mathrm{formed}}=2(0.0100) \] \[ n(\mathrm{H_2O})_{\mathrm{formed}}=0.0200\ \mathrm{mol} \] The final liquid water amount is: \[ n(\mathrm{H_2O})_{P,f}=0.400+0.0200 \] \[ n(\mathrm{H_2O})_{P,f}=0.420\ \mathrm{mol} \] \(\textbf{Calculate the final acid mole fraction in P:}\) \[ x_{\mathrm{acid},P}=\frac{n(\mathrm{H_2SO_4})_{P,f}}{n(\mathrm{H_2SO_4})_{P,f}+n(\mathrm{H_2O})_{P,f}} \] \[ x_{\mathrm{acid},P}=\frac{0.0800}{0.0800+0.420} \] \[ x_{\mathrm{acid},P}=0.160 \] \(\textbf{Construct the dilution condition for Q:}\) No electrode reaction occurs in Q. Therefore, the acid amount remains: \[ n(\mathrm{H_2SO_4})_Q=0.100\ \mathrm{mol} \] Let \(w\) be the moles of pure water added. The final water amount is: \[ n(\mathrm{H_2O})_Q=0.400+w \] Equal acid mole fractions require: \[ \frac{0.100}{0.100+0.400+w}=0.160 \] \(\textbf{Solve the water-addition amount:}\) \[ 0.100=0.160(0.500+w) \] \[ 0.100=0.0800+0.160w \] \[ 0.0200=0.160w \] \[ w=0.125\ \mathrm{mol} \] \(\textbf{Determine lead sulphate formed in P:}\) The discharge reaction produces two moles of PbSO4 per mole of reaction. Therefore: \[ n(\mathrm{PbSO_4})_P=2\xi_P \] \[ n(\mathrm{PbSO_4})_P=2(0.0100) \] \[ n(\mathrm{PbSO_4})_P=0.0200\ \mathrm{mol} \] No additional PbSO4 is formed in Q because no discharge reaction occurs there. \(\textbf{Interpret the equal-composition observation:}\) P reaches its final electrolyte composition through electrochemical consumption of acid and formation of water. Q reaches the same acid mole fraction through dilution alone. The chemical states of their plates are different despite the equal final electrolyte mole fractions. An electrolyte-composition measurement by itself therefore cannot establish that two accumulators have undergone the same amount of discharge when dilution or electrolyte replacement is possible. The electrode-material balance provides independent information about actual chemical conversion. \(\textbf{Final answer:}\) \[ \boxed{\left(n(\mathrm{H_2O})_{\mathrm{added\ to\ Q}},n(\mathrm{PbSO_4})_{\mathrm{formed\ in\ P}}\right)=(0.125,\ 0.0200)\ \mathrm{mol}} \]

Explanation: \(\textbf{Identify the activity controlling the hydrogen reaction:}\) The hydrogen electrode responds to hydrogen partial pressure, not to the total pressure of an inert-containing gas mixture alone. Because the gas is ideal: \[ p_{\mathrm{H_2}}=x_{\mathrm{H_2}}p_{\mathrm{total}} \] The total pressure equals the standard reference pressure, so: \[ \frac{p_{\mathrm{H_2}}}{p^\circ}=x_{\mathrm{H_2}} \] \(\textbf{Construct the full fuel-cell quotient:}\) For: \[ \mathrm{2H_2+O_2\rightarrow2H_2O(l)} \] with unit water activity: \[ Q_r=\frac1{(p_{\mathrm{H_2}}/p^\circ)^2(p_{\mathrm{O_2}}/p^\circ)} \] Therefore: \[ Q_r=\frac1{x_{\mathrm{H_2}}^2(0.250)} \] \(\textbf{Determine the measured decrease in reversible emf:}\) \[ E^\circ-E=1.23000-1.20045 \] \[ E^\circ-E=0.02955\ \mathrm V \] \(\textbf{Use the Nernst equation to infer the quotient:}\) At \(298\ \mathrm K\): \[ E=E^\circ-\frac{0.0591}{4}\log Q_r \] Thus: \[ 0.02955=\frac{0.0591}{4}\log Q_r \] \[ \log Q_r=\frac{4(0.02955)}{0.0591} \] \[ \log Q_r=2 \] \[ Q_r=100 \] \(\textbf{Recover hydrogen partial pressure from the quotient:}\) \[ 100=\frac1{0.250x_{\mathrm{H_2}}^2} \] Rearranging: \[ 0.250x_{\mathrm{H_2}}^2=0.0100 \] \[ x_{\mathrm{H_2}}^2=0.0400 \] \[ x_{\mathrm{H_2}}=0.200 \] Only the positive root is physically meaningful for a gas mole fraction. \(\textbf{Determine the inert-gas mole fraction:}\) The mixture contains only H2 and N2. Therefore: \[ x_{\mathrm{N_2}}=1-x_{\mathrm{H_2}} \] \[ x_{\mathrm{N_2}}=1-0.200 \] \[ x_{\mathrm{N_2}}=0.800 \] \(\textbf{Verify the inferred reaction quotient:}\) \[ Q_r=\frac1{(0.200)^2(0.250)} \] \[ Q_r=\frac1{0.0100} \] \[ Q_r=100 \] The inferred composition reproduces the measured Nernst correction. \(\textbf{Interpret the reverse measurement:}\) Nitrogen does not appear in the overall reaction quotient because it is stipulated to be chemically inert. Nevertheless, replacing part of the hydrogen with nitrogen lowers the hydrogen partial pressure at fixed total pressure. That changes the reversible electrochemical potential. The measured emf therefore provides indirect information about the hydrogen fraction when oxygen pressure and product activity are known. A total gas pressure measurement alone could not make this distinction. The calculation integrates gas-mixture composition with the stoichiometric powers of the complete fuel-cell Nernst quotient. \(\textbf{Final answer:}\) \[ \boxed{(x_{\mathrm{H_2}},x_{\mathrm{N_2}})=(0.200,\ 0.800)} ] \]

Explanation: \(\textbf{Identify the mass change per mole of useful plate reaction:}\) The negative-plate reaction converts one mole of Pb into one mole of PbSO4. The entire PbSO4 product remains attached to the measured plate. Therefore, the solid mass gain per mole converted is: \[ \Delta M=M(\mathrm{PbSO_4})-M(\mathrm{Pb}) \] \[ \Delta M=303-207 \] \[ \Delta M=96.0\ \mathrm{g\,mol^{-1}} \] \(\textbf{Infer the amount of lead converted from the observed mass:}\) \[ n(\mathrm{Pb})_{\mathrm{converted}}=\frac{\Delta m}{\Delta M} \] \[ n(\mathrm{Pb})_{\mathrm{converted}}=\frac{0.576}{96.0} \] \[ n(\mathrm{Pb})_{\mathrm{converted}}=0.00600\ \mathrm{mol} \] \(\textbf{Determine the useful electron amount:}\) Each oxidised Pb atom releases two electrons. Therefore: \[ n(e^-)_{\mathrm{useful}}=2(0.00600) \] \[ n(e^-)_{\mathrm{useful}}=0.0120\ \mathrm{mol} \] \(\textbf{Calculate the total external electron amount:}\) \[ n(e^-)_{\mathrm{total}}=\frac QF \] \[ n(e^-)_{\mathrm{total}}=\frac{1930}{96500} \] \[ n(e^-)_{\mathrm{total}}=0.0200\ \mathrm{mol} \] \(\textbf{Determine the useful charge fraction:}\) \[ f_{\mathrm{useful}}=\frac{n(e^-)_{\mathrm{useful}}}{n(e^-)_{\mathrm{total}}} \] \[ f_{\mathrm{useful}}=\frac{0.0120}{0.0200} \] \[ f_{\mathrm{useful}}=0.600 \] \[ f_{\mathrm{useful}}\times100=60.0\% \] \(\textbf{Calculate sulphuric acid consumption:}\) The intended complete discharge reaction is: \[ \mathrm{Pb+PbO_2+2H_2SO_4\rightarrow2PbSO_4+2H_2O} \] It consumes two moles of H2SO4 while transferring two moles of electrons. Hence: \[ n(\mathrm{H_2SO_4})_{\mathrm{consumed}}=n(e^-)_{\mathrm{useful}} \] \[ n(\mathrm{H_2SO_4})_{\mathrm{consumed}}=0.0120\ \mathrm{mol} \] \(\textbf{Check the non-useful electron amount:}\) \[ n(e^-)_{\mathrm{side}}=0.0200-0.0120 \] \[ n(e^-)_{\mathrm{side}}=0.00800\ \mathrm{mol} \] The stipulated competing process accounts for this portion without producing the measured plate-mass gain. \(\textbf{Interpret the measurement:}\) The external charge records all electrons transferred through the circuit. The plate mass change records only those electrons associated with the specified Pb-to-PbSO4 conversion. Because the two measurements respond to different aspects of cell operation, comparing them can reveal that not all charge contributes to the intended discharge chemistry. The acid consumption is governed by the useful reaction extent, not by the total current when an additional process is present. The mass-gain measurement therefore supplies the information needed to distinguish useful electrochemical conversion from competing charge transfer. \(\textbf{Final answer:}\) \[ \boxed{\left(f_{\mathrm{useful}},n(\mathrm{H_2SO_4})_{\mathrm{consumed}}\right)=(60.0\%,\ 0.0120\ \mathrm{mol})} \]

Explanation: \(\textbf{Statement I — Correct:}\) A working fuel cell requires oxidation and reduction at physically separated electrode regions. Hydrogen oxidation releases electrons into the external electronic circuit. Oxygen reduction consumes electrons arriving through that circuit. Meanwhile, appropriate ions must move through the electrolyte to prevent uncontrolled accumulation of electrical charge in the separated compartments. Separator P provides the stipulated ionic pathway without directly short-circuiting the electrode materials. It can therefore support the intended coupled operation. \(\textbf{Statement II — Incorrect:}\) Electronic and ionic conduction play different roles in an electrochemical cell. Electrons transport charge through metallic wires and electronically conducting electrode structures. Ions transport charge through the electrolyte. A separator that conducts electrons but blocks every ion cannot simply replace the ionic pathway while preserving the same separated electrode processes. An electronic short can also reduce the amount of current available to an external load. \(\textbf{Statement III — Correct:}\) Continuous gas supply ensures the availability of chemical reactants. It does not eliminate the requirement for electrical charge balance. If the compartments have no permitted ionic pathway, the oxidation and reduction reactions rapidly encounter charge-separation constraints. A finite transient involving interfacial charging is distinct from sustained electrochemical operation. Under the specified sustained-state model, separator Q cannot support the intended ionic charge transport. \(\textbf{Statement IV — Correct:}\) An electronic connection between the electrode bodies provides a possible internal route for electrons. Such a connection may bypass the external load, so energy can be lost without useful external electrical delivery. It also does not perform the required transport of appropriate electrolyte ions between reaction regions. These limitations remain even when adequate hydrogen and oxygen are continuously available. \(\textbf{Integrate the transport mechanisms:}\) Fuel-cell operation joins two half-reactions through an external electronic pathway and an internal ionic pathway. These pathways complete different parts of the electrical circuit. The reactant-supply system provides chemical fuel and oxidant but cannot replace either charge-transport requirement. A suitable separator must therefore be judged by its ionic conductivity and its ability to prevent unwanted electronic short-circuiting. The correct statements are I, III and IV.

Explanation: \(\textbf{Identify both oxidation centres in pyrite:}\) Iron changes from \(+2\) in FeS2 to \(+3\) in Fe3+. Each iron atom therefore loses one electron. Sulphur changes from \(-1\) in the disulphide unit to \(+6\) in sulphate. Each sulphur atom loses seven electrons. Because one FeS2 unit contains two sulphur atoms, its total electron loss is: \[ n(e^-)_{\mathrm{lost}}=1+2(7) \] \[ n(e^-)_{\mathrm{lost}}=15 \] \(\textbf{Identify the nitrate reduction:}\) Nitrogen changes from \(+5\) in nitrate to \(+2\) in NO. Each nitrate ion therefore accepts three electrons. To accept the fifteen electrons released by one FeS2 unit: \[ b=\frac{15}{3} \] \[ b=5 \] \(\textbf{Balance the oxidation half-reaction in acid:}\) First balance the iron and sulphur atoms: \[ \mathrm{FeS_2\rightarrow Fe^{3+}+2SO_4^{2-}} \] Eight oxygen atoms are required on the product side, so introduce eight water molecules on the reactant side. Then balance hydrogen using sixteen hydrogen ions on the product side. Finally, balance charge using fifteen electrons: \[ \mathrm{FeS_2+8H_2O\rightarrow Fe^{3+}+2SO_4^{2-}+16H^++15e^-} \] \(\textbf{Construct the nitrate reduction half-reaction:}\) In acidic medium: \[ \mathrm{NO_3^-+4H^++3e^-\rightarrow NO+2H_2O} \] Multiply this equation by five: \[ \mathrm{5NO_3^-+20H^++15e^-\rightarrow5NO+10H_2O} \] \(\textbf{Combine the half-reactions:}\) Adding the two half-reactions cancels fifteen electrons. Sixteen hydrogen ions cancel from the opposite sides. Eight water molecules also cancel, leaving two water molecules on the product side. The simplified overall equation is: \[ \mathrm{FeS_2+5NO_3^-+4H^+\rightarrow Fe^{3+}+2SO_4^{2-}+5NO+2H_2O} \] \(\textbf{Verify atom and charge conservation:}\) The oxygen count is: \[ 5(3)=2(4)+5(1)+2 \] \[ 15=15 \] The hydrogen count is four on both sides. The total ionic charge on the reactant side is: \[ 5(-1)+4(+1)=-1 \] On the product side: \[ (+3)+2(-2)=-1 \] The equation conserves both atoms and electrical charge. \(\textbf{Interpret the balancing trap:}\) Counting only the oxidation of iron would miss fourteen of the fifteen electrons released by FeS2. Treating sulphur as having oxidation state \(-2\) would also give an incorrect electron balance because the compound contains a disulphide unit. The hydrogen-ion coefficient cannot be obtained from electron counting alone; it follows after oxygen and hydrogen are balanced in acidic medium. The smallest integer coefficients give the required ratios. \(\textbf{Final answer:}\) \[ \boxed{\left(\frac ba,\frac ca\right)=(5,\ 4)} \]

Explanation: \(\textbf{Identify the changing oxidation states:}\) Chromium changes from \(+3\) in Cr(OH)3 to \(+6\) in chromate. Each chromium atom loses three electrons. Chlorine changes from \(+1\) in ClO− to \(-1\) in Cl−. Each hypochlorite ion accepts two electrons. The least common electron count is six. \(\textbf{Balance chromium oxidation in alkaline medium:}\) The balanced oxidation half-reaction is: \[ \mathrm{Cr(OH)_3+5OH^-\rightarrow CrO_4^{2-}+4H_2O+3e^-} \] Check the electrical charge: \[ -5=-2-3 \] The chromium, oxygen and hydrogen atoms also balance. \(\textbf{Balance hypochlorite reduction:}\) The reduction half-reaction is: \[ \mathrm{ClO^-+H_2O+2e^-\rightarrow Cl^-+2OH^-} \] Multiply chromium oxidation by two and hypochlorite reduction by three. \(\textbf{Combine and simplify the half-reactions:}\) After cancelling six electrons, common water molecules and hydroxide ions: \[ \mathrm{2Cr(OH)_3+3ClO^-+4OH^-\rightarrow2CrO_4^{2-}+3Cl^-+5H_2O} \] This equation requires four hydroxide ions for every two chromium(III) hydroxide formula units. \(\textbf{Compare the available reactant extents:}\) For chromium(III) hydroxide: \[ \xi_{\mathrm{Cr(OH)_3}}=\frac{0.0200}{2} \] \[ \xi_{\mathrm{Cr(OH)_3}}=0.0100\ \mathrm{mol} \] For hypochlorite: \[ \xi_{\mathrm{ClO^-}}=\frac{0.0500}{3} \] \[ \xi_{\mathrm{ClO^-}}\approx0.0167\ \mathrm{mol} \] For hydroxide: \[ \xi_{\mathrm{OH^-}}=\frac{0.0300}{4} \] \[ \xi_{\mathrm{OH^-}}=0.00750\ \mathrm{mol} \] \(\textbf{Identify the limiting reactant:}\) The hydroxide inventory permits the smallest reaction extent. Therefore: \[ \xi=0.00750\ \mathrm{mol} \] \(\textbf{Determine the chromium compound amounts:}\) Chromium(III) hydroxide consumed: \[ n(\mathrm{Cr(OH)_3})_{\mathrm{consumed}}=2(0.00750) \] \[ n(\mathrm{Cr(OH)_3})_{\mathrm{consumed}}=0.0150\ \mathrm{mol} \] Chromium(III) hydroxide remaining: \[ n(\mathrm{Cr(OH)_3})_{\mathrm{remaining}}=0.0200-0.0150 \] \[ n(\mathrm{Cr(OH)_3})_{\mathrm{remaining}}=0.00500\ \mathrm{mol} \] Chromate produced: \[ n(\mathrm{CrO_4^{2-}})=2(0.00750) \] \[ n(\mathrm{CrO_4^{2-}})=0.0150\ \mathrm{mol} \] \(\textbf{Check the unused hypochlorite:}\) \[ n(\mathrm{ClO^-})_{\mathrm{remaining}}=0.0500-3(0.00750) \] \[ n(\mathrm{ClO^-})_{\mathrm{remaining}}=0.0275\ \mathrm{mol} \] Both chromium(III) hydroxide and hypochlorite remain, confirming that neither is limiting. \(\textbf{Interpret the alkaline-medium requirement:}\) The hydroxide coefficient is not determined merely by inspecting chromium and chlorine oxidation numbers. It emerges from balancing oxygen, hydrogen and charge in the alkaline half-reactions. The reaction mixture contains enough oxidising and reducing agents for greater conversion, but insufficient hydroxide for that conversion under the specified reaction-only model. The endpoint must therefore be determined from the complete balanced equation, including the medium-dependent reactant. \(\textbf{Final answer:}\) \[ \boxed{\left(n(\mathrm{Cr(OH)_3})_{\mathrm{remaining}},n(\mathrm{CrO_4^{2-}})_{\mathrm{formed}}\right)=(0.00500,\ 0.0150)\ \mathrm{mol}} \]

Explanation: \(\textbf{Statement I — Correct:}\) Chlorine has oxidation state \(+5\) in chlorate: \[ x+3(-2)=-1 \] \[ x=+5 \] In chloride, its oxidation state is \(-1\). The decrease of six units corresponds to acceptance of six electrons. Sulphur changes from \(+4\) in sulphite to \(+6\) in sulphate and therefore loses two electrons. \(\textbf{Statement II — Correct:}\) One chlorate ion must accept the electrons released by three sulphite ions: \[ 3(2)=6 \] The smallest electron-balanced reactant ratio is consequently: \[ \mathrm{ClO_3^-}:\mathrm{SO_3^{2-}}=1:3 \] The corresponding product ratio also follows from conservation of chlorine and sulphur atoms. \(\textbf{Statement III — Incorrect:}\) Hydroxide and water may appear in intermediate half-reactions during the alkaline ion–electron method. However, species appearing on both sides of the combined equation must be cancelled. Their presence in an intermediate balancing step does not require their presence in the final net reaction. \(\textbf{Statement IV — Correct:}\) The correctly simplified overall equation is: \[ \mathrm{ClO_3^-+3SO_3^{2-}\rightarrow Cl^-+3SO_4^{2-}} \] The number of oxygen atoms on the left is: \[ 3+3(3)=12 \] The number on the right is: \[ 3(4)=12 \] The total ionic charge on each side is: \[ -1+3(-2)=-7 \] \[ -1+3(-2)=-7 \] No hydrogen-containing balancing species is required in the simplified equation. \(\textbf{Verify by the alkaline half-reactions:}\) Chlorate reduction can be written as: \[ \mathrm{ClO_3^-+3H_2O+6e^-\rightarrow Cl^-+6OH^-} \] Sulphite oxidation is: \[ \mathrm{SO_3^{2-}+2OH^-\rightarrow SO_4^{2-}+H_2O+2e^-} \] Multiplying the second equation by three cancels six electrons. All six hydroxide ions and all three water molecules then cancel as well. \(\textbf{Interpret the balancing distinction:}\) A reaction medium determines the permitted balancing species and the appropriate half-reaction procedure. It does not determine whether those species survive in the final net equation. The complete cancellation here is possible because the oxygen atoms already balance between the oxidised and reduced ionic species after electron coefficients are fixed. This is a useful check against inserting unnecessary H+ or OH− merely because a balancing method has been specified. Statements I, II and IV are correct.

Explanation: \(\textbf{Identify the two chemical roles of iodide:}\) Copper changes from oxidation state \(+2\) in Cu2+ to \(+1\) in CuI. It therefore undergoes reduction and accepts one electron per copper ion. Some iodide ions undergo oxidation from \(-1\) to zero in I2. Other iodide ions enter the insoluble CuI product without changing oxidation state. Consequently, not every iodide ion in the overall reaction acts as a reducing agent. \(\textbf{Check the electron-balanced coefficients:}\) The reduction of two copper(II) ions consumes: \[ 2(1)=2 \] electrons. Oxidation of two iodide ions produces one iodine molecule and releases two electrons: \[ \mathrm{2I^-\rightarrow I_2+2e^-} \] A further two iodide ions are needed to form the two CuI formula units. The complete four-iodide stoichiometry is therefore essential. \(\textbf{Determine the copper-supported reaction extent:}\) \[ \xi_{\mathrm{Cu^{2+}}}=\frac{0.0300}{2} \] \[ \xi_{\mathrm{Cu^{2+}}}=0.0150\ \mathrm{mol} \] \(\textbf{Determine the iodide-supported reaction extent:}\) \[ \xi_{\mathrm{I^-}}=\frac{0.0500}{4} \] \[ \xi_{\mathrm{I^-}}=0.0125\ \mathrm{mol} \] Iodide is limiting. \(\textbf{Calculate iodine production:}\) One mole of I2 is produced per mole of balanced reaction. Therefore: \[ n(\mathrm{I_2})=\xi \] \[ n(\mathrm{I_2})=0.0125\ \mathrm{mol} \] \(\textbf{Calculate copper(II) consumption:}\) \[ n(\mathrm{Cu^{2+}})_{\mathrm{consumed}}=2(0.0125) \] \[ n(\mathrm{Cu^{2+}})_{\mathrm{consumed}}=0.0250\ \mathrm{mol} \] \(\textbf{Determine copper(II) remaining:}\) \[ n(\mathrm{Cu^{2+}})_{\mathrm{remaining}}=0.0300-0.0250 \] \[ n(\mathrm{Cu^{2+}})_{\mathrm{remaining}}=0.00500\ \mathrm{mol} \] \(\textbf{Verify the precipitate and iodine inventories:}\) Copper(I) iodide formed: \[ n(\mathrm{CuI})=2(0.0125) \] \[ n(\mathrm{CuI})=0.0250\ \mathrm{mol} \] Iodide incorporated into CuI: \[ 0.0250\ \mathrm{mol} \] Iodide oxidised into I2: \[ 2(0.0125)=0.0250\ \mathrm{mol} \] Their sum is the available \(0.0500\ \mathrm{mol}\) of iodide. \(\textbf{Interpret the integrated redox–precipitation process:}\) Copper(II) is reduced to copper(I), while iodide is partly oxidised to iodine. Formation of insoluble CuI accounts for the remaining iodide consumption. Using only the electron-balanced iodine oxidation would miss the additional iodide incorporated into the precipitate. The limiting-reagent decision must therefore use the complete balanced reaction, not merely the redox-active fraction of the iodide. \(\textbf{Final answer:}\) \[ \boxed{\left(n(\mathrm{I_2}),n(\mathrm{Cu^{2+}})_{\mathrm{remaining}}\right)=(0.0125,\ 0.00500)\ \mathrm{mol}} \]

Explanation: \(\textbf{Identify nitrogen's initial oxidation state:}\) In the ammonium ion, hydrogen has oxidation state \(+1\). Therefore: \[ x+4(+1)=+1 \] \[ x=-3 \] Nitrogen has oxidation state \(-3\) in NH4+. It becomes elemental nitrogen with oxidation state zero in N2. \(\textbf{Calculate nitrogen's electron loss:}\) Each nitrogen atom loses three electrons. There are two nitrogen atoms per ammonium dichromate formula unit. Thus, the electron loss per formula unit is: \[ n(e^-)_{\mathrm{lost}}=2(3) \] \[ n(e^-)_{\mathrm{lost}}=6 \] \(\textbf{Identify chromium's reduction:}\) Chromium has oxidation state \(+6\) in dichromate. In Cr2O3: \[ 2x+3(-2)=0 \] \[ x=+3 \] Each chromium atom gains three electrons. Two chromium atoms consequently gain six electrons. \(\textbf{Verify the internal electron balance:}\) Electron loss by nitrogen: \[ 2(3)=6 \] Electron gain by chromium: \[ 2(3)=6 \] The oxidation and reduction changes are equal for every formula unit decomposed. \(\textbf{Determine total internal electron transfer:}\) \[ n(e^-)_{\mathrm{transferred}}=6n(\mathrm{(NH_4)_2Cr_2O_7}) \] \[ n(e^-)_{\mathrm{transferred}}=6(0.0500) \] \[ n(e^-)_{\mathrm{transferred}}=0.300\ \mathrm{mol} \] \(\textbf{Calculate nitrogen formation:}\) The balanced equation produces one N2 molecule per ammonium dichromate formula unit. Therefore: \[ n(\mathrm{N_2})=0.0500\ \mathrm{mol} \] \(\textbf{Check the accompanying products:}\) One mole of Cr2O3 forms per mole of reactant: \[ n(\mathrm{Cr_2O_3})=0.0500\ \mathrm{mol} \] Four moles of water form per mole of reactant: \[ n(\mathrm{H_2O})=4(0.0500) \] \[ n(\mathrm{H_2O})=0.200\ \mathrm{mol} \] \(\textbf{Interpret the intramolecular redox process:}\) Ammonium dichromate contains both the nitrogen species undergoing oxidation and the chromium species undergoing reduction. No separate reducing reagent needs to be added for the displayed decomposition. The electron amount is an oxidation-number-accounting quantity within the chemical reaction, not a charge necessarily delivered through an external circuit. Counting only the three-electron change of one chromium or nitrogen atom would miss the presence of two atoms of each changing element. The correct electron total follows from oxidation-state change multiplied by atomic multiplicity and reaction amount. \(\textbf{Final answer:}\) \[ \boxed{\left(n(e^-)_{\mathrm{transferred}},n(\mathrm{N_2})\right)=(0.300,\ 0.0500)\ \mathrm{mol}} \]

Explanation: \(\textbf{Determine the oxidation-state changes:}\) Arsenic has oxidation state \(+3\) in arsenite: \[ x+3(-2)=-3 \] \[ x=+3 \] It becomes \(+5\) in arsenate and therefore loses two electrons. Manganese changes from \(+7\) in permanganate to \(+4\) in MnO2. It therefore gains three electrons. \(\textbf{Balance arsenite oxidation in alkaline medium:}\) The oxidation half-reaction is: \[ \mathrm{AsO_3^{3-}+2OH^-\rightarrow AsO_4^{3-}+H_2O+2e^-} \] Both atoms and charge are conserved. \(\textbf{Balance permanganate reduction:}\) The alkaline-medium reduction half-reaction is: \[ \mathrm{MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-} \] \(\textbf{Equalise the electron amounts:}\) Multiply arsenite oxidation by three: \[ \mathrm{3AsO_3^{3-}+6OH^-\rightarrow3AsO_4^{3-}+3H_2O+6e^-} \] Multiply permanganate reduction by two: \[ \mathrm{2MnO_4^-+4H_2O+6e^-\rightarrow2MnO_2+8OH^-} \] \(\textbf{Combine and cancel balancing species:}\) Cancel six electrons. Cancel six of the eight hydroxide ions and three of the four water molecules. The net equation is: \[ \mathrm{3AsO_3^{3-}+2MnO_4^-+H_2O\rightarrow3AsO_4^{3-}+2MnO_2+2OH^-} \] \(\textbf{Identify the limiting reactant:}\) For arsenite: \[ \xi_{\mathrm{arsenite}}=\frac{0.0600}{3} \] \[ \xi_{\mathrm{arsenite}}=0.0200\ \mathrm{mol} \] For permanganate: \[ \xi_{\mathrm{permanganate}}=\frac{0.0300}{2} \] \[ \xi_{\mathrm{permanganate}}=0.0150\ \mathrm{mol} \] Permanganate is limiting. \(\textbf{Calculate the product amounts:}\) Arsenate formed: \[ n(\mathrm{AsO_4^{3-}})=3(0.0150) \] \[ n(\mathrm{AsO_4^{3-}})=0.0450\ \mathrm{mol} \] Hydroxide formed: \[ n(\mathrm{OH^-})=2(0.0150) \] \[ n(\mathrm{OH^-})=0.0300\ \mathrm{mol} \] \(\textbf{Check net charge conservation:}\) The net ionic charge on the left is: \[ 3(-3)+2(-1)=-11 \] The charge on the right is: \[ 3(-3)+2(-1)=-11 \] The equation is electrically balanced. \(\textbf{Interpret the changed-medium inference:}\) Alkaline-medium balancing does not imply that every completed redox reaction must consume hydroxide ions. Here, hydroxide appears as a net product after the correctly balanced half-reactions are combined. Its amount depends on the reaction extent determined by permanganate availability. Counting electrons without completing the water and hydroxide balance would miss this result. \(\textbf{Final answer:}\) \[ \boxed{\left(n(\mathrm{AsO_4^{3-}}),n(\mathrm{OH^-})_{\mathrm{formed}}\right)=(0.0450,\ 0.0300)\ \mathrm{mol}} \]

Explanation: \(\textbf{Preserve the stated voltage-sign convention:}\) Every measurement is reported as the right-hand reduction potential minus the left-hand reduction potential. A negative signed reading is allowed and does not establish that an experiment is invalid. The reference electrode can appear on either side. \(\textbf{Extract the standard reduction potential of A:}\) For the first cell: \[ E_{\mathrm{right}}-E_{\mathrm{left}}=E^\circ_{\mathrm{A^+/A}}-E^\circ_{\mathrm{SHE}} \] Substitute the measured value: \[ +0.360=E^\circ_{\mathrm{A^+/A}}-0 \] \[ E^\circ_{\mathrm{A^+/A}}=+0.360\ \mathrm V \] \(\textbf{Interpret the second cell's reversed reference orientation:}\) The SHE is on the right in the second measurement. Therefore: \[ E_{\mathrm{right}}-E_{\mathrm{left}}=E^\circ_{\mathrm{SHE}}-E^\circ_{\mathrm{B^{2+}/B}} \] The reported value is negative: \[ -0.200=0-E^\circ_{\mathrm{B^{2+}/B}} \] Hence: \[ E^\circ_{\mathrm{B^{2+}/B}}=+0.200\ \mathrm V \] \(\textbf{Choose the spontaneous reduction electrode:}\) The standard reduction potentials are: \[ E^\circ_{\mathrm{A^+/A}}=+0.360\ \mathrm V \] \[ E^\circ_{\mathrm{B^{2+}/B}}=+0.200\ \mathrm V \] A has the greater reduction potential. It therefore acts as the cathode in the spontaneous A–B cell. B undergoes oxidation at the anode. \(\textbf{Calculate the standard cell emf:}\) \[ E^\circ_{\mathrm{cell}}=E^\circ_{\mathrm{cathode}}-E^\circ_{\mathrm{anode}} \] \[ E^\circ_{\mathrm{cell}}=0.360-0.200 \] \[ E^\circ_{\mathrm{cell}}=+0.160\ \mathrm V \] \(\textbf{Verify with the balanced overall reaction:}\) The reduction at A is: \[ \mathrm{A^++e^-\rightarrow A} \] The oxidation at B is: \[ \mathrm{B\rightarrow B^{2+}+2e^-} \] Balancing electrons gives: \[ \mathrm{B+2A^+\rightarrow B^{2+}+2A} \] The positive standard emf confirms that this direction is spontaneous under standard conditions. \(\textbf{Interpret the sign trap:}\) The second negative reading is caused by the chosen left-to-right reference orientation. It does not mean that B has a negative standard reduction potential. Failing to preserve the electrode order would incorrectly infer \(E^\circ_{\mathrm B}=-0.200\ \mathrm V\), producing an erroneous cell emf of \(0.560\ \mathrm V\). The reliable procedure is to translate each signed reading into an explicit right-minus-left equation before comparing the two electrodes. \(\textbf{Final answer:}\) The A electrode is the spontaneous cathode of the A–B cell, and its standard emf is \(0.160\ \mathrm V\).

Explanation: \(\textbf{Identify why the two voltage readings differ:}\) A voltmeter with finite input resistance permits a small current to flow. The measured terminal voltage can consequently be less than the cell's ideal open-circuit emf. A higher input resistance usually produces a smaller loading current. The two readings can be used together to recover both \(E\) and \(r\). \(\textbf{Write the circuit equation for a loaded measurement:}\) For a voltmeter acting as the external resistance: \[ I=\frac{E}{R+r} \] The voltmeter reads: \[ V=IR \] Therefore: \[ V=\frac{ER}{R+r} \] \(\textbf{Apply the first voltmeter reading:}\) \[ 0.960=\frac{800E}{800+r} \] Rearranging: \[ E=0.960\left(1+\frac r{800}\right) \] \[ E=0.960+0.00120r \] \(\textbf{Apply the second voltmeter reading:}\) \[ 1.152=\frac{4800E}{4800+r} \] Rearranging: \[ E=1.152\left(1+\frac r{4800}\right) \] \[ E=1.152+0.000240r \] \(\textbf{Eliminate the unknown emf:}\) Equating the two expressions: \[ 0.960+0.00120r=1.152+0.000240r \] \[ 0.000960r=0.192 \] \[ r=200\ \Omega \] \(\textbf{Recover the open-circuit emf:}\) Use the first measurement: \[ E=0.960+0.00120(200) \] \[ E=0.960+0.240 \] \[ E=1.200\ \mathrm V \] \(\textbf{Verify both observed readings:}\) With \(R_1=800\ \Omega\): \[ V_1=\frac{(1.200)(800)}{800+200} \] \[ V_1=0.960\ \mathrm V \] With \(R_2=4800\ \Omega\): \[ V_2=\frac{(1.200)(4800)}{4800+200} \] \[ V_2=1.152\ \mathrm V \] Both observations are reproduced. \(\textbf{Interpret the measurement principle:}\) An emf is ideally measured without drawing appreciable current from the cell. A finite-resistance voltmeter is only an approximation to that condition. Its measured voltage includes the effect of current flowing through the cell's internal resistance. The larger-input-resistance meter gives a value closer to the true emf, but the reading is still lower in this model. Using two known loading resistances allows the emf to be inferred without assuming that either meter is perfectly non-invasive. \(\textbf{Final answer:}\) \[ \boxed{(E,r)=(1.200\ \mathrm V,\ 200\ \Omega)} \]

Explanation: \(\textbf{Identify the required role of an inert redox electrode:}\) Both Fe3+ and Fe2+ are dissolved ions. The redox pair needs an electrically conducting surface through which electrons can enter or leave the solution. An inert platinum surface can fulfil this role without being one of the species in the overall iron half-reaction. The relevant reduction is: \[ \mathrm{Fe^{3+}+e^-\rightarrow Fe^{2+}} \] \(\textbf{Test whether copper is chemically inert in the same solution:}\) Copper can itself undergo oxidation: \[ \mathrm{Cu(s)\rightarrow Cu^{2+}+2e^-} \] Fe3+ can accept electrons: \[ \mathrm{Fe^{3+}+e^-\rightarrow Fe^{2+}} \] The higher standard reduction potential belongs to the iron couple. Therefore, Fe3+ can act as the oxidising agent against copper metal under standard conditions. \(\textbf{Balance the resulting chemical reaction:}\) Two iron(III) ions accept the two electrons released by one copper atom: \[ \mathrm{Cu(s)+2Fe^{3+}\rightarrow Cu^{2+}+2Fe^{2+}} \] \(\textbf{Calculate its standard cell potential:}\) \[ E^\circ_{\mathrm{reaction}}=E^\circ_{\mathrm{Fe^{3+}/Fe^{2+}}}-E^\circ_{\mathrm{Cu^{2+}/Cu}} \] \[ E^\circ_{\mathrm{reaction}}=0.77-0.34 \] \[ E^\circ_{\mathrm{reaction}}=+0.43\ \mathrm V \] \(\textbf{Interpret the standard thermodynamic sign:}\) The positive value indicates a negative standard Gibbs energy change for the written reaction. Thus, copper can chemically reduce Fe3+ to Fe2+ while itself forming Cu2+. It cannot automatically be treated as a spectator conducting surface in this environment. \(\textbf{Distinguish conductivity from chemical suitability:}\) A material may conduct electrons well but still be unsuitable as an inert electrode because it reacts with the solution. Platinum supplies the required electronic pathway without appearing in the iron redox equation under the stated assumptions. Copper would introduce a second electroactive couple and alter the intended solution composition. The choice of electrode material must therefore consider chemical compatibility, not merely electrical conductivity. The correct interpretation is option B.