Redox Reactions and Electrochemistry MCQs with Answers – Part 5 (NEET Chemistry)

Explanation: \(\textbf{Interpret the Nernst graph at unknown temperature:}\) The graph is linear in \(\log Q_r\). Its vertical intercept represents the standard emf at the temperature of the measurement. Its slope is: \[ m=-\frac{2.303RT}{nF} \] The slope can therefore be used to infer temperature when the electron-transfer number is known. \(\textbf{Determine the graph intercept:}\) At: \[ \log Q_r=0 \] the reaction quotient equals unity. The graph gives: \[ E^\circ(T)=0.8000\ \mathrm V \] This value is specific to the unknown measurement temperature. \(\textbf{Calculate the observed slope:}\) \[ m=\frac{0.7306-0.8000}{2-0} \] \[ m=\frac{-0.0694}{2} \] \[ m=-0.0347\ \mathrm V \] per unit change in \(\log Q_r\). \(\textbf{Infer the temperature:}\) For \(n=2\): \[ 0.0347=\frac{2.303(8.314)T}{2(96500)} \] Rearranging: \[ T=\frac{0.0347(2)(96500)}{2.303(8.314)} \] \[ T\approx349.7\ \mathrm K \] \[ T\approx350\ \mathrm K \] \(\textbf{Determine the logarithm for the requested composition:}\) At: \[ Q_r=0.100 \] \[ \log Q_r=-1 \] The potential is found from the same graph equation: \[ E=0.8000-0.0347\log Q_r \] \(\textbf{Calculate the potential:}\) \[ E=0.8000-0.0347(-1) \] \[ E=0.8347\ \mathrm V \] \(\textbf{Verify the sign of the potential shift:}\) For: \[ Q_r\lt 1 \] the logarithm is negative. The Nernst term therefore increases the forward potential relative to the standard emf. This agrees with the predicted value: \[ 0.8347\gt 0.8000 \] \(\textbf{Interpret the temperature dependence:}\) The familiar \(0.0591/n\) coefficient applies approximately at \(298\ \mathrm K\). At another temperature, the factor must be calculated from \(2.303RT/nF\). The graph slope measures this factor directly. Although the standard emf itself can depend on temperature, the supplied graph intercept already provides its value at the measurement temperature. The temperature inference therefore comes from the slope rather than from assuming that \(E^\circ\) is independent of temperature. \(\textbf{Final answer:}\) \[ \boxed{(T,E_{Q_r=0.100})=(350\ \mathrm K,\ 0.8347\ \mathrm V)} \]

Explanation: \(\textbf{Determine the standard cell emf:}\) Iron(III) is reduced, while iodide is oxidised. Therefore: \[ E^\circ_{\mathrm{cell}}=0.77-0.54 \] \[ E^\circ_{\mathrm{cell}}=0.2300\ \mathrm V \] Two iron(III) ions accept the two electrons released by two iodide ions. Thus: \[ n=2 \] \(\textbf{Construct the quotient using the balanced coefficients:}\) Solid iodine has unit activity and is omitted from the quotient. For the stated reaction: \[ Q_r=\frac{[\mathrm{Fe^{2+}}]^2}{[\mathrm{Fe^{3+}}]^2[\mathrm{I^-}]^2} \] The powers cannot be cancelled merely because all concentrations initially have the same numerical value. The reactant side contains four ionic formula units, whereas the dissolved product side contains two. \(\textbf{Calculate the initial reaction quotient:}\) \[ Q_i=\frac{(0.100)^2}{(0.100)^2(0.100)^2} \] \[ Q_i=\frac{10^{-2}}{10^{-4}} \] \[ Q_i=100 \] \(\textbf{Calculate the initial potential:}\) \[ E_i=0.2300-\frac{0.0591}{2}\log(100) \] \[ E_i=0.2300-0.0591 \] \[ E_i=0.1709\ \mathrm V \] \(\textbf{Apply the common dilution factor:}\) After dilution: \[ [\mathrm{Fe^{3+}}]=[\mathrm{Fe^{2+}}]=[\mathrm{I^-}]=0.0100\ \mathrm{mol\,L^{-1}} \] Therefore: \[ Q_f=\frac{(0.0100)^2}{(0.0100)^2(0.0100)^2} \] \[ Q_f=\frac{10^{-4}}{10^{-8}} \] \[ Q_f=10^4 \] \(\textbf{Calculate the new emf:}\) \[ E_f=0.2300-\frac{0.0591}{2}\log(10^4) \] \[ E_f=0.2300-(0.02955)(4) \] \[ E_f=0.1118\ \mathrm V \] \(\textbf{Determine the effect of dilution:}\) \[ \Delta E=0.1118-0.1709 \] \[ \Delta E=-0.0591\ \mathrm V \] \(\textbf{Interpret the non-cancelling concentration change:}\) A common dilution factor cancels from a reaction quotient only when the total stoichiometric powers of the dissolved species in its numerator and denominator are equal. They are not equal for this reaction. Dilution therefore increases the quotient even though all three ionic concentrations are changed identically. This raises the Nernst correction and lowers the forward potential. The cell remains capable of operating spontaneously in the written direction because its final emf is still positive. \(\textbf{Final answer:}\) \[ \boxed{(E_i,E_f)=(0.1709\ \mathrm V,\ 0.1118\ \mathrm V)} \]

Explanation: \(\textbf{Identify the two reactant activities:}\) The oxygen reduction reaction consumes both gaseous oxygen and hydrogen ions. Its reaction quotient is: \[ Q_r=\frac{a_{\mathrm{H_2O}}^2}{(p_{\mathrm{O_2}}/p^\circ)a_{\mathrm{H^+}}^4} \] Because water activity is unity: \[ Q_r=\frac1{(p_{\mathrm{O_2}}/p^\circ)a_{\mathrm{H^+}}^4} \] \(\textbf{Determine the initial quotient:}\) Initially: \[ a_{\mathrm{H^+}}=0.100,\qquad p_{\mathrm{O_2}}/p^\circ=1 \] Therefore: \[ Q_i=\frac1{(1)(0.100)^4} \] \[ Q_i=10^4 \] \(\textbf{Determine the changed hydrogen-ion activity:}\) Increasing pH by one unit lowers hydrogen-ion activity by a factor of ten. Thus: \[ a_{\mathrm{H^+},f}=0.0100 \] The oxygen relative pressure increases to: \[ p_{\mathrm{O_2},f}/p^\circ=10^4 \] \(\textbf{Calculate the final reaction quotient:}\) \[ Q_f=\frac1{(10^4)(0.0100)^4} \] \[ Q_f=\frac1{10^4\times10^{-8}} \] \[ Q_f=10^4 \] \(\textbf{Compare the two Nernst corrections:}\) The Nernst equation is: \[ E=E^\circ-\frac{0.0591}{4}\log Q_r \] Because: \[ Q_f=Q_i \] the Nernst correction is the same before and after the adjustments. Therefore: \[ E_f=E_i \] \(\textbf{Interpret the compensation:}\) Decreasing hydrogen-ion activity makes the reduction less favourable when oxygen pressure is unchanged. Increasing oxygen pressure makes the same reduction more favourable when hydrogen-ion activity is unchanged. The hydrogen-ion activity is raised to the fourth power in the quotient. The four-order increase in oxygen relative pressure exactly offsets the four-order decrease produced by the new hydrogen-ion activity term. Thus, the individual chemical changes are substantial, but their combined effect on the ideal reversible electrode potential is zero. The result depends on the complete reaction quotient rather than on either changed variable in isolation.

Explanation: \(\textbf{Identify the standard electrode directions:}\) Chlorine has the higher standard reduction potential. Therefore, chlorine undergoes reduction: \[ \mathrm{Cl_2+2e^-\rightarrow2Cl^-} \] Bromide undergoes oxidation: \[ \mathrm{2Br^-\rightarrow Br_2(l)+2e^-} \] The balanced cell reaction transfers: \[ n=2 \] \(\textbf{Calculate the standard emf:}\) \[ E^\circ_{\mathrm{cell}}=1.36-1.07 \] \[ E^\circ_{\mathrm{cell}}=0.290\ \mathrm V \] \(\textbf{Construct the reaction quotient:}\) Liquid bromine is a pure liquid, so its activity is unity. Chlorine is a gaseous reactant. The reaction quotient is: \[ Q_r=\frac{[\mathrm{Cl^-}]^2}{(p_{\mathrm{Cl_2}}/p^\circ)[\mathrm{Br^-}]^2} \] \(\textbf{Calculate the initial quotient:}\) \[ Q_i=\frac{(0.100)^2}{(1.00)(0.0100)^2} \] \[ Q_i=\frac{0.0100}{1.00\times10^{-4}} \] \[ Q_i=100 \] \(\textbf{Calculate the initial cell emf:}\) \[ E_i=E^\circ-\frac{0.0591}{2}\log Q_i \] \[ E_i=0.290-\frac{0.0591}{2}\log(100) \] \[ E_i=0.290-0.0591 \] \[ E_i=0.2309\ \mathrm V \] \[ E_i\approx0.231\ \mathrm V \] \(\textbf{Relate a pressure increase to the reaction quotient:}\) With ionic concentrations unchanged: \[ Q_r\propto\frac1{p_{\mathrm{Cl_2}}} \] If chlorine pressure increases by a factor \(f\): \[ Q_f=\frac{Q_i}{f} \] The change in potential is: \[ E_f-E_i=-\frac{0.0591}{2}\log\left(\frac{Q_f}{Q_i}\right) \] \[ \Delta E=\frac{0.0591}{2}\log f \] \(\textbf{Determine the required pressure factor:}\) The desired increase is: \[ \Delta E=0.0591\ \mathrm V \] Thus: \[ 0.0591=\frac{0.0591}{2}\log f \] \[ \log f=2 \] \[ f=100 \] \(\textbf{Interpret the gas-pressure dependence:}\) Increasing chlorine pressure favours chlorine reduction because chlorine is a reactant. However, a tenfold pressure increase changes the cell potential by only \(0.02955\ \mathrm V\) for this two-electron reaction. A hundredfold increase is required for a \(0.0591\ \mathrm V\) increase. The pure bromine phase does not enter the quotient as a variable concentration. The pressure dependence is therefore fixed by the chlorine coefficient and the two-electron Nernst denominator. \(\textbf{Final answer:}\) \[ \boxed{(E_i,f)=(0.231\ \mathrm V,\ 100)} \]

Explanation: \(\textbf{Identify the different electron-transfer numbers:}\) In Reaction P, nitrogen changes from oxidation state \(+5\) in nitrate to \(+4\) in nitrogen dioxide. The reduction therefore requires: \[ n_P=1 \] In Reaction Q, nitrogen changes from \(+5\) in nitrate to \(+2\) in nitric oxide. Therefore: \[ n_Q=3 \] \(\textbf{Relate pH to the Nernst equation:}\) For a reduction half-reaction consuming \(m\) hydrogen ions and \(n\) electrons, with other activities fixed: \[ E=E^\circ-\frac{0.0591m}{n}\mathrm{pH}+\text{constant terms} \] Therefore, a pH increase \(\Delta\mathrm{pH}\) changes the reduction potential by: \[ \Delta E=-\frac{0.0591m}{n}\Delta\mathrm{pH} \] The standard potential and unchanged activity terms cancel in the potential difference. \(\textbf{Determine the pH sensitivity of Reaction P:}\) Reaction P consumes two hydrogen ions and one electron: \[ m_P=2,\qquad n_P=1 \] Thus: \[ \frac{\Delta E_P}{\Delta\mathrm{pH}}=-\frac{0.0591(2)}{1} \] \[ \frac{\Delta E_P}{\Delta\mathrm{pH}}=-0.1182\ \mathrm{V\ per\ pH\ unit} \] \(\textbf{Calculate the change for the two-unit pH increase:}\) \[ \Delta\mathrm{pH}=3.00-1.00 \] \[ \Delta\mathrm{pH}=2.00 \] Therefore: \[ \Delta E_P=-0.1182(2.00) \] \[ \Delta E_P=-0.2364\ \mathrm V \] \(\textbf{Determine the pH sensitivity of Reaction Q:}\) Reaction Q consumes four hydrogen ions and three electrons: \[ m_Q=4,\qquad n_Q=3 \] Therefore: \[ \frac{\Delta E_Q}{\Delta\mathrm{pH}}=-\frac{0.0591(4)}{3} \] \[ \frac{\Delta E_Q}{\Delta\mathrm{pH}}=-0.0788\ \mathrm{V\ per\ pH\ unit} \] \(\textbf{Calculate its potential change:}\) \[ \Delta E_Q=-0.0788(2.00) \] \[ \Delta E_Q=-0.1576\ \mathrm V \] \(\textbf{Compare the two half-reactions:}\) Both reductions consume hydrogen ions, so increasing pH lowers both reduction potentials. However, their pH sensitivities differ. Reaction P consumes two hydrogen ions per electron. Reaction Q consumes four hydrogen ions per three electrons. The relevant ratio is the hydrogen-ion coefficient divided by the electron-transfer number, not simply the total hydrogen-ion coefficient. \(\textbf{Interpret the chemical conditions:}\) The stipulated comparison holds while each reduction continues to give its specified product. The result does not establish which nitrate-reduction product will actually dominate in an uncontrolled experiment. Product identity also depends on reaction conditions and kinetics. The calculation compares the thermodynamic pH responses of two explicitly defined reversible half-reactions. \(\textbf{Final answer:}\) \[ \boxed{(|\Delta E_P|,|\Delta E_Q|)=(0.2364,\ 0.1576)\ \mathrm V} \]

Explanation: \(\textbf{Identify the reversible electrode process:}\) The reduction half-reaction is: \[ \mathrm{M^{z+}+ze^-\rightarrow M(s)} \] The metal is a pure solid and has unit activity. The unknown ionic charge also specifies the number of electrons required to deposit one metal atom. \(\textbf{Construct the potential-dependence relation:}\) For a metal-ion electrode at \(298\ \mathrm K\): \[ E=E^\circ+\frac{0.0591}{z}\log a_{\mathrm{M^{z+}}} \] When the ionic activity changes, the standard potential cancels from the difference: \[ E_2-E_1=\frac{0.0591}{z}\log\left(\frac{a_2}{a_1}\right) \] \(\textbf{Evaluate the imposed activity ratio:}\) \[ \frac{a_2}{a_1}=\frac{1.00\times10^{-1}}{1.00\times10^{-3}} \] \[ \frac{a_2}{a_1}=100 \] Therefore: \[ \log\left(\frac{a_2}{a_1}\right)=2 \] \(\textbf{Infer the ionic charge number:}\) The measured potential increase is \(0.0394\ \mathrm V\). Thus: \[ 0.0394=\frac{0.0591}{z}(2) \] Rearranging: \[ z=\frac{2(0.0591)}{0.0394} \] \[ z=3 \] The participating ion is therefore \(\mathrm{M^{3+}}\). \(\textbf{Determine the electron amount in the separate electrolysis:}\) \[ n(e^-)=\frac{Q}{F} \] \[ n(e^-)=\frac{2895}{96500} \] \[ n(e^-)=0.0300\ \mathrm{mol} \] \(\textbf{Apply the inferred deposition stoichiometry:}\) The electrolytic reduction is: \[ \mathrm{M^{3+}+3e^-\rightarrow M} \] Three electrons deposit one metal atom. Therefore: \[ n(\mathrm M)=\frac{0.0300}{3} \] \[ n(\mathrm M)=0.0100\ \mathrm{mol} \] \(\textbf{Calculate the deposited mass:}\) \[ m(\mathrm M)=nM \] \[ m(\mathrm M)=(0.0100)(60.0) \] \[ m(\mathrm M)=0.600\ \mathrm g \] \(\textbf{Interpret the independent measurements:}\) The electrode-potential experiment determines the ionic charge number through the dependence on activity. The electrolysis experiment then uses that charge number to relate electrical charge to the amount of deposited metal. The two experiments examine different manifestations of the same electron stoichiometry. Treating the observed \(0.0394\ \mathrm V\) change as a standard potential would be incorrect because only the ionic activity was altered. Similarly, dividing the electrolysis charge by one Faraday per mole of metal would ignore the inferred trivalent ion. \(\textbf{Final answer:}\) \[ \boxed{(z,m(\mathrm M))=(3,\ 0.600\ \mathrm g)} \]

Explanation: \(\textbf{Distinguish standard and operating potentials:}\) The standard emf applies when the reaction quotient equals unity. The actual experiment maintains a different quotient: \[ Q_r=10^4 \] Therefore, the reversible operating potential must first be determined using the Nernst equation. \(\textbf{Calculate the non-standard cell potential:}\) \[ E=E^\circ-\frac{0.0591}{n}\log Q_r \] With \(n=2\): \[ E=0.1773-\frac{0.0591}{2}\log(10^4) \] \[ E=0.1773-(0.02955)(4) \] \[ E=0.1773-0.1182 \] \[ E=0.0591\ \mathrm V \] \(\textbf{Connect cell potential with reversible electrical work:}\) For one mole of the balanced reaction, the Gibbs energy change is: \[ \Delta_rG=-nFE \] The maximum electrical work delivered by a reversible galvanic cell is the magnitude of this negative Gibbs energy change. Thus: \[ w_{\mathrm{elec,max}}=nFE \] per mole of reaction at the specified fixed conditions. \(\textbf{Calculate the work for the actual reaction extent:}\) The stated extent is: \[ \xi=0.0300\ \mathrm{mol} \] Therefore: \[ W_{\mathrm{actual,max}}=\xi nFE \] \[ W_{\mathrm{actual,max}}=(0.0300)(2)(96500)(0.0591) \] \[ W_{\mathrm{actual,max}}=342.189\ \mathrm J \] \[ W_{\mathrm{actual,max}}\approx0.342\ \mathrm{kJ} \] \(\textbf{Calculate the corresponding standard-state work:}\) For the same extent at standard conditions: \[ W^\circ_{\mathrm{max}}=\xi nFE^\circ \] \[ W^\circ_{\mathrm{max}}=(0.0300)(2)(96500)(0.1773) \] \[ W^\circ_{\mathrm{max}}=1026.567\ \mathrm J \] \[ W^\circ_{\mathrm{max}}\approx1.027\ \mathrm{kJ} \] \(\textbf{Determine the work ratio:}\) Because both experiments involve the same reaction and extent: \[ \frac{W_{\mathrm{actual,max}}}{W^\circ_{\mathrm{max}}}=\frac{E}{E^\circ} \] \[ \frac{W_{\mathrm{actual,max}}}{W^\circ_{\mathrm{max}}}=\frac{0.0591}{0.1773} \] \[ \frac{W_{\mathrm{actual,max}}}{W^\circ_{\mathrm{max}}}=\frac13 \] \(\textbf{Check the thermodynamic direction:}\) The actual cell potential remains positive. Therefore: \[ \Delta_rG\lt 0 \] and the forward reaction can deliver electrical work. The available reversible work is smaller than under standard conditions because the non-standard reaction quotient reduces the forward driving force. \(\textbf{Interpret the reservoir condition:}\) In an ordinary closed cell, the reaction quotient changes as reactants are consumed and products form. The reservoir-controlled condition is specified so that the operating potential remains fixed during the stated reaction extent. This allows the electrical work to be calculated using one constant potential. Without that condition, determining the total work over a finite reaction extent could require accounting for the changing potential. The comparison illustrates that electrical charge and reaction extent alone do not determine the maximum work without the operating emf. \(\textbf{Final answer:}\) \[ \boxed{\left(W_{\mathrm{actual,max}},W_{\mathrm{actual,max}}:W^\circ_{\mathrm{max}}\right)=(0.342\ \mathrm{kJ},\ 1:3)} \]

Explanation: \(\textbf{Identify the initial electrode directions:}\) Both electrodes involve the same reduction couple: \[ \mathrm{Cu^{2+}+2e^-\rightleftharpoons Cu} \] The more concentrated solution has the greater reduction potential. Therefore, Q initially acts as the cathode. P acts as the anode. \(\textbf{Identify how discharge changes the two compartments:}\) At P: \[ \mathrm{Cu(s)\rightarrow Cu^{2+}+2e^-} \] Copper ions enter the initially dilute solution. At Q: \[ \mathrm{Cu^{2+}+2e^-\rightarrow Cu(s)} \] Copper ions are removed from the initially concentrated solution. The cell therefore tends to equalise the two copper-ion concentrations. \(\textbf{Calculate the initial copper-ion inventories:}\) For P: \[ n_{P,i}=c_PV_P \] \[ n_{P,i}=(0.0100)(0.100) \] \[ n_{P,i}=0.00100\ \mathrm{mol} \] For Q: \[ n_{Q,i}=(0.100)(0.300) \] \[ n_{Q,i}=0.0300\ \mathrm{mol} \] \(\textbf{Express the concentration changes through one reaction extent:}\) Let \(x\) moles of copper dissolve from the P electrode. The same amount of copper is deposited at Q. Therefore: \[ n_{P,f}=0.00100+x \] \[ n_{Q,f}=0.0300-x \] \(\textbf{Apply the zero-emf condition:}\) For a concentration cell involving identical electrodes, the reversible emf becomes zero when the two relevant ionic activities are equal. Under the stated concentration approximation: \[ [\mathrm{Cu^{2+}}]_{P,f}=[\mathrm{Cu^{2+}}]_{Q,f} \] Using the different compartment volumes: \[ \frac{0.00100+x}{0.100}=\frac{0.0300-x}{0.300} \] \(\textbf{Solve for the copper-transfer amount:}\) Cross-multiplying: \[ 0.300(0.00100+x)=0.100(0.0300-x) \] \[ 0.000300+0.300x=0.00300-0.100x \] \[ 0.400x=0.00270 \] \[ x=0.00675\ \mathrm{mol} \] \(\textbf{Determine the equilibrium concentration:}\) For P: \[ [\mathrm{Cu^{2+}}]_{\mathrm{eq}}=\frac{0.00100+0.00675}{0.100} \] \[ [\mathrm{Cu^{2+}}]_{\mathrm{eq}}=0.0775\ \mathrm{mol\,L^{-1}} \] For Q: \[ [\mathrm{Cu^{2+}}]_{\mathrm{eq}}=\frac{0.0300-0.00675}{0.300} \] \[ [\mathrm{Cu^{2+}}]_{\mathrm{eq}}=0.0775\ \mathrm{mol\,L^{-1}} \] Both compartments reach the same concentration. \(\textbf{Calculate the charge delivered:}\) Each copper atom oxidised at P releases two electrons. Thus: \[ n(e^-)=2x \] \[ n(e^-)=2(0.00675) \] \[ n(e^-)=0.01350\ \mathrm{mol} \] Therefore: \[ Q=n(e^-)F \] \[ Q=(0.01350)(96500) \] \[ Q=1302.75\ \mathrm C \] \[ Q\approx1303\ \mathrm C \] \(\textbf{Verify the total dissolved copper inventory:}\) Initially: \[ n_{\mathrm{total}}=0.00100+0.0300 \] \[ n_{\mathrm{total}}=0.0310\ \mathrm{mol} \] At equilibrium: \[ n_{\mathrm{total}}=(0.100+0.300)(0.0775) \] \[ n_{\mathrm{total}}=0.0310\ \mathrm{mol} \] The dissolved copper inventory is conserved across the two compartments under the specified electrode reactions. \(\textbf{Interpret the effect of unequal volumes:}\) Equal final concentrations do not require equal final amounts of copper ions because the compartment volumes differ. The initially dilute compartment needs only a limited amount of copper dissolution to reach the common concentration. The initially concentrated compartment loses that same amount through deposition. The discharge extent, and hence the electrical charge, is determined by both the initial ionic inventories and the unequal solution volumes. \(\textbf{Final answer:}\) \[ \boxed{\left(Q,[\mathrm{Cu^{2+}}]_{\mathrm{eq}}\right)=(1303\ \mathrm C,\ 0.0775\ \mathrm{mol\,L^{-1}})} \]

Explanation: \(\textbf{Identify the electron-transfer number:}\) Hydrogen oxidation is: \[ \mathrm{2H_2\rightarrow4H^++4e^-} \] Oxygen reduction consumes the same four electrons: \[ \mathrm{O_2+4H^++4e^-\rightarrow2H_2O} \] The overall reaction therefore transfers: \[ n=4 \] electrons as written. \(\textbf{Construct the complete reaction quotient:}\) The overall reaction is: \[ \mathrm{2H_2+O_2\rightarrow2H_2O} \] Pure liquid water has unit activity. Therefore: \[ Q_r=\frac1{(p_{\mathrm{H_2}}/p^\circ)^2(p_{\mathrm{O_2}}/p^\circ)} \] The hydrogen pressure is squared because two hydrogen molecules occur in the balanced reaction. \(\textbf{Evaluate the initial quotient:}\) Initially, both relative gas pressures are unity: \[ Q_i=\frac1{(1.00)^2(1.00)} \] \[ Q_i=1 \] Thus: \[ E_i=E^\circ=1.2300\ \mathrm V \] \(\textbf{Evaluate the changed gas-pressure product:}\) After the adjustment: \[ (p_{\mathrm{H_2}}/p^\circ)^2(p_{\mathrm{O_2}}/p^\circ)=(4.00)^2(0.250) \] \[ =16.0(0.250) \] \[ =4.00 \] Therefore: \[ Q_f=\frac1{4.00} \] \[ Q_f=0.250 \] \(\textbf{Calculate the final cell potential:}\) The Nernst equation gives: \[ E_f=1.23-\frac{0.0591}{4}\log(0.250) \] Using: \[ \log(0.250)=-0.60206 \] \[ E_f=1.23-\frac{0.0591}{4}(-0.60206) \] \[ E_f=1.23+0.00890 \] \[ E_f\approx1.2389\ \mathrm V \] \(\textbf{Determine the potential change:}\) \[ \Delta E=E_f-E_i \] \[ \Delta E=1.2389-1.2300 \] \[ \Delta E=+0.00890\ \mathrm V \] \(\textbf{Interpret the opposing pressure adjustments:}\) Increasing hydrogen pressure favours its consumption in the forward reaction. Decreasing oxygen pressure opposes the forward reaction. These two effects must be combined through the stoichiometrically weighted gas-pressure product. The fourfold hydrogen-pressure increase contributes a factor of sixteen because hydrogen has coefficient two. The fourfold oxygen-pressure decrease contributes a factor of one quarter. Their product therefore increases by a factor of four, lowering the reaction quotient and increasing the emf. The two pressure changes do not cancel merely because one is a fourfold increase and the other a fourfold decrease. \(\textbf{Final answer:}\) \[ \boxed{(\Delta E,E_f)=(+0.00890\ \mathrm V,\ 1.2389\ \mathrm V)} \]

Explanation: \(\textbf{Identify what equality of actual and standard emf implies:}\) At \(298\ \mathrm K\), the cell potential is: \[ E=E^\circ-\frac{0.0591}{2}\log Q_r \] The observation states: \[ E=E^\circ \] Therefore, the Nernst correction must vanish: \[ \log Q_r=0 \] \[ Q_r=1 \] This does not mean the cell has reached equilibrium, because its emf is still \(1.10\ \mathrm V\). \(\textbf{Construct the reaction quotient:}\) For: \[ \mathrm{Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu} \] the pure metal activities are unity. Hence: \[ Q_r=\frac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]} \] The condition \(Q_r=1\) requires: \[ [\mathrm{Zn^{2+}}]=[\mathrm{Cu^{2+}}] \] \(\textbf{Represent discharge by its reaction extent:}\) Let \(x\) moles of zinc undergo oxidation. At the anode: \[ \mathrm{Zn\rightarrow Zn^{2+}+2e^-} \] At the cathode: \[ \mathrm{Cu^{2+}+2e^-\rightarrow Cu} \] Therefore, \(x\) moles of zinc ions are formed while \(x\) moles of copper ions are consumed. \(\textbf{Express the final ionic concentrations:}\) Because both compartment volumes remain \(1.00\ \mathrm L\): \[ [\mathrm{Zn^{2+}}]_f=0.0100+x \] \[ [\mathrm{Cu^{2+}}]_f=0.100-x \] with numerical concentrations in \(\mathrm{mol\,L^{-1}}\). \(\textbf{Apply the measured-emf condition:}\) \[ 0.0100+x=0.100-x \] \[ 2x=0.0900 \] \[ x=0.0450\ \mathrm{mol} \] \(\textbf{Verify the resulting concentrations:}\) \[ [\mathrm{Zn^{2+}}]_f=0.0100+0.0450 \] \[ [\mathrm{Zn^{2+}}]_f=0.0550\ \mathrm{mol\,L^{-1}} \] \[ [\mathrm{Cu^{2+}}]_f=0.100-0.0450 \] \[ [\mathrm{Cu^{2+}}]_f=0.0550\ \mathrm{mol\,L^{-1}} \] Their ratio is unity, as required. \(\textbf{Calculate the charge transferred:}\) Each mole of the balanced reaction transfers two moles of electrons: \[ n(e^-)=2x \] \[ n(e^-)=2(0.0450) \] \[ n(e^-)=0.0900\ \mathrm{mol} \] Therefore: \[ Q=n(e^-)F \] \[ Q=(0.0900)(96500) \] \[ Q=8685\ \mathrm C \] \(\textbf{Determine copper deposited:}\) The amount of copper metal formed is: \[ n(\mathrm{Cu})=x=0.0450\ \mathrm{mol} \] Its mass is: \[ m(\mathrm{Cu})=nM \] \[ m(\mathrm{Cu})=0.0450(63.5) \] \[ m(\mathrm{Cu})=2.8575\ \mathrm g \] \[ m(\mathrm{Cu})\approx2.858\ \mathrm g \] \(\textbf{Interpret the significant potential condition:}\) The cell begins with more copper(II) than zinc(II) ions, so its initial reaction quotient is less than unity and its initial emf exceeds \(E^\circ\). During discharge, zinc(II) accumulates and copper(II) is consumed. Eventually, their concentrations become equal, making the Nernst correction zero. The actual emf then equals the standard emf even though neither ionic concentration equals the standard value. The equality depends on their ratio, not on each concentration separately. The measured potential condition therefore permits the extent of discharge to be inferred from the electrolyte inventories. \(\textbf{Final answer:}\) \[ \boxed{(Q,m(\mathrm{Cu}))=(8685\ \mathrm C,\ 2.858\ \mathrm g)} \]

Explanation: \(\textbf{Identify the two consecutive iron reductions:}\) Iron(III) is first reduced to iron(II): \[ \mathrm{Fe^{3+}+e^-\rightarrow Fe^{2+}} \] This step transfers one electron. Iron(II) is then reduced to iron metal: \[ \mathrm{Fe^{2+}+2e^-\rightarrow Fe} \] This step transfers two electrons. Adding the balanced half-reactions cancels the intermediate iron(II) species: \[ \mathrm{Fe^{3+}+3e^-\rightarrow Fe} \] \(\textbf{Identify the thermodynamic quantity that can be added:}\) Standard electrode potentials cannot be added directly to obtain the potential for the overall reduction. The appropriate additive quantity is the standard Gibbs energy change: \[ \Delta_rG^\circ=-nFE^\circ \] Each reduction step contributes a Gibbs energy change determined by its own electron-transfer number. \(\textbf{Determine the Gibbs energy contribution of the first step:}\) \[ \Delta_rG_1^\circ=-1F(0.77) \] \[ \Delta_rG_1^\circ=-0.77F \] \(\textbf{Determine the Gibbs energy contribution of the second step:}\) \[ \Delta_rG_2^\circ=-2F(-0.44) \] \[ \Delta_rG_2^\circ=+0.88F \] \(\textbf{Add the stepwise Gibbs energy changes:}\) \[ \Delta_rG_{\mathrm{overall}}^\circ=\Delta_rG_1^\circ+\Delta_rG_2^\circ \] \[ \Delta_rG_{\mathrm{overall}}^\circ=-0.77F+0.88F \] \[ \Delta_rG_{\mathrm{overall}}^\circ=+0.11F \] \(\textbf{Recover the overall iron reduction potential:}\) The complete iron(III)-to-iron reduction transfers three electrons. Therefore: \[ \Delta_rG_{\mathrm{overall}}^\circ=-3FE^\circ_{\mathrm{Fe^{3+}/Fe}} \] Equating: \[ -3FE^\circ_{\mathrm{Fe^{3+}/Fe}}=+0.11F \] Cancel Faraday's constant: \[ E^\circ_{\mathrm{Fe^{3+}/Fe}}=-\frac{0.11}{3} \] \[ E^\circ_{\mathrm{Fe^{3+}/Fe}}=-0.0367\ \mathrm V \] \(\textbf{Determine the favourable electrode directions with copper:}\) Copper(II)/copper has: \[ E^\circ_{\mathrm{Cu^{2+}/Cu}}=+0.34\ \mathrm V \] This is greater than the overall iron(III)/iron reduction potential. Therefore, copper(II) is reduced at the cathode. Iron metal is oxidised to iron(III) at the anode. \(\textbf{Construct the balanced combined reaction:}\) Iron oxidation: \[ \mathrm{2Fe\rightarrow2Fe^{3+}+6e^-} \] Copper reduction: \[ \mathrm{3Cu^{2+}+6e^-\rightarrow3Cu} \] Adding gives: \[ \mathrm{2Fe+3Cu^{2+}\rightarrow2Fe^{3+}+3Cu} \] The complete cell reaction transfers six electrons. \(\textbf{Calculate the standard cell emf:}\) \[ E^\circ_{\mathrm{cell}}=E^\circ_{\mathrm{cathode}}-E^\circ_{\mathrm{anode}} \] \[ E^\circ_{\mathrm{cell}}=0.34-(-0.0367) \] \[ E^\circ_{\mathrm{cell}}=+0.3767\ \mathrm V \] \(\textbf{Interpret the weighted-potential relationship:}\) The two stepwise iron reductions do not transfer the same number of electrons. Their potentials therefore contribute unequally to the overall iron reduction potential. The correct overall value is obtained from the electron-weighted Gibbs energy changes rather than by taking an ordinary arithmetic mean of the two potentials. The subsequent galvanic emf depends on the difference between copper's reduction potential and the derived overall iron reduction potential. The calculation links sequential oxidation-state changes with thermodynamic additivity and full-cell electron balance. \(\textbf{Final answer:}\) \[ \boxed{\left(E^\circ_{\mathrm{Fe^{3+}/Fe}},E^\circ_{\mathrm{cell}}\right)=(-0.0367\ \mathrm V,\ +0.3767\ \mathrm V)} \]

Explanation: \(\textbf{Recognise the concentration-cell arrangement:}\) Both electrodes involve the same silver(I)/silver reduction couple. Their standard reduction potentials are therefore identical. The measured emf arises from the difference between silver-ion activities in the two compartments. For the spontaneous orientation, P has the higher reduction potential. \(\textbf{Apply the concentration-cell Nernst equation:}\) The silver reduction involves one electron: \[ \mathrm{Ag^++e^-\rightarrow Ag} \] Therefore: \[ E_{\mathrm{cell}}=0.0591\log\left(\frac{a_{\mathrm{Ag^+},P}}{a_{\mathrm{Ag^+},Q}}\right) \] Substituting the measured emf: \[ 0.4137=0.0591\log\left(\frac{0.0100}{[\mathrm{Ag^+}]_Q}\right) \] \(\textbf{Infer the unknown silver-ion concentration:}\) \[ \log\left(\frac{0.0100}{[\mathrm{Ag^+}]_Q}\right)=\frac{0.4137}{0.0591} \] \[ \log\left(\frac{0.0100}{[\mathrm{Ag^+}]_Q}\right)=7 \] Therefore: \[ \frac{0.0100}{[\mathrm{Ag^+}]_Q}=10^7 \] \[ [\mathrm{Ag^+}]_Q=\frac{0.0100}{10^7} \] \[ [\mathrm{Ag^+}]_Q=1.00\times10^{-9}\ \mathrm{mol\,L^{-1}} \] \(\textbf{Connect the electrode activity to AgCl solubility:}\) Solid AgCl establishes: \[ \mathrm{AgCl(s)\rightleftharpoons Ag^++Cl^-} \] Its solubility product is: \[ K_{sp}=[\mathrm{Ag^+}]_Q[\mathrm{Cl^-}]_Q \] Substituting: \[ K_{sp}=(1.00\times10^{-9})(0.100) \] \[ K_{sp}=1.00\times10^{-10} \] \(\textbf{Identify the reaction at electrode Q:}\) P is specified to be the cathode, where silver(I) undergoes reduction: \[ \mathrm{Ag^++e^-\rightarrow Ag} \] Q must therefore operate as the anode. The oxidation process at its silver surface is: \[ \mathrm{Ag\rightarrow Ag^++e^-} \] In the presence of chloride ions, the produced silver ions can precipitate as AgCl. Combining these processes gives the effective anodic reaction: \[ \mathrm{Ag(s)+Cl^-(aq)\rightarrow AgCl(s)+e^-} \] \(\textbf{Interpret the coupled equilibria:}\) The cell potential determines the ratio of free silver-ion activities. The chloride concentration and AgCl dissolution equilibrium then convert that electrical information into a solubility product. Although both electrodes are silver, their surrounding chemical environments establish different reduction potentials. At Q, spontaneous oxidation rather than silver deposition supplies electrons to the external circuit. The result therefore combines concentration-cell thermodynamics with precipitation equilibrium. \(\textbf{Final answer:}\) The solubility product is \(K_{sp}(\mathrm{AgCl})=1.00\times10^{-10}\). Silver is oxidised at electrode Q, the anode.

Explanation: \(\textbf{Statement I — Correct:}\) The reversible cell emf describes the thermodynamic driving force for the balanced cell reaction at the specified chemical activities. Kinetic polarisation is associated with additional potential losses when electrode reactions proceed at a finite rate. Improving the electrode surfaces can reduce these losses and improve the voltage observed under load. However, a kinetic improvement does not alter the Gibbs energy difference between unchanged reactant and product states. The reversible equilibrium emf therefore remains unchanged. \(\textbf{Statement II — Correct:}\) The salt bridge supplies a pathway for ionic charge transport between the half-cells. A bridge with greater ionic conductivity can have a lower resistance when its geometry is otherwise suitable. Under current flow, this can reduce the internal ohmic potential drop. The external terminal voltage may consequently improve. The improvement concerns electrical transport through the operating cell, not a change in its standard thermodynamic properties. \(\textbf{Statement III — Incorrect:}\) The standard equilibrium constant is related to the standard cell potential: \[ E^\circ_{\mathrm{cell}}=\frac{0.0591}{n}\log K \] at \(298\ \mathrm K\). Neither electrode kinetics nor the resistance of an inert salt bridge appears in this equilibrium relation. Changing the ease of current flow does not, by itself, change the standard Gibbs energy of the overall chemical reaction. Therefore, the standard equilibrium constant is not necessarily altered by either modification. \(\textbf{Statement IV — Correct:}\) The Daniell-cell reaction is: \[ \mathrm{Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu} \] Its reversible emf depends on: \[ Q_r=\frac{a_{\mathrm{Zn^{2+}}}}{a_{\mathrm{Cu^{2+}}}} \] Changing this quotient changes the Nernst correction at fixed temperature. Thus, a chemical concentration change can alter the reversible open-circuit potential even though a purely kinetic or resistive improvement does not. \(\textbf{Integrate the experimental distinctions:}\) The open-circuit reversible emf, the operating terminal voltage and the standard equilibrium constant are related but distinct quantities. Thermodynamic activities determine the reversible emf. Kinetic polarisation and resistance affect performance when current flows. An improved current-delivery capability must not automatically be interpreted as a change in reaction spontaneity or equilibrium. Statements I, II and IV are correct.

Explanation: \(\textbf{Determine hydrogen-ion activity from weak-acid equilibrium:}\) The analytical acid concentration is: \[ c=0.0100\ \mathrm{mol\,L^{-1}} \] Let \(x\) be the equilibrium hydrogen-ion concentration. For the monoprotic acid: \[ K_a=\frac{x^2}{c-x} \] Substituting: \[ 1.111\times10^{-4}=\frac{x^2}{0.0100-x} \] \(\textbf{Solve the equilibrium relation:}\) \[ x^2+(1.111\times10^{-4})x-1.111\times10^{-6}=0 \] The positive root is approximately: \[ x=1.00\times10^{-3}\ \mathrm{mol\,L^{-1}} \] This corresponds to approximately \(10.0\%\) acid dissociation. \(\textbf{Calculate the hydrogen-electrode potential:}\) The reduction half-reaction is: \[ \mathrm{2H^++2e^-\rightarrow H_2} \] Its Nernst equation at \(298\ \mathrm K\) is: \[ E_{\mathrm H}=0-\frac{0.0591}{2}\log\left(\frac{p_{\mathrm{H_2}}/p^\circ}{a_{\mathrm{H^+}}^2}\right) \] With unit relative hydrogen pressure: \[ E_{\mathrm H}=0.0591\log a_{\mathrm{H^+}} \] Therefore: \[ E_{\mathrm H}=0.0591\log(10^{-3}) \] \[ E_{\mathrm H}=-0.1773\ \mathrm V \] \(\textbf{Calculate the zinc-electrode potential:}\) For: \[ \mathrm{Zn^{2+}+2e^-\rightarrow Zn} \] the reduction potential is: \[ E_{\mathrm{Zn}}=-0.76+\frac{0.0591}{2}\log[\mathrm{Zn^{2+}}] \] Substituting: \[ E_{\mathrm{Zn}}=-0.76+\frac{0.0591}{2}\log(0.0100) \] \[ E_{\mathrm{Zn}}=-0.76-0.0591 \] \[ E_{\mathrm{Zn}}=-0.8191\ \mathrm V \] \(\textbf{Assign the electrode roles:}\) The hydrogen electrode has the more positive reduction potential: \[ -0.1773\gt -0.8191 \] It is therefore the cathode. Zinc undergoes oxidation at the anode. \(\textbf{Calculate the complete cell emf:}\) \[ E_{\mathrm{cell}}=E_{\mathrm{cathode}}-E_{\mathrm{anode}} \] \[ E_{\mathrm{cell}}=-0.1773-(-0.8191) \] \[ E_{\mathrm{cell}}=0.6418\ \mathrm V \] \(\textbf{Interpret the weak-acid condition:}\) The analytical concentration of HA is not the same as the hydrogen-ion concentration because HA is only partially dissociated. Using \(0.0100\ \mathrm{mol\,L^{-1}}\) directly as the hydrogen-ion activity would give an incorrect hydrogen-electrode potential. The weak-acid equilibrium must be solved before the Nernst equation is applied. The zinc-ion concentration must also be included in the zinc reduction potential. The spontaneous cell direction follows from comparing the two actual, rather than standard, half-cell potentials. \(\textbf{Final answer:}\) The hydrogen electrode is the cathode, and \(E_{\mathrm{cell}}\approx0.6418\ \mathrm V\).

Explanation: \(\textbf{Identify which reduction responds to pH:}\) Permanganate reduction consumes eight hydrogen ions for every five electrons. Its reduction potential therefore depends strongly on hydrogen-ion activity. The supplied chlorine reduction does not contain hydrogen ions. With chlorine pressure and chloride activity fixed, its reduction potential remains constant. \(\textbf{Construct the permanganate pH relation:}\) For the specified fixed manganese-ion activity ratio: \[ Q_{\mathrm{Mn}}=\frac1{a_{\mathrm{H^+}}^8} \] At \(298\ \mathrm K\): \[ E_{\mathrm{Mn}}=1.51-\frac{0.0591}{5}\log Q_{\mathrm{Mn}} \] Using: \[ \log Q_{\mathrm{Mn}}=8\,\mathrm{pH} \] gives: \[ E_{\mathrm{Mn}}=1.51-\frac{8(0.0591)}5\mathrm{pH} \] \[ E_{\mathrm{Mn}}=1.51-0.09456\,\mathrm{pH} \] \(\textbf{Compare the couples at pH 1.00:}\) \[ E_{\mathrm{Mn},1}=1.51-0.09456(1.00) \] \[ E_{\mathrm{Mn},1}=1.41544\ \mathrm V \] The chlorine reduction potential is: \[ E_{\mathrm{Cl}}=1.3600\ \mathrm V \] Therefore: \[ E_{\mathrm{Mn},1}\gt E_{\mathrm{Cl}} \] Permanganate has the greater thermodynamic tendency to accept electrons at this pH. \(\textbf{Compare the couples at pH 2.00:}\) \[ E_{\mathrm{Mn},2}=1.51-0.09456(2.00) \] \[ E_{\mathrm{Mn},2}=1.32088\ \mathrm V \] Now: \[ E_{\mathrm{Cl}}\gt E_{\mathrm{Mn},2} \] The ordering has reversed. \(\textbf{Identify the crossover condition:}\) The two potentials are equal when: \[ 1.51-0.09456\,\mathrm{pH}=1.36 \] Therefore: \[ \mathrm{pH}_{\mathrm{cross}}=\frac{0.150}{0.09456} \] \[ \mathrm{pH}_{\mathrm{cross}}\approx1.59 \] The supplied pH values lie on opposite sides of this crossover. \(\textbf{Interpret the redox consequence:}\) Standard reduction potentials compare couples under standard conditions. The actual oxidising strength also depends on the chemical activities appearing in the Nernst equation. Hydrogen ions are involved in one reduction but not the other. Consequently, changing pH modifies their relative reduction potentials. The conclusion concerns thermodynamic preference under the explicitly maintained half-reaction conditions; it does not establish reaction rate or product selectivity in an uncontrolled chemical mixture. The correct comparison is given in option D.

Explanation: \(\textbf{Identify the spontaneous direction under standard conditions:}\) Silver(I) has the higher standard reduction potential. Therefore, the standard spontaneous direction involves silver deposition and copper oxidation. For the specified forward reaction: \[ E^\circ_{\mathrm{cell}}=0.80-0.34 \] \[ E^\circ_{\mathrm{cell}}=+0.460\ \mathrm V \] \(\textbf{Construct the actual reaction quotient:}\) The balanced forward reaction transfers two electrons: \[ n=2 \] Pure copper and silver solids have unit activities. Therefore: \[ Q_r=\frac{[\mathrm{Cu^{2+}}]}{[\mathrm{Ag^+}]^2} \] \(\textbf{Evaluate the non-standard composition:}\) \[ Q_r=\frac{0.100}{(0.0100)^2} \] \[ Q_r=\frac{0.100}{1.00\times10^{-4}} \] \[ Q_r=1000 \] \(\textbf{Calculate the spontaneous forward emf:}\) \[ E_{\mathrm{forward}}=E^\circ-\frac{0.0591}{2}\log Q_r \] \[ E_{\mathrm{forward}}=0.460-\frac{0.0591}{2}\log(1000) \] \[ E_{\mathrm{forward}}=0.460-(0.02955)(3) \] \[ E_{\mathrm{forward}}=0.37135\ \mathrm V \] \(\textbf{Determine the reverse-reaction voltage requirement:}\) The desired electrolytic reaction is: \[ \mathrm{Cu^{2+}+2Ag\rightarrow Cu+2Ag^+} \] This reaction is opposite to the spontaneous direction at the specified ionic activities. Its reversible cell potential is: \[ E_{\mathrm{reverse}}=-0.37135\ \mathrm V \] Therefore, the external source must oppose the spontaneous tendency by at least \(0.37135\ \mathrm V\) in the ideal reversible limit. \(\textbf{Test the lower applied voltage:}\) \[ V_1=0.350\ \mathrm V \] Since: \[ 0.350\lt 0.37135 \] the applied voltage is insufficient to drive the reverse electrochemical reaction under the stated ideal conditions. \(\textbf{Test the higher applied voltage:}\) \[ V_2=0.400\ \mathrm V \] Since: \[ 0.400\gt 0.37135 \] this applied voltage exceeds the reversible opposing emf and can drive the specified electrolytic direction. \(\textbf{Interpret the distinction from standard emf:}\) The required opposing voltage is determined by the actual reversible cell emf, not automatically by the standard emf. The ionic concentrations increase the forward reaction quotient and reduce its emf below \(0.460\ \mathrm V\). The resulting reversible threshold is \(0.37135\ \mathrm V\). In a real electrolytic system, additional overpotential and resistive losses could require a larger applied voltage. Those effects are explicitly excluded here so that the thermodynamic threshold can be identified. \(\textbf{Final answer:}\) The reversible opposing-voltage threshold is \(0.37135\ \mathrm V\). An applied \(0.350\ \mathrm V\) is insufficient, whereas \(0.400\ \mathrm V\) is sufficient under the stated ideal conditions.

Explanation: \(\textbf{Determine which half-cell initially undergoes reduction:}\) For the reduction: \[ \mathrm{Fe^{3+}+e^-\rightarrow Fe^{2+}} \] the actual electrode potential is: \[ E=0.77+0.0591\log\left(\frac{[\mathrm{Fe^{3+}}]}{[\mathrm{Fe^{2+}}]}\right) \] The initial iron(III)-to-iron(II) ratio in Q is greater than that in P. Therefore, Q has the greater reduction potential and acts as the cathode. P acts as the anode. \(\textbf{Identify how discharge changes both iron-ion populations:}\) At P: \[ \mathrm{Fe^{2+}\rightarrow Fe^{3+}+e^-} \] At Q: \[ \mathrm{Fe^{3+}+e^-\rightarrow Fe^{2+}} \] Each electron transferred changes the oxidation state of one iron ion in each compartment. Neither electrode deposits iron metal. \(\textbf{Represent the discharge extent:}\) Let \(x\) moles of electrons pass through the external wire. The final inventories in P are: \[ n_P(\mathrm{Fe^{3+}})=0.0300+x \] \[ n_P(\mathrm{Fe^{2+}})=0.0700-x \] The final inventories in Q are: \[ n_Q(\mathrm{Fe^{3+}})=0.0900-x \] \[ n_Q(\mathrm{Fe^{2+}})=0.0100+x \] \(\textbf{Apply the zero-emf equilibrium condition:}\) The two electrodes involve the same standard couple. Their potentials become equal when their iron(III)-to-iron(II) activity ratios become equal. Thus: \[ \frac{0.0300+x}{0.0700-x}=\frac{0.0900-x}{0.0100+x} \] \(\textbf{Solve the composition-balance equation:}\) Cross-multiplying: \[ (0.0300+x)(0.0100+x)=(0.0900-x)(0.0700-x) \] Expanding: \[ 0.000300+0.0400x+x^2=0.006300-0.1600x+x^2 \] The quadratic terms cancel: \[ 0.2000x=0.006000 \] \[ x=0.0300\ \mathrm{mol} \] \(\textbf{Calculate the charge delivered:}\) \[ Q=n(e^-)F \] \[ Q=(0.0300)(96500) \] \[ Q=2895\ \mathrm C \] \(\textbf{Determine the common final ionic ratio:}\) For P: \[ n_P(\mathrm{Fe^{3+}})=0.0300+0.0300 \] \[ n_P(\mathrm{Fe^{3+}})=0.0600\ \mathrm{mol} \] \[ n_P(\mathrm{Fe^{2+}})=0.0700-0.0300 \] \[ n_P(\mathrm{Fe^{2+}})=0.0400\ \mathrm{mol} \] Therefore: \[ \left(\frac{[\mathrm{Fe^{3+}}]}{[\mathrm{Fe^{2+}}]}\right)_P=\frac{0.0600}{0.0400} \] \[ =1.50 \] For Q: \[ n_Q(\mathrm{Fe^{3+}})=0.0900-0.0300=0.0600\ \mathrm{mol} \] \[ n_Q(\mathrm{Fe^{2+}})=0.0100+0.0300=0.0400\ \mathrm{mol} \] Its ratio is also \(1.50\). \(\textbf{Verify the zero-emf result:}\) Both final electrode potentials are: \[ E=0.77+0.0591\log(1.50) \] Because their actual potentials are equal: \[ E_{\mathrm{cell}}=E_Q-E_P=0 \] \(\textbf{Interpret the difference from a metal-ion concentration cell:}\) Here, two dissolved oxidation states participate at each electrode. The equilibrium condition is equality of their activity ratios, rather than equality of a single metal-ion concentration alone. Both the numerator and denominator of each ratio change during discharge. Consequently, the electron amount must be obtained using two coupled oxidation-state inventories. The final ratios determine the potential difference, while Faraday's law connects the reaction extent to the delivered charge. \(\textbf{Final answer:}\) \[ \boxed{\left(Q,\frac{[\mathrm{Fe^{3+}}]}{[\mathrm{Fe^{2+}}]}\right)=(2895\ \mathrm C,\ 1.50)} \]

Explanation: \(\textbf{Use Gibbs energy to infer electron stoichiometry:}\) The standard Gibbs energy of an electrochemical reaction satisfies: \[ \Delta_rG^\circ=-nFE^\circ \] Rearranging: \[ n=-\frac{\Delta_rG^\circ}{FE^\circ} \] Convert the given Gibbs energy into joules: \[ \Delta_rG^\circ=-17109.45\ \mathrm{J\,mol^{-1}} \] \(\textbf{Calculate the electron-transfer number:}\) \[ n=\frac{17109.45}{(96500)(0.0591)} \] \[ n=\frac{17109.45}{5703.15} \] \[ n=3 \] The balanced reaction therefore transfers three moles of electrons per mole of reaction as written. \(\textbf{Recover the equilibrium constant:}\) At \(298\ \mathrm K\): \[ E^\circ=\frac{0.0591}{n}\log K \] Therefore: \[ \log K=\frac{nE^\circ}{0.0591} \] Substituting: \[ \log K=\frac{3(0.0591)}{0.0591} \] \[ \log K=3 \] \[ K=10^3 \] \(\textbf{Calculate the non-standard cell potential:}\) The reaction quotient is: \[ Q_r=100=10^2 \] Using the Nernst equation: \[ E=E^\circ-\frac{0.0591}{n}\log Q_r \] \[ E=0.0591-\frac{0.0591}{3}\log(10^2) \] \[ E=0.0591-(0.0197)(2) \] \[ E=0.0197\ \mathrm V \] \(\textbf{Verify the actual Gibbs energy sign:}\) The non-standard Gibbs energy change is: \[ \Delta_rG=-nFE \] \[ \Delta_rG=-3(96500)(0.0197) \] \[ \Delta_rG=-5703.15\ \mathrm{J\,mol^{-1}} \] Its negative sign confirms that the forward direction remains thermodynamically favourable. \(\textbf{Check against the equilibrium criterion:}\) The supplied reaction quotient satisfies: \[ Q_r=10^2\lt K=10^3 \] Therefore, the mixture is on the reactant-favoured side relative to equilibrium and can still progress in the forward direction. This agrees with the positive cell potential. \(\textbf{Interpret the three independent constraints:}\) The ratio of standard Gibbs energy to standard emf determines the electron-transfer number. That electron number and standard emf establish the equilibrium constant. Finally, the difference between the equilibrium constant and actual reaction quotient determines the direction and magnitude of the non-standard driving force. The three calculations must remain thermodynamically consistent. The correct result is the combination stated in option A. \(\textbf{Final answer:}\) \[ \boxed{(n,K,E)=(3,\ 10^3,\ +0.0197\ \mathrm V)} \]

Explanation: \(\textbf{Statement I — Correct:}\) The original reaction transfers two electrons because zinc changes from oxidation state zero to \(+2\). Tripling every reaction coefficient triples the electron-transfer number represented by the balanced equation. Therefore: \[ n_R=2,\qquad n_{3R}=6 \] The new equation represents three times the original chemical reaction extent. \(\textbf{Statement II — Correct:}\) The original quotient is: \[ Q_R=\frac{a_{\mathrm{Zn^{2+}}}}{a_{\mathrm{Cu^{2+}}}} \] For the tripled reaction, the ionic stoichiometric powers also triple: \[ Q_{3R}=\frac{a_{\mathrm{Zn^{2+}}}^3}{a_{\mathrm{Cu^{2+}}}^3} \] Therefore: \[ Q_{3R}=Q_R^3 \] The physical activities have not changed; only their powers in the written reaction quotient have changed. \(\textbf{Statement III — Correct:}\) For R: \[ E_R=E_R^\circ-\frac{0.0591}{2}\log Q_R \] For 3R: \[ E_{3R}=E_{3R}^\circ-\frac{0.0591}{6}\log(Q_R^3) \] Because electrode potential is intensive: \[ E_{3R}^\circ=E_R^\circ \] Also: \[ \log(Q_R^3)=3\log Q_R \] Hence: \[ E_{3R}=E_R^\circ-\frac{0.0591}{6}(3\log Q_R) \] \[ E_{3R}=E_R^\circ-\frac{0.0591}{2}\log Q_R \] Therefore, both representations give the same actual emf. \(\textbf{Statement IV — Correct:}\) The Gibbs energy change satisfies: \[ \Delta_rG=-nFE \] The cell emf remains unchanged, but the electron-transfer number triples. Consequently: \[ \Delta_rG_{3R}=3\Delta_rG_R \] This is consistent with Gibbs energy being extensive with respect to reaction stoichiometry. \(\textbf{Statement V — Incorrect:}\) Standard cell emf does not scale with the number of times a balanced chemical equation is written. It is the Gibbs energy change per unit transferred electrical charge. Tripling the reaction triples both the Gibbs energy change and the associated electron charge. Their ratio, which determines the emf, remains unchanged. \(\textbf{Integrate the thermodynamic invariance:}\) The quotient, electron-transfer number and Gibbs energy change all depend on the normalisation of the written chemical reaction. The reversible cell potential does not. The Nernst equation preserves that invariance because tripling the logarithmic reaction quotient is exactly offset by tripling the electron-transfer number. Statements I, II, III and IV are correct.

Explanation: \(\textbf{Distinguish the two half-cell quotients:}\) The platinum electrode involves reduction of dissolved iron(III): \[ \mathrm{Fe^{3+}+e^-\rightarrow Fe^{2+}} \] Its potential depends on the ratio: \[ \frac{[\mathrm{Fe^{3+}}]}{[\mathrm{Fe^{2+}}]} \] The silver electrode involves: \[ \mathrm{Ag^++e^-\rightarrow Ag(s)} \] Its reduction potential depends on the activity of silver(I) alone because the silver solid has unit activity. A common dilution therefore affects the two electrodes differently. \(\textbf{Calculate the initial platinum-electrode potential:}\) \[ E_{\mathrm{Fe},i}=0.77+0.0591\log\left(\frac{0.100}{0.100}\right) \] \[ E_{\mathrm{Fe},i}=0.7700\ \mathrm V \] \(\textbf{Calculate the initial silver-electrode potential:}\) \[ E_{\mathrm{Ag},i}=0.80+0.0591\log(0.100) \] \[ E_{\mathrm{Ag},i}=0.80-0.0591 \] \[ E_{\mathrm{Ag},i}=0.7409\ \mathrm V \] \(\textbf{Determine the initial cell emf:}\) The iron(III)/iron(II) electrode has the higher reduction potential. Therefore, platinum is the cathode and silver is the anode. The initial emf is: \[ E_i=0.7700-0.7409 \] \[ E_i=0.0291\ \mathrm V \] \(\textbf{Apply the common dilution:}\) After tenfold dilution: \[ [\mathrm{Fe^{3+}}]=[\mathrm{Fe^{2+}}]=0.0100\ \mathrm{mol\,L^{-1}} \] Their ratio remains unity. Therefore: \[ E_{\mathrm{Fe},f}=0.77+0.0591\log(1) \] \[ E_{\mathrm{Fe},f}=0.7700\ \mathrm V \] \(\textbf{Determine the changed silver potential:}\) The silver-ion concentration decreases tenfold: \[ E_{\mathrm{Ag},f}=0.80+0.0591\log(0.0100) \] \[ E_{\mathrm{Ag},f}=0.80-0.1182 \] \[ E_{\mathrm{Ag},f}=0.6818\ \mathrm V \] \(\textbf{Calculate the new cell emf:}\) Platinum remains the cathode: \[ E_f=0.7700-0.6818 \] \[ E_f=0.0882\ \mathrm V \] The emf change is: \[ \Delta E=0.0882-0.0291 \] \[ \Delta E=+0.0591\ \mathrm V \] \(\textbf{Interpret the unequal electrochemical responses:}\) Both iron oxidation states decrease in concentration by the same factor, so their ratio and platinum-electrode potential remain unchanged. Only the silver-ion concentration appears in the silver-electrode Nernst correction. Its decrease lowers the silver reduction potential. The difference between the electrode potentials therefore becomes larger. Diluting every dissolved ionic species by the same factor does not guarantee an unchanged cell emf when the half-cell reactions have different reaction quotients. The correct comparison is given in option B.