Redox Reactions and Electrochemistry MCQs with Answers – Part 3 (NEET Chemistry)

Explanation: \(\textbf{Determine the transferred electron amount:}\) The charge carried by one mole of electrons is one faraday. Therefore: \[ n(e^-)=\frac{Q}{F} \] Substituting: \[ n(e^-)=\frac{1930\ \mathrm C}{96500\ \mathrm{C\,mol^{-1}}} \] \[ n(e^-)=0.0200\ \mathrm{mol} \] \(\textbf{Analyse the zinc half-cell:}\) Zinc undergoes oxidation: \[ \mathrm{Zn\rightarrow Zn^{2+}+2e^-} \] Two moles of electrons are released per mole of zinc dissolved. The amount of zinc ions produced is: \[ n(\mathrm{Zn^{2+}})=\frac{0.0200}{2} \] \[ n(\mathrm{Zn^{2+}})=0.0100\ \mathrm{mol} \] Each zinc ion introduces two positive charge equivalents into the zinc compartment. Therefore, the uncompensated positive charge-equivalent amount would be: \[ n_{\mathrm{positive\ equivalents}}=2(0.0100) \] \[ =0.0200\ \mathrm{mol} \] \(\textbf{Identify bridge-ion migration at the anode:}\) Nitrate ions are monovalent anions. Under the stated simplified charge-compensation model, they migrate into the zinc half-cell to compensate its increasing positive charge. Hence: \[ n(\mathrm{NO_3^-})_{\mathrm{to\ Zn}}=0.0200\ \mathrm{mol} \] \(\textbf{Analyse the copper half-cell:}\) Copper(II) ions are reduced: \[ \mathrm{Cu^{2+}+2e^-\rightarrow Cu} \] The amount removed from solution is: \[ n(\mathrm{Cu^{2+}})_{\mathrm{consumed}}=\frac{0.0200}{2} \] \[ n(\mathrm{Cu^{2+}})_{\mathrm{consumed}}=0.0100\ \mathrm{mol} \] Removing this amount of copper(II) produces a deficit of \(0.0200\ \mathrm{mol}\) of positive charge equivalents. \(\textbf{Identify bridge-ion migration at the cathode:}\) Potassium ions migrate towards the copper compartment to compensate the loss of positive ionic charge. Because potassium ions are monovalent: \[ n(\mathrm{K^+})_{\mathrm{to\ Cu}}=0.0200\ \mathrm{mol} \] \(\textbf{Verify the electrical accounting:}\) Each compartment requires compensation corresponding to \(0.0200\ \mathrm{mol}\) of elementary charges. This equals the electron amount transferred through the external circuit. The zinc and copper metal amounts changing are only \(0.0100\ \mathrm{mol}\) each because each metal-ion conversion involves two electrons. The salt bridge transports ions, not electrons. The bridge-ion amounts are therefore governed by charge equivalents rather than simply by the number of moles of metal dissolved or deposited. \(\textbf{Final answer:}\) \[ \boxed{\left(n(\mathrm{NO_3^-})_{\mathrm{to\ Zn}},n(\mathrm{K^+})_{\mathrm{to\ Cu}}\right)=(0.0200,\ 0.0200)\ \mathrm{mol}} \]

Explanation: \(\textbf{Determine the charge in electron-mole units:}\) \[ n(e^-)=\frac{Q}{F} \] \[ n(e^-)=\frac{19300}{96500} \] \[ n(e^-)=0.200\ \mathrm{mol} \] \(\textbf{Evaluate the common cathode reaction:}\) At either copper cathode: \[ \mathrm{Cu^{2+}+2e^-\rightarrow Cu} \] Two electrons are needed per copper(II) ion reduced. Therefore: \[ n(\mathrm{Cu^{2+}})_{\mathrm{cathode}}=\frac{0.200}{2} \] \[ =0.100\ \mathrm{mol} \] Both cathodes remove \(0.100\ \mathrm{mol}\) of copper(II) ions from solution. \(\textbf{Evaluate the inert anode in P:}\) The specified oxygen-evolution half-reaction is: \[ \mathrm{2H_2O\rightarrow O_2+4H^++4e^-} \] Four moles of electrons are released per mole of oxygen produced. Therefore: \[ n(\mathrm{O_2})_{\mathrm P}=\frac{0.200}{4} \] \[ n(\mathrm{O_2})_{\mathrm P}=0.0500\ \mathrm{mol} \] The platinum anode supplies no copper(II) ions. Thus, the net amount of dissolved copper(II) decreases by \(0.100\ \mathrm{mol}\). \(\textbf{Evaluate the active copper anode in Q:}\) At the copper anode: \[ \mathrm{Cu\rightarrow Cu^{2+}+2e^-} \] The copper(II) amount released is: \[ n(\mathrm{Cu^{2+}})_{\mathrm{anode}}=\frac{0.200}{2} \] \[ n(\mathrm{Cu^{2+}})_{\mathrm{anode}}=0.100\ \mathrm{mol} \] The same amount is removed at the cathode. Therefore: \[ \Delta n(\mathrm{Cu^{2+}})_{\mathrm Q}=0.100-0.100 \] \[ \Delta n(\mathrm{Cu^{2+}})_{\mathrm Q}=0 \] \(\textbf{Interpret electrode-material dependence:}\) The cathode process is identical in both cells, but the anode chemistry differs. In P, copper(II) is consumed while oxygen evolves. In Q, copper is transferred from the anode to the cathode without a net change in dissolved copper(II), under the stated assumptions. An anode is defined by oxidation, not by whether it is chemically inert. This is why the identity of the electrode material matters when predicting electrolysis products.

Explanation: \(\textbf{Determine the common electrical charge:}\) Charge passed through an electrode is: \[ Q=It \] Substituting: \[ Q=(1.93\ \mathrm A)(500\ \mathrm s) \] \[ Q=965\ \mathrm C \] \(\textbf{Convert charge into electron amount:}\) \[ n(e^-)=\frac{Q}{F} \] \[ n(e^-)=\frac{965}{96500} \] \[ n(e^-)=0.0100\ \mathrm{mol} \] Because the cells are in series, the same electron-mole equivalent passes through both. \(\textbf{Determine silver deposition:}\) The cathode reaction is: \[ \mathrm{Ag^++e^-\rightarrow Ag} \] One mole of electrons deposits one mole of silver. Therefore: \[ n(\mathrm{Ag})=0.0100\ \mathrm{mol} \] The mass is: \[ m(\mathrm{Ag})=nM \] \[ m(\mathrm{Ag})=0.0100(108) \] \[ m(\mathrm{Ag})=1.080\ \mathrm g \] \(\textbf{Determine copper deposition:}\) Copper(II) requires two electrons per deposited atom: \[ \mathrm{Cu^{2+}+2e^-\rightarrow Cu} \] Thus: \[ n(\mathrm{Cu})=\frac{0.0100}{2} \] \[ n(\mathrm{Cu})=0.00500\ \mathrm{mol} \] The copper mass is: \[ m(\mathrm{Cu})=0.00500(63.5) \] \[ m(\mathrm{Cu})=0.3175\ \mathrm g \] \[ m(\mathrm{Cu})\approx0.318\ \mathrm g \] \(\textbf{Verify the relative deposited masses:}\) Equal charges do not imply equal masses of metal deposited. The deposited amount depends on the electron coefficient in the half-reaction. Here: \[ \frac{m(\mathrm{Ag})}{m(\mathrm{Cu})}=\frac{108/1}{63.5/2} \] The electron stoichiometry and molar masses together determine the mass ratio. The electrical charge is the same in the two series cells, while the required electrons per metal atom differ. \(\textbf{Final answer:}\) \[ \boxed{\left(m(\mathrm{Ag}),m(\mathrm{Cu})\right)=(1.080,\ 0.318)\ \mathrm g} \]

Explanation: \(\textbf{Compare the supplied reduction tendencies:}\) A more positive standard reduction potential indicates the greater tendency to undergo reduction under compatible standard conditions. The iron(III)/iron(II) reduction potential is \(+0.77\ \mathrm V\). The zinc(II)/zinc reduction potential is \(-0.76\ \mathrm V\). The iron(III) couple is therefore chosen as the cathode process in the spontaneous galvanic direction. \(\textbf{Write the cathode reaction:}\) At the platinum electrode: \[ \mathrm{Fe^{3+}+e^-\rightarrow Fe^{2+}} \] Platinum serves as an inert electrical conductor. It is not required to enter the reaction as a platinum ion. \(\textbf{Determine the anode reaction:}\) The zinc reduction half-reaction must be reversed: \[ \mathrm{Zn\rightarrow Zn^{2+}+2e^-} \] Zinc is therefore the negative anode during galvanic operation. Electrons move from zinc through the external circuit to the platinum cathode. \(\textbf{Balance the cell reaction:}\) Multiply the iron half-reaction by two to cancel the two electrons released by zinc: \[ \mathrm{2Fe^{3+}+2e^-\rightarrow 2Fe^{2+}} \] Adding gives: \[ \mathrm{Zn+2Fe^{3+}\rightarrow Zn^{2+}+2Fe^{2+}} \] \(\textbf{Calculate the standard emf:}\) Using reduction potentials: \[ E^\circ_{\mathrm{cell}}=E^\circ_{\mathrm{cathode}}-E^\circ_{\mathrm{anode}} \] \[ E^\circ_{\mathrm{cell}}=0.77-(-0.76) \] \[ E^\circ_{\mathrm{cell}}=1.53\ \mathrm V \] \(\textbf{Distinguish electron balancing from potential scaling:}\) The iron half-reaction is multiplied by two to balance electrons. Its reduction potential is not multiplied by two. Electrode potential measures an electrical driving force per unit charge, not total charge or total reaction energy. Therefore, the alternative calculation that doubles \(+0.77\ \mathrm V\) is incorrect. \(\textbf{Final answer:}\) \[ E^\circ_{\mathrm{cell}}=1.53\ \mathrm V \]

Explanation: \(\textbf{Identify the discharge half-reactions:}\) Zinc undergoes oxidation: \[ \mathrm{Zn\rightarrow Zn^{2+}+2e^-} \] Silver ions undergo reduction: \[ \mathrm{2Ag^++2e^-\rightarrow 2Ag} \] The zinc compartment is the anode half-cell. The silver compartment is the cathode half-cell. \(\textbf{Examine charge development at the anode:}\) Oxidation releases zinc(II) ions into the zinc solution. This tends to increase positive ionic charge in that compartment. Anions from the salt bridge must migrate towards the anode compartment to preserve electrical neutrality. \(\textbf{Examine charge development at the cathode:}\) Silver ions are removed from solution as silver metal is deposited. This tends to leave an excess of negative charge from the accompanying anions. Cations from the salt bridge migrate towards the cathode compartment to maintain electrical neutrality. \(\textbf{Compare the two proposed bridge electrolytes:}\) Potassium nitrate supplies \(\mathrm{K^+}\) and \(\mathrm{NO_3^-}\). Under the stipulated conditions, these ions can perform the required ionic charge compensation without introducing the precipitation problem associated with chloride. Potassium chloride supplies chloride ions that can react with silver ions: \[ \mathrm{Ag^++Cl^-\rightarrow AgCl(s)} \] Silver chloride precipitation would change the participating silver-ion concentration and disturb the intended silver electrode system. Therefore, KCl is unsuitable under these conditions. \(\textbf{Assign the correct migration directions:}\) With KNO3, nitrate ions migrate towards the zinc anode compartment. Potassium ions migrate towards the silver cathode compartment. Neither species is intended to participate in the electrode redox half-reactions. \(\textbf{Interpret the experimental requirement:}\) A salt bridge is not merely any conducting solution placed between two half-cells. Its ions should maintain ionic conduction and charge balance without interfering significantly with the intended electrode chemistry. The selection of the bridge electrolyte and the migration directions must therefore be considered together. Potassium nitrate is preferable in the stated zinc–silver cell.

Explanation: \(\textbf{Arrangement P — zinc metal and zinc ions:}\) A zinc electrode immersed in a solution containing zinc(II) ions forms a metal–metal-ion electrode. Its reduction half-reaction is: \[ \mathrm{Zn^{2+}+2e^-\rightleftharpoons Zn} \] The metal itself is one of the participating forms of the redox couple. Therefore, P matches description 3. \(\textbf{Arrangement Q — dissolved iron redox couple:}\) Both iron(II) and iron(III) are present as dissolved ions. Their electron exchange is represented by: \[ \mathrm{Fe^{3+}+e^-\rightleftharpoons Fe^{2+}} \] Neither form is a solid iron electrode in the described arrangement. Platinum provides an electronically conducting surface at which electron transfer can occur. It is not consumed as a platinum ion in the specified half-reaction. Therefore, Q matches description 1. \(\textbf{Arrangement R — hydrogen gas electrode:}\) The hydrogen electrode involves hydrogen gas and hydrogen ions in contact with a suitable platinum surface. Its reversible reduction half-reaction is: \[ \mathrm{2H^++2e^-\rightleftharpoons H_2} \] Under standard-state conditions, this arrangement forms the basis of the standard hydrogen electrode. Therefore, R matches description 2. \(\textbf{Arrangement S — chlorine gas electrode:}\) The chlorine electrode contains molecular chlorine and chloride ions. Platinum supplies the conducting surface. The reversible reduction half-reaction is: \[ \mathrm{Cl_2+2e^-\rightleftharpoons 2Cl^-} \] Therefore, S matches description 4. \(\textbf{Interpret the common principle:}\) A half-cell must provide both the chemical species of the redox couple and an electronic connection to the external circuit. A solid metal can itself serve as the electrode when it participates in the redox equilibrium. When both redox forms are dissolved or a gaseous reactant is involved, an inert conducting material may provide the necessary surface. The presence of platinum does not imply that platinum must undergo oxidation or reduction. The correct matching is P–3, Q–1, R–2, S–4.

Explanation: \(\textbf{Choose the electrodes for maximum potential difference:}\) For a galvanic cell under standard conditions: \[ E^\circ_{\mathrm{cell}}=E^\circ_{\mathrm{cathode}}-E^\circ_{\mathrm{anode}} \] The cathode should have the higher reduction potential. To maximise the difference, select the most positive reduction potential as the cathode and the most negative as the anode. Silver has the highest listed reduction potential: \[ E^\circ_{\mathrm{Ag^+/Ag}}=+0.80\ \mathrm V \] Zinc has the lowest: \[ E^\circ_{\mathrm{Zn^{2+}/Zn}}=-0.76\ \mathrm V \] Therefore, the maximum-emf pair is zinc and silver. \(\textbf{Determine the maximum standard emf:}\) \[ E^\circ_{\mathrm{cell,max}}=0.80-(-0.76) \] \[ E^\circ_{\mathrm{cell,max}}=1.56\ \mathrm V \] \(\textbf{Identify zinc oxidation:}\) At the anode: \[ \mathrm{Zn\rightarrow Zn^{2+}+2e^-} \] Each mole of zinc oxidised releases two moles of electrons. For the stated zinc consumption: \[ n(e^-)_{\mathrm{released}}=2(0.100) \] \[ n(e^-)_{\mathrm{released}}=0.200\ \mathrm{mol} \] \(\textbf{Identify silver-ion reduction:}\) At the cathode: \[ \mathrm{Ag^++e^-\rightarrow Ag} \] One electron deposits one silver atom. Therefore: \[ n(\mathrm{Ag})_{\mathrm{deposited}}=0.200\ \mathrm{mol} \] \(\textbf{Confirm using the balanced cell reaction:}\) \[ \mathrm{Zn+2Ag^+\rightarrow Zn^{2+}+2Ag} \] One mole of zinc supplies electrons for the reduction of two moles of silver ions. Thus, \(0.100\ \mathrm{mol}\) zinc corresponds to \(0.200\ \mathrm{mol}\) silver deposited. \(\textbf{Distinguish potential from electron stoichiometry:}\) The emf is obtained from the difference of reduction potentials. The silver amount is obtained from the electron coefficients in the half-reactions. Multiplying the silver half-reaction by two does not multiply its reduction potential. These two independent principles determine the correct combined result. \(\textbf{Final answer:}\) \[ \boxed{\left(E^\circ_{\mathrm{cell,max}},n(\mathrm{Ag})\right)=\left(1.56\ \mathrm V,\ 0.200\ \mathrm{mol}\right)} \]

Explanation: \(\textbf{Determine the role of SHE with zinc:}\) The standard reduction potential of zinc is \(-0.76\ \mathrm V\). The standard reduction potential of the hydrogen electrode is \(0.00\ \mathrm V\). The hydrogen couple therefore has the higher reduction potential. It undergoes reduction at the cathode: \[ \mathrm{2H^++2e^-\rightarrow H_2} \] Zinc undergoes oxidation at the anode: \[ \mathrm{Zn\rightarrow Zn^{2+}+2e^-} \] Hydrogen gas is consequently produced in the zinc–SHE galvanic cell. Its standard emf is: \[ E^\circ_{\mathrm{cell}}=0.00-(-0.76) \] \[ E^\circ_{\mathrm{cell}}=+0.76\ \mathrm V \] \(\textbf{Determine the role of SHE with copper:}\) Copper has standard reduction potential \(+0.34\ \mathrm V\). This is higher than the hydrogen electrode's assigned zero value. Copper(II) ions therefore undergo reduction: \[ \mathrm{Cu^{2+}+2e^-\rightarrow Cu} \] The hydrogen electrode now acts as the anode. Its reaction is reversed: \[ \mathrm{H_2\rightarrow2H^++2e^-} \] Hydrogen gas is consumed rather than formed. The standard emf is: \[ E^\circ_{\mathrm{cell}}=0.34-0.00 \] \[ E^\circ_{\mathrm{cell}}=+0.34\ \mathrm V \] \(\textbf{Interpret the reference-electrode convention:}\) Assigning zero standard reduction potential to SHE does not make it permanently an anode or permanently a cathode. Its role depends on the reduction potential of the other electrode. Platinum supplies the electronic conducting surface, while the hydrogen/hydrogen-ion couple participates in electron transfer. The same reference electrode can therefore undergo opposite reactions when coupled with different metal electrodes.

Explanation: The Assertion is true because a galvanic cell requires both electronic and ionic charge-transport pathways for sustained operation. The zinc electrode undergoes oxidation: \[ \mathrm{Zn\rightarrow Zn^{2+}+2e^-} \] This reaction introduces additional positive ions into the zinc half-cell. At the copper electrode: \[ \mathrm{Cu^{2+}+2e^-\rightarrow Cu} \] Copper(II) ions are removed from the solution. Their removal leaves the cathode compartment with a tendency towards excess negative ionic charge. The salt bridge permits appropriate ionic migration between the half-cell regions. Anions generally migrate towards the anode compartment, while cations migrate towards the cathode compartment. These movements help compensate for the local charge changes caused by the electrode reactions. If the salt bridge is removed and no alternative ionic connection remains, charge separation develops. The resulting electrical opposition prevents sustained continuation of the coupled electrode reactions. Zinc metal and copper(II) ions may still be chemically available, but availability of reactants alone cannot maintain a continuous electrical current. The Reason is true and supplies the mechanism that explains the Assertion. The salt bridge does not need to transfer electrons between the electrodes. Electrons are carried through the external metallic circuit, while ions carry charge through the internal conducting pathway. The interruption of current is therefore a consequence of incomplete charge transport, not necessarily exhaustion of the redox reactants.

Explanation: \(\textbf{Determine the spontaneous electrode direction:}\) Silver has the slightly higher standard reduction potential. Therefore, silver ions undergo reduction at the cathode: \[ \mathrm{Ag^++e^-\rightarrow Ag} \] Iron(II) undergoes oxidation at the platinum anode: \[ \mathrm{Fe^{2+}\rightarrow Fe^{3+}+e^-} \] Platinum transfers electrons without being consumed as a reactant. \(\textbf{Write the complete cell reaction:}\) Both half-reactions involve one electron, so their electron coefficients are already equal. Adding them gives: \[ \mathrm{Fe^{2+}+Ag^+\rightarrow Fe^{3+}+Ag} \] The reaction is in the galvanic direction under the specified standard initial conditions. \(\textbf{Calculate the initial standard emf:}\) \[ E^\circ_{\mathrm{cell}}=E^\circ_{\mathrm{cathode}}-E^\circ_{\mathrm{anode}} \] \[ E^\circ_{\mathrm{cell}}=0.80-0.77 \] \[ E^\circ_{\mathrm{cell}}=0.03\ \mathrm V \] The small positive value is the initial standard emf. It must not be interpreted as a guarantee that the cell potential remains unchanged as the solution compositions change during operation. \(\textbf{Convert charge into electron amount:}\) \[ n(e^-)=\frac{Q}{F} \] \[ n(e^-)=\frac{9650}{96500} \] \[ n(e^-)=0.100\ \mathrm{mol} \] \(\textbf{Calculate silver deposition:}\) One silver ion requires one electron. Therefore: \[ n(\mathrm{Ag})_{\mathrm{deposited}}=n(e^-) \] \[ n(\mathrm{Ag})_{\mathrm{deposited}}=0.100\ \mathrm{mol} \] \(\textbf{Check the anode stoichiometry:}\) Iron(II) releases one electron per iron ion oxidised. Thus, the same charge corresponds to oxidation of \(0.100\ \mathrm{mol}\) of iron(II) to iron(III). The electron losses and gains are equal. The emf is determined by the difference between the two reduction potentials, whereas deposited silver is determined by charge and electron stoichiometry. A small cell emf does not mean that each silver ion requires a fractional electron. \(\textbf{Final answer:}\) \[ \boxed{\left(E^\circ_{\mathrm{cell}},n(\mathrm{Ag})\right)=\left(0.03\ \mathrm V,\ 0.100\ \mathrm{mol}\right)} \]

Explanation: \(\textbf{Determine hydrogen formation from cathode stoichiometry:}\) At the cathode: \[ \mathrm{2H^++2e^-\rightarrow H_2} \] Two electrons form one molecule of hydrogen. Therefore: \[ n(\mathrm{H_2})=\frac{0.400}{2} \] \[ n(\mathrm{H_2})=0.200\ \mathrm{mol} \] \(\textbf{Determine oxygen formation from anode stoichiometry:}\) At the anode: \[ \mathrm{2H_2O\rightarrow O_2+4H^++4e^-} \] Four electrons correspond to one oxygen molecule evolved. Therefore: \[ n(\mathrm{O_2})=\frac{0.400}{4} \] \[ n(\mathrm{O_2})=0.100\ \mathrm{mol} \] \(\textbf{Combine the half-reactions:}\) Multiply the hydrogen-evolution half-reaction by two: \[ \mathrm{4H^++4e^-\rightarrow2H_2} \] Adding to the oxygen-evolution half-reaction: \[ \mathrm{2H_2O+4H^++4e^-\rightarrow O_2+4H^++4e^-+2H_2} \] After cancelling hydrogen ions and electrons: \[ \mathrm{2H_2O\rightarrow2H_2+O_2} \] \(\textbf{Calculate the amount of water decomposed:}\) Two moles of water form two moles of hydrogen. Thus: \[ n(\mathrm{H_2O})_{\mathrm{decomposed}}=n(\mathrm{H_2}) \] \[ n(\mathrm{H_2O})_{\mathrm{decomposed}}=0.200\ \mathrm{mol} \] \(\textbf{Verify gas and atom conservation:}\) The gas amount ratio is: \[ n(\mathrm{H_2}):n(\mathrm{O_2})=0.200:0.100 \] \[ =2:1 \] The produced gases contain \(0.400\ \mathrm{mol}\) of hydrogen atoms and \(0.200\ \mathrm{mol}\) of oxygen atoms. These amounts correspond to decomposition of \(0.200\ \mathrm{mol}\) of water. The hydrogen ions appearing separately in the electrode equations cancel from the overall reaction. The acid provides ionic conductivity but is not consumed in the balanced net decomposition reaction under the stated conditions. \(\textbf{Final answer:}\) \[ \boxed{\left(n(\mathrm{H_2}),n(\mathrm{O_2}),n(\mathrm{H_2O})_{\mathrm{decomposed}}\right)=(0.200,\ 0.100,\ 0.200)\ \mathrm{mol}} \]

Explanation: \(\textbf{Identify the potentiometer measurement principle:}\) A potentiometer compares potential differences using a known potential gradient along a uniform conducting wire. At the balance point, no current is drawn from the cell being measured through the balancing branch. This null condition avoids a measurement error arising from a discharge current through the unknown cell. \(\textbf{Establish the emf–length relationship:}\) When the potential gradient remains unchanged: \[ E\propto l \] For a standard cell and an unknown cell: \[ \frac{E_{\mathrm{unknown}}}{E_{\mathrm{standard}}}=\frac{l_{\mathrm{unknown}}}{l_{\mathrm{standard}}} \] \(\textbf{Substitute the measured balance lengths:}\) \[ \frac{E_{\mathrm{unknown}}}{1.10}=\frac{82.5}{55.0} \] \[ \frac{82.5}{55.0}=1.50 \] Therefore: \[ E_{\mathrm{unknown}}=1.10(1.50) \] \[ E_{\mathrm{unknown}}=1.65\ \mathrm V \] \(\textbf{Distinguish emf from loaded terminal potential:}\) Emf refers to the open-circuit driving potential of the cell under specified conditions. When the cell supplies current through an external resistance, its terminal voltage can be lower because some potential is lost across its internal resistance. For a simple discharging-cell model: \[ V_{\mathrm{terminal}}=E-Ir \] where \(r\) is the internal resistance and \(I\) is the discharge current. \(\textbf{Interpret the measurement conditions:}\) At the potentiometer null point, the unknown cell does not deliver current through its balancing branch. The measured balance length therefore corresponds to its emf under the stated conditions. When connected to a load, current flows and an internal potential drop may develop. Thus, emf and loaded terminal potential are not necessarily identical. The distinction is experimental: obtaining a null balance avoids drawing the measurement current that could alter the terminal reading. \(\textbf{Final answer:}\) \[ E_{\mathrm{unknown}}=1.65\ \mathrm V \]

Explanation: \(\textbf{Use the hydrogen electrode as a reference:}\) The standard reduction potential of the hydrogen electrode is assigned: \[ E^\circ(\mathrm{H^+/H_2})=0.00\ \mathrm V \] The unknown metal electrode is explicitly observed to be the positive cathode when connected to SHE. It therefore undergoes reduction: \[ \mathrm{M^{2+}+2e^-\rightarrow M} \] The hydrogen electrode acts as the anode. \(\textbf{Infer the metal's standard reduction potential:}\) For the SHE–M cell: \[ E^\circ_{\mathrm{cell}}=E^\circ_{\mathrm{M^{2+}/M}}-E^\circ_{\mathrm{H^+/H_2}} \] Substituting: \[ 0.42=E^\circ_{\mathrm{M^{2+}/M}}-0.00 \] \[ E^\circ_{\mathrm{M^{2+}/M}}=+0.42\ \mathrm V \] \(\textbf{Compare M with silver:}\) The silver electrode has: \[ E^\circ_{\mathrm{Ag^+/Ag}}=+0.80\ \mathrm V \] This is more positive than the metal electrode's reduction potential. Consequently, silver ions undergo reduction in the spontaneous galvanic direction. M must undergo oxidation: \[ \mathrm{M\rightarrow M^{2+}+2e^-} \] Thus, M is the anode and silver is the cathode. \(\textbf{Balance the overall cell reaction:}\) The silver reduction reaction requires one electron per silver ion. Multiply it by two: \[ \mathrm{2Ag^++2e^-\rightarrow2Ag} \] Combining with metal oxidation: \[ \mathrm{M+2Ag^+\rightarrow M^{2+}+2Ag} \] \(\textbf{Calculate the second cell emf:}\) \[ E^\circ_{\mathrm{cell}}=E^\circ_{\mathrm{Ag^+/Ag}}-E^\circ_{\mathrm{M^{2+}/M}} \] \[ E^\circ_{\mathrm{cell}}=0.80-0.42 \] \[ E^\circ_{\mathrm{cell}}=0.38\ \mathrm V \] \(\textbf{Interpret the changed electrode role:}\) The unknown metal is a cathode against the hydrogen electrode but an anode against silver. This does not indicate that its standard reduction potential has changed. Electrode roles depend on the relative reduction potentials of the two coupled half-cells. A positive reduction potential relative to SHE does not mean that an electrode must act as a cathode against every other electrode. \(\textbf{Final answer:}\) M has \(E^\circ_{\mathrm{M^{2+}/M}}=+0.42\ \mathrm V\). In the silver-containing cell, M is the anode and \(E^\circ_{\mathrm{cell}}=+0.38\ \mathrm V\).

Explanation: \(\textbf{Identify the oxidation half-reaction:}\) Tin changes from oxidation number \(+2\) to \(+4\). Therefore, tin(II) ions undergo oxidation: \[ \mathrm{Sn^{2+}\rightarrow Sn^{4+}+2e^-} \] This process occurs at the anode. In conventional galvanic-cell notation, the anode half-cell is written on the left. \(\textbf{Identify the reduction half-reaction:}\) Iron changes from \(+3\) to \(+2\). Each iron(III) ion accepts one electron: \[ \mathrm{Fe^{3+}+e^-\rightarrow Fe^{2+}} \] Two iron(III) ions accept the two electrons supplied by one tin(II) ion. This reduction occurs at the cathode, conventionally written on the right. \(\textbf{Choose a suitable conducting surface:}\) Both tin species are dissolved ions. Neither tin metal nor solid iron participates in the specified overall reaction. Consequently, platinum can provide an inert conducting surface for both half-cells. The electrode reaction involves exchange of electrons between ionic oxidation states, not deposition of elemental tin or iron. \(\textbf{Interpret the notation symbols:}\) A single vertical line indicates a boundary between different phases, such as platinum metal and an aqueous solution. A comma between dissolved species indicates that both belong to the same solution phase. A double vertical line represents the connection between the two half-cells through the salt bridge or liquid junction. \(\textbf{Evaluate the proposed representations:}\) Option A incorrectly introduces elemental tin and elemental iron as though they were the participating metal electrodes. Option B reverses the oxidation and reduction half-cells relative to the specified galvanic reaction. Option C correctly places the tin ionic redox couple on the left and the iron ionic redox couple on the right, with inert platinum conductors. Option D incorrectly uses phase-boundary symbols to separate ionic species that are present together in the same aqueous phase. The correct representation must be consistent simultaneously with the electron-transfer direction, electrode materials and phase notation. Option C satisfies all three requirements.

Explanation: \(\textbf{Interpret the original galvanic reaction:}\) Zinc undergoes oxidation: \[ \mathrm{Zn\rightarrow Zn^{2+}+2e^-} \] Silver ions undergo reduction: \[ \mathrm{2Ag^++2e^-\rightarrow2Ag} \] The reaction is spontaneous under the specified standard conditions. Its standard emf is positive: \[ E^\circ_{\mathrm{forward}}=+1.56\ \mathrm V \] \(\textbf{Analyse Reaction P:}\) Reaction P is obtained by multiplying every stoichiometric coefficient of the original reaction by two. The chemical identities and electrode couples are unchanged. The standard reduction potentials of those electrode couples also remain unchanged. A standard cell potential is a potential difference, representing the driving force per unit charge. Doubling the reaction doubles the number of electrons transferred for the stoichiometric equation as written. It does not double the potential difference. Therefore: \[ E^\circ_{\mathrm P}=+1.56\ \mathrm V \] \(\textbf{Analyse Reaction Q:}\) Reaction Q is the exact reverse of the original spontaneous cell reaction. The original cathode process becomes oxidation in the reverse direction. The original anode process becomes reduction. Thus, the electrode roles exchange when the reaction direction is reversed. The standard potential for the written reverse reaction changes sign: \[ E^\circ_{\mathrm Q}=-E^\circ_{\mathrm{forward}} \] \[ E^\circ_{\mathrm Q}=-1.56\ \mathrm V \] \(\textbf{Distinguish the two mathematical operations:}\) Scaling a balanced reaction changes the amount of chemical transformation described by the equation. Reversing a reaction changes the direction of the transformation. The first operation leaves the standard potential unchanged. The second operation reverses its sign. The negative potential of Q indicates that the reverse reaction is not spontaneous under the same standard conditions. Therefore, option D correctly distinguishes reaction scaling from reaction reversal.

Explanation: \(\textbf{Translate Cell P into a potential difference:}\) Y acts as the anode, while X acts as the cathode. Using standard reduction potentials: \[ E^\circ_{\mathrm P}=E^\circ_{\mathrm X}-E^\circ_{\mathrm Y} \] The measured value is: \[ E^\circ_{\mathrm X}-E^\circ_{\mathrm Y}=0.42\ \mathrm V \] Therefore, X has a more positive standard reduction potential than Y. \(\textbf{Translate Cell Q into a second difference:}\) Z is the anode and Y is the cathode. Therefore: \[ E^\circ_{\mathrm Q}=E^\circ_{\mathrm Y}-E^\circ_{\mathrm Z} \] The measured value gives: \[ E^\circ_{\mathrm Y}-E^\circ_{\mathrm Z}=0.28\ \mathrm V \] Thus, Y has a more positive reduction potential than Z. \(\textbf{Establish the reduction-potential order:}\) Combining the observations: \[ E^\circ_{\mathrm X}\gt E^\circ_{\mathrm Y}\gt E^\circ_{\mathrm Z} \] Therefore, when X and Z are connected, X is the cathode and Z is the anode in the spontaneous direction. \(\textbf{Infer the missing potential difference:}\) The required standard emf is: \[ E^\circ_{\mathrm{X-Z}}=E^\circ_{\mathrm X}-E^\circ_{\mathrm Z} \] Insert the intermediate Y potential: \[ E^\circ_{\mathrm X}-E^\circ_{\mathrm Z} =(E^\circ_{\mathrm X}-E^\circ_{\mathrm Y})+(E^\circ_{\mathrm Y}-E^\circ_{\mathrm Z}) \] Substitute the experimental differences: \[ E^\circ_{\mathrm{X-Z}}=0.42+0.28 \] \[ E^\circ_{\mathrm{X-Z}}=0.70\ \mathrm V \] \(\textbf{Determine the electrode half-reactions:}\) At the Z anode: \[ \mathrm{Z\rightarrow Z^{2+}+2e^-} \] At the X cathode: \[ \mathrm{X^{2+}+2e^-\rightarrow X} \] Both processes involve two electrons, so their half-reactions combine directly. \(\textbf{Construct the spontaneous cell reaction:}\) \[ \mathrm{Z+X^{2+}\rightarrow Z^{2+}+X} \] \(\textbf{Verify the physical interpretation:}\) The X–Z cell spans both measured potential differences. Its standard emf is consequently greater than either \(0.42\ \mathrm V\) or \(0.28\ \mathrm V\). The individual absolute reduction potentials of X, Y and Z are unnecessary because only their differences enter the cell-emf calculation. The result follows from a consistent reduction-potential ordering and compatible electrode stoichiometry. \(\textbf{Final answer:}\) The spontaneous reaction is \(\mathrm{Z+X^{2+}\rightarrow Z^{2+}+X}\), with \(E^\circ_{\mathrm{cell}}=+0.70\ \mathrm V\).

Explanation: \(\textbf{Identify the preferred reduction:}\) The \(\mathrm{Fe^{3+}/Fe^{2+}}\) couple has a higher standard reduction potential than the \(\mathrm{Cu^{2+}/Cu}\) couple. Therefore, iron(III) ions are reduced in the spontaneous galvanic direction. Copper metal supplies the electrons by undergoing oxidation. \(\textbf{Construct the electrode half-reactions:}\) At the copper anode: \[ \mathrm{Cu\rightarrow Cu^{2+}+2e^-} \] At the platinum cathode: \[ \mathrm{Fe^{3+}+e^-\rightarrow Fe^{2+}} \] The iron half-reaction must be multiplied by two for electron cancellation: \[ \mathrm{2Fe^{3+}+2e^-\rightarrow2Fe^{2+}} \] Adding gives Reaction P. \(\textbf{Determine the standard emf of P:}\) Using standard reduction potentials: \[ E^\circ_{\mathrm P}=E^\circ_{\mathrm{cathode}}-E^\circ_{\mathrm{anode}} \] \[ E^\circ_{\mathrm P}=0.77-0.34 \] \[ E^\circ_{\mathrm P}=+0.43\ \mathrm V \] A positive standard emf supports spontaneous operation in this direction under standard conditions. \(\textbf{Examine the reverse reaction:}\) Reaction Q reverses both electrode processes. Copper(II) ions would have to undergo reduction while iron(II) ions underwent oxidation. The corresponding standard potential is: \[ E^\circ_{\mathrm Q}=0.34-0.77 \] \[ E^\circ_{\mathrm Q}=-0.43\ \mathrm V \] The reverse reaction is therefore not spontaneous under the same standard conditions. \(\textbf{Distinguish potential from reaction coefficients:}\) Multiplying the iron half-reaction by two balances electrons but does not double its electrode potential. The two reactions involve the same couples, yet reversing their direction reverses the sign of the cell potential. Platinum provides an inert conducting surface and does not appear in either net ionic reaction. The correct comparison is a positive emf for P and an equally large negative emf for Q.

Explanation: \(\textbf{Separate the two oxidation-state changes:}\) Copper initially has oxidation number \(+1\). In the disproportionation products, some copper has oxidation number \(+2\), while the remaining copper has oxidation number zero. The same initial species therefore undergoes oxidation and reduction. \(\textbf{Identify the cathode process:}\) Reduction of copper(I) to copper metal is: \[ \mathrm{Cu^++e^-\rightarrow Cu} \] Its standard reduction potential is: \[ E^\circ_{\mathrm{cathode}}=+0.52\ \mathrm V \] \(\textbf{Identify the anode process:}\) The other copper(I) ion must undergo oxidation: \[ \mathrm{Cu^+\rightarrow Cu^{2+}+e^-} \] This is the reverse of the supplied \(\mathrm{Cu^{2+}/Cu^+}\) reduction half-reaction. The corresponding standard oxidation potential is: \[ E^\circ_{\mathrm{ox,anode}}=-0.15\ \mathrm V \] \(\textbf{Calculate the standard cell emf:}\) The cell potential can be calculated by adding the cathode reduction potential and anode oxidation potential: \[ E^\circ_{\mathrm{cell}}=E^\circ_{\mathrm{red,cathode}}+E^\circ_{\mathrm{ox,anode}} \] \[ E^\circ_{\mathrm{cell}}=0.52+(-0.15) \] \[ E^\circ_{\mathrm{cell}}=+0.37\ \mathrm V \] \(\textbf{Verify the electron balance:}\) One copper(I) ion accepts one electron. The other copper(I) ion releases one electron. Adding the two half-reactions gives: \[ \mathrm{2Cu^+\rightarrow Cu^{2+}+Cu} \] No electron multiplier is necessary. \(\textbf{Interpret the thermodynamic inference:}\) The positive standard emf indicates that the disproportionation direction is thermodynamically favoured under the stated standard conditions. Adding the magnitudes \(0.52\) and \(0.15\) without reversing the sign of the latter would incorrectly give \(0.67\ \mathrm V\). Reversing the disproportionation reaction would instead give a standard emf of \(-0.37\ \mathrm V\). The anode is always the oxidation site, even when the same oxidation-state species participates in both electrode half-reactions. \(\textbf{Final answer:}\) The disproportionation emf is \(+0.37\ \mathrm V\). At the anode, \(\mathrm{Cu^+\rightarrow Cu^{2+}+e^-}\) is the oxidation half-reaction.

Explanation: \(\textbf{Statement I: Correct.}\) The standard hydrogen electrode is defined using the hydrogen/hydrogen-ion redox couple under specified standard-state conditions. At \(298\ \mathrm K\), hydrogen at \(1\ \mathrm{bar}\) and unit hydrogen-ion activity constitute the stated reference arrangement. Its standard reduction potential is assigned: \[ E^\circ(\mathrm{H^+/H_2})=0.00\ \mathrm V \] Platinised platinum supplies a conducting surface and facilitates the electrode process. \(\textbf{Statement II: Incorrect.}\) Arrangement Q has hydrogen pressure \(2\ \mathrm{bar}\), not the standard gas pressure. The platinum surface is inert, but this does not force the hydrogen electrode potential to remain exactly zero at every pressure. Electrode potential depends on the chemical conditions of the participating redox couple. The zero value is a reference convention for the standard hydrogen electrode, not a universal potential imposed by platinum. \(\textbf{Statement III: Correct.}\) Arrangement R has the standard hydrogen pressure but hydrogen-ion activity \(0.10\), rather than unity. Therefore, its chemical conditions differ from those of the standard hydrogen electrode. Correct gas pressure alone does not establish standard-state conditions for the entire half-cell. \(\textbf{Statement IV: Correct.}\) The reference value zero is assigned to the standard hydrogen electrode. Other hydrogen electrodes may be operated at non-standard pressures, ion activities or temperatures. Their potentials must be interpreted under those actual conditions. The statement is therefore consistent with the distinction between standard and non-standard electrode potential. \(\textbf{Interpret arrangement S:}\) Arrangement S illustrates that temperature is also part of the experimental specification. Although its gas pressure and hydrogen-ion activity equal the stated standard values, it is not the conventional reference arrangement specifically described at \(298\ \mathrm K\). It is unnecessary to calculate a non-standard potential to identify the incorrect assertion about Q. The decisive distinction is between a conventional reference value and the conditions under which an actual electrode operates. The correct combination is I, III and IV.

Explanation: \(\textbf{Convert electrical charge into electron amount:}\) \[ n(e^-)=\frac{Q}{F} \] Substituting: \[ n(e^-)=\frac{3860\ \mathrm C}{96500\ \mathrm{C\,mol^{-1}}} \] \[ n(e^-)=0.0400\ \mathrm{mol} \] \(\textbf{Calculate chlorine formation at the anode:}\) The oxidation half-reaction is: \[ \mathrm{2Cl^-\rightarrow Cl_2+2e^-} \] One mole of molecular chlorine corresponds to the release of two moles of electrons. Therefore: \[ n(\mathrm{Cl_2})=\frac{0.0400}{2} \] \[ n(\mathrm{Cl_2})=0.0200\ \mathrm{mol} \] \(\textbf{Analyse hydrogen formation and hydroxide production:}\) At the cathode: \[ \mathrm{2H_2O+2e^-\rightarrow H_2+2OH^-} \] Two electrons produce two hydroxide ions. Thus, the mole ratio of hydroxide formed to electrons consumed is \(1:1\). \[ n(\mathrm{OH^-})=0.0400\ \mathrm{mol} \] Sodium ions remain in the electrolyte as the positive counter-ions. The hydroxide produced therefore corresponds to formation of: \[ n(\mathrm{NaOH})=0.0400\ \mathrm{mol} \] \(\textbf{Confirm with the overall reaction:}\) Adding the half-reactions gives: \[ \mathrm{2NaCl+2H_2O\rightarrow2NaOH+H_2+Cl_2} \] The reaction produces two moles of NaOH per mole of chlorine. Therefore: \[ n(\mathrm{NaOH})=2n(\mathrm{Cl_2}) \] \[ n(\mathrm{NaOH})=2(0.0200) \] \[ n(\mathrm{NaOH})=0.0400\ \mathrm{mol} \] \(\textbf{Distinguish ion migration from electrode discharge:}\) Sodium ions migrate towards the negative cathode, but sodium metal is not deposited under the specified aqueous conditions. Instead, water is reduced to hydrogen. Chloride ions are oxidised at the anode, while the remaining sodium ions provide charge balance for the hydroxide-rich solution. The specified product selection is essential: molten NaCl would give a different cathode product. The amount of NaOH follows from the electron stoichiometry and the net reaction, not from assuming every transported sodium ion must receive an electron. \(\textbf{Final answer:}\) \[ \boxed{\left(n(\mathrm{Cl_2}),n(\mathrm{NaOH})\right)=(0.0200,\ 0.0400)\ \mathrm{mol}} \]