101. Two standard cells use the \(Ag^+/Ag\) half-cell as the cathode. Cell P uses \(Cu^{2+}/Cu\) as the anode, while Cell Q uses \(Zn^{2+}/Zn\) as the anode. Given \(E^\circ_{Ag^+/Ag}=+0.80\,V\), \(E^\circ_{Cu^{2+}/Cu}=+0.34\,V\), and \(E^\circ_{Zn^{2+}/Zn}=-0.76\,V\), the difference \(E^\circ_Q-E^\circ_P\) is:
ⓐ. \(+1.10\,V\)
ⓑ. \(+0.46\,V\)
ⓒ. \(+1.56\,V\)
ⓓ. \(-1.10\,V\)
Correct Answer: \(+1.10\,V\)
Explanation: \( \textbf{Cell P potential:} \)
\[
E^\circ_P=(+0.80\,V)-(+0.34\,V)=+0.46\,V
\]
\( \textbf{Cell Q potential:} \)
\[
E^\circ_Q=(+0.80\,V)-(-0.76\,V)=+1.56\,V
\]
Zinc has the more negative reduction potential, so its oxidation produces the larger potential difference against the silver cathode.
\( \textbf{Required difference:} \)
\[
E^\circ_Q-E^\circ_P
\]
\( \textbf{Substitution:} \)
\[
E^\circ_Q-E^\circ_P=1.56\,V-0.46\,V
\]
\[
E^\circ_Q-E^\circ_P=+1.10\,V
\]
The common silver-cathode contribution cancels when the two cell potentials are compared.
\( \textbf{Final answer:} \) Cell Q exceeds Cell P by \(1.10\,V\).
102. Consider the following statements about an electrochemical series.
Statement I: The oxidised species of a couple with a more positive \(E^\circ\) is generally a stronger oxidising agent.
Statement II: The reduced species of a couple with a more negative \(E^\circ\) is generally a stronger reducing agent.
Statement III: Standard electrode potentials alone always determine the observable reaction rate.
ⓐ. Statements I and III only
ⓑ. Statements I and II only
ⓒ. Statements II and III only
ⓓ. Statements I, II and III
Correct Answer: Statements I and II only
Explanation: A more positive reduction potential indicates a greater tendency of the oxidised species to accept electrons. Such a species acts as a stronger oxidising agent. A more negative reduction potential means the reverse oxidation of the reduced form is relatively favourable, so that form is commonly a stronger reducing agent. Standard potentials describe thermodynamic tendency under specified conditions. Reaction rates can remain slow because of activation barriers, surface films, or other kinetic limitations, making Statement III too absolute.
103. A copper vessel is proposed for storing \(AgNO_3(aq)\). Given \(E^\circ_{Ag^+/Ag}=+0.80\,V\) and \(E^\circ_{Cu^{2+}/Cu}=+0.34\,V\), the proposal is unsuitable because:
ⓐ. copper ions will reduce deposited silver back to \(Ag^+\)
ⓑ. nitrate ions will oxidise copper while leaving silver ions unchanged
ⓒ. \(Ag^+\) can oxidise copper and deposit silver on the vessel
ⓓ. copper and silver have identical reduction tendencies
Correct Answer: \(Ag^+\) can oxidise copper and deposit silver on the vessel
Explanation: Silver ions have a more positive standard reduction potential than copper ions. They can therefore accept electrons from copper metal. Copper is oxidised to \(Cu^{2+}\), while \(Ag^+\) is reduced and deposited as silver. The standard cell potential is \(+0.80\,V-(+0.34\,V)=+0.46\,V\), so the displacement is thermodynamically favourable. The vessel would gradually corrode and contaminate the stored solution with copper ions.
104. The table lists metal-displacement situations under standard conditions. Select the row with the consistent prediction.
| Row | System | Prediction |
|---|
| P | \(Zn(s)\) placed in \(CuSO_4(aq)\) | Copper deposits and zinc dissolves |
| Q | \(Cu(s)\) placed in \(ZnSO_4(aq)\) | Zinc deposits and copper remains unchanged |
| R | \(Ag(s)\) placed in \(CuSO_4(aq)\) | Copper deposits and silver dissolves spontaneously |
| S | \(Cu(s)\) placed in \(AgNO_3(aq)\) | No redox change occurs |
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row P
Explanation: Zinc has a more negative reduction potential than copper, so zinc metal is more readily oxidised. When zinc is placed in \(CuSO_4(aq)\), \(Cu^{2+}\) accepts electrons and copper deposits. Copper cannot spontaneously reduce \(Zn^{2+}\) under standard conditions. Silver is less readily oxidised than copper, so it does not displace copper from \(CuSO_4(aq)\). Copper does react with \(Ag^+\), making the prediction in Row S unsuitable.
105. Assertion: In a galvanic cell made from two metal-metal-ion half-cells, the metal belonging to the more negative standard reduction potential usually acts as the anode.
Reason: The reduced form of that couple has a greater tendency to lose electrons.
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Assertion is false, but Reason is true
ⓓ. Both Assertion and Reason are true, and Reason explains Assertion
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: The anode is the site of oxidation. A more negative reduction potential indicates that reduction of the metal ion is relatively less favourable. The reverse oxidation of the metal is therefore comparatively favourable. That metal supplies electrons to the external circuit and becomes the anode in the spontaneous cell. The Reason directly connects the potential order with the electron-loss process described in the Assertion.
106. Two standard half-cells have \(E^\circ_{X^{2+}/X}=+0.20\,V\) and \(E^\circ_{Y^{2+}/Y}=-0.50\,V\). When they form a spontaneous galvanic cell:
ⓐ. X is oxidised at the negative electrode
ⓑ. Y is oxidised at the negative electrode
ⓒ. \(Y^{2+}\) is reduced at the positive electrode
ⓓ. both metal ions are reduced simultaneously
Correct Answer: Y is oxidised at the negative electrode
Explanation: The \(X^{2+}/X\) couple has the more positive standard reduction potential. \(X^{2+}\) therefore gains electrons at the cathode. Metal Y has the stronger tendency to undergo oxidation because its couple has the more negative reduction potential. Y consequently acts as the negative anode and supplies electrons. The standard cell potential is \(+0.20\,V-(-0.50\,V)=+0.70\,V\), supporting this spontaneous direction.
107. A reaction has \(E^\circ_{\mathrm{cell}}=-0.05\,V\), but it is examined under strongly non-standard concentrations. The most appropriate conclusion is:
ⓐ. use the non-standard potential to find the actual reaction direction
ⓑ. the negative standard potential fixes the forward direction at every concentration
ⓒ. the standard value remains the exact potential at all concentrations
ⓓ. concentration changes affect reaction rate but never affect cell potential
Correct Answer: use the non-standard potential to find the actual reaction direction
Explanation: A standard cell potential applies when the reacting species are in their standard states. Changing activities or concentrations changes the electrode potentials and therefore the actual cell potential. A small negative \(E^\circ_{\mathrm{cell}}\) does not prove that the reaction is impossible under every composition. The Nernst equation is used to evaluate the non-standard potential through the reaction quotient. Thermodynamic prediction must be tied to the conditions under which the cell is operating.
108. Aluminium has a strongly negative standard reduction potential, yet an aluminium surface may react slowly with an acid under some conditions. The best explanation is:
ⓐ. aluminium has no thermodynamic tendency to undergo oxidation
ⓑ. the standard hydrogen-electrode potential becomes negative near aluminium
ⓒ. a protective oxide film can create a kinetic barrier to the reaction
ⓓ. aluminium ions cannot exist in aqueous solution
Correct Answer: a protective oxide film can create a kinetic barrier to the reaction
Explanation: The negative standard reduction potential of aluminium indicates a strong thermodynamic tendency for aluminium metal to undergo oxidation. Observable reaction rate, however, is not determined by potential alone. Aluminium commonly develops a compact oxide layer that restricts contact between the underlying metal and the solution. This passivation can greatly slow hydrogen liberation even when the overall redox reaction is thermodynamically favourable. Electrochemical-series predictions should therefore be separated from kinetic and surface effects.
109. A zinc strip is proposed as an inert conducting electrode in a half-cell containing \(Cu^{2+}(aq)\). The arrangement is unsuitable because:
ⓐ. zinc cannot conduct electrons
ⓑ. zinc can reduce \(Cu^{2+}\) and become coated with copper
ⓒ. \(Cu^{2+}\) cannot accept electrons at a solid surface
ⓓ. zinc and copper have equal standard reduction potentials
Correct Answer: zinc can reduce \(Cu^{2+}\) and become coated with copper
Explanation: Zinc is chemically active toward \(Cu^{2+}\). It can lose electrons to form \(Zn^{2+}\), while copper ions gain electrons and deposit as copper metal. The zinc surface would therefore change composition rather than behave as an inert conductor. This direct displacement would alter both the electrode and the electrolyte. Platinum or graphite is preferred when an inert conducting surface is required and is chemically compatible with the half-cell species.
110. Given \(E^\circ_{H^+/H_2}=0.00\,V\), \(E^\circ_{Zn^{2+}/Zn}=-0.76\,V\), and \(E^\circ_{Cu^{2+}/Cu}=+0.34\,V\), the standard thermodynamic prediction for zinc and copper placed separately in a non-oxidising acid is:
ⓐ. both zinc and copper liberate \(H_2(g)\) from the acid
ⓑ. copper liberates \(H_2(g)\), whereas zinc does not
ⓒ. neither zinc nor copper liberates \(H_2(g)\)
ⓓ. zinc liberates \(H_2(g)\), whereas copper does not
Correct Answer: zinc liberates \(H_2(g)\), whereas copper does not
Explanation: \( \textbf{Hydrogen reduction at the cathode:} \)
\[
2H^+(aq)+2e^-\rightarrow H_2(g)
\]
\( \textbf{For zinc oxidation:} \)
\[
E^\circ_{\mathrm{cell}}=0.00\,V-(-0.76\,V)=+0.76\,V
\]
The positive value predicts that zinc can reduce \(H^+\) to hydrogen.
\( \textbf{For copper oxidation:} \)
\[
E^\circ_{\mathrm{cell}}=0.00\,V-(+0.34\,V)=-0.34\,V
\]
The negative value shows that copper oxidation by \(H^+\) is not favoured under standard conditions.
These conclusions apply to a non-oxidising acid and do not describe reactions caused by other oxidising species.
\( \textbf{Final answer:} \) Zinc can liberate hydrogen, while copper cannot under the stated standard conditions.
111. Under standard conditions with kinetic effects neglected, metal M displaces N and P from solutions containing \(N^{2+}\) and \(P^{2+}\). Metal Q also displaces M from a solution containing \(M^{2+}\). The relation consistent with these observations is:
ⓐ. \(E^\circ_{M^{2+}/M}\gt E^\circ_{N^{2+}/N}\gt E^\circ_{Q^{2+}/Q}\) and \(E^\circ_{M^{2+}/M}\gt E^\circ_{P^{2+}/P}\gt E^\circ_{Q^{2+}/Q}\)
ⓑ. \(E^\circ_{Q^{2+}/Q}\gt E^\circ_{M^{2+}/M}\gt E^\circ_{N^{2+}/N}\) and \(E^\circ_{Q^{2+}/Q}\gt E^\circ_{M^{2+}/M}\gt E^\circ_{P^{2+}/P}\)
ⓒ. \(E^\circ_{N^{2+}/N}\gt E^\circ_{M^{2+}/M}\gt E^\circ_{Q^{2+}/Q}\) and \(E^\circ_{P^{2+}/P}\gt E^\circ_{M^{2+}/M}\gt E^\circ_{Q^{2+}/Q}\)
ⓓ. \(E^\circ_{N^{2+}/N}=E^\circ_{M^{2+}/M}=E^\circ_{P^{2+}/P}=E^\circ_{Q^{2+}/Q}\), so none of the stated displacements is favoured
Correct Answer: \(E^\circ_{N^{2+}/N}\gt E^\circ_{M^{2+}/M}\gt E^\circ_{Q^{2+}/Q}\) and \(E^\circ_{P^{2+}/P}\gt E^\circ_{M^{2+}/M}\gt E^\circ_{Q^{2+}/Q}\)
Explanation: When M displaces N, metal M is oxidised and \(N^{2+}\) is reduced. The reduction potential of \(N^{2+}/N\) must therefore be greater than that of \(M^{2+}/M\). The same reasoning gives \(E^\circ_{P^{2+}/P}\gt E^\circ_{M^{2+}/M}\). Because Q displaces M, \(M^{2+}\) is reduced while Q is oxidised; therefore \(E^\circ_{M^{2+}/M}\gt E^\circ_{Q^{2+}/Q}\). These three comparisons give the ordering stated in option C.
112. The general Nernst equation for a cell reaction is:
ⓐ. \(E=E^\circ+\frac{nF}{RT}\ln Q\)
ⓑ. \(E=E^\circ-\frac{nF}{RT}\ln Q\)
ⓒ. \(E=E^\circ+\frac{RT}{nF}\ln Q\)
ⓓ. \(E=E^\circ-\frac{RT}{nF}\ln Q\)
Correct Answer: \(E=E^\circ-\frac{RT}{nF}\ln Q\)
Explanation: The Nernst equation relates the electrode or cell potential under non-standard conditions to its standard potential. The reaction quotient \(Q\) represents the activities of products relative to reactants for the balanced reaction. The correction term is \(\frac{RT}{nF}\ln Q\), and it is subtracted from \(E^\circ\). The electron number \(n\) must correspond to the balanced overall reaction. If \(Q\) increases while the other quantities remain fixed, the potential for the forward reaction decreases.
113. In the Nernst equation, the symbol \(n\) represents:
ⓐ. the electron stoichiometry of the balanced reaction
ⓑ. the total number of ions present in the electrolyte
ⓒ. the number of electrode compartments in the cell
ⓓ. the concentration exponent of every reactant
Correct Answer: the electron stoichiometry of the balanced reaction
Explanation: The value of \(n\) is obtained by balancing the oxidation and reduction half-reactions. It gives the stoichiometric number of moles of electrons exchanged when the overall reaction proceeds as written. It is not the number of ions, electrodes, or phases in the cell. Multiplying the entire balanced reaction changes the written value of \(n\), although it does not change the cell potential itself. Using an electron coefficient from only one unbalanced half-reaction can give an incorrect Nernst correction.
114. The activity-based reaction quotient for the balanced reaction \(aA+bB\rightarrow cC+dD\) is:
ⓐ. \(Q=\frac{a(A)^a a(B)^b}{a(C)^c a(D)^d}\)
ⓑ. \(Q=\frac{a(C)+a(D)}{a(A)+a(B)}\)
ⓒ. \(Q=\frac{a(C)^c a(D)^d}{a(A)^a a(B)^b}\)
ⓓ. \(Q=\frac{a(CD)^{c+d}}{a(AB)^{a+b}}\)
Correct Answer: \(Q=\frac{a(C)^c a(D)^d}{a(A)^a a(B)^b}\)
Explanation: The reaction quotient is written as the product of product activities divided by the product of reactant activities. Each activity is raised to its stoichiometric coefficient in the balanced equation. Addition of activities is not used in constructing \(Q\). The expression must correspond to the reaction direction exactly as written. Reversing the reaction replaces \(Q\) by \(\frac{1}{Q}\).
115. The Daniell-cell reaction \(Zn(s)+Cu^{2+}(aq)\rightarrow Zn^{2+}(aq)+Cu(s)\) has the reaction quotient:
ⓐ. \(Q=\frac{[Cu^{2+}]}{[Zn^{2+}]}\)
ⓑ. \(Q=[Zn^{2+}][Cu^{2+}]\)
ⓒ. \(Q=\frac{[Zn^{2+}][Cu(s)]}{[Zn(s)][Cu^{2+}]}\)
ⓓ. \(Q=\frac{[Zn^{2+}]}{[Cu^{2+}]}\)
Correct Answer: \(Q=\frac{[Zn^{2+}]}{[Cu^{2+}]}\)
Explanation: Pure solids have unit activity and are omitted from the reaction quotient. The only variable activities in the Daniell-cell reaction are those of \(Zn^{2+}\) and \(Cu^{2+}\). \(Zn^{2+}\) is a product, so it appears in the numerator. \(Cu^{2+}\) is a reactant, so it appears in the denominator. Including solid zinc or copper as concentration terms would incorrectly treat pure phases as composition-dependent quantities.
116. Pure solids and pure liquids are normally omitted from \(Q\) because:
ⓐ. they never participate in electrochemical reactions
ⓑ. their activities are unity in their pure standard states
ⓒ. they always appear with zero stoichiometric coefficients
ⓓ. their masses remain constant during every cell reaction
Correct Answer: their activities are unity in their pure standard states
Explanation: A pure solid or pure liquid has an activity conventionally taken as unity at a specified temperature and pressure. Multiplying or dividing by unity does not alter the numerical value of \(Q\). Such species may still be consumed or formed during the reaction. Their omission does not mean that they are chemically inactive or that their masses cannot change. Only their activities are excluded from the variable reaction-quotient expression.
117. Assertion: The overall cell reaction must be balanced before the Nernst equation is applied.
Reason: The balanced coefficients determine both the reaction-quotient powers and the electron number \(n\).
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Both Assertion and Reason are true, and Reason explains Assertion
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: The reaction quotient is constructed using stoichiometric coefficients as exponents. The balanced half-reactions also establish the number of electrons transferred in the overall process. An unbalanced equation can therefore produce errors in both \(Q\) and \(n\). Since both quantities enter the Nernst equation, balancing is not merely a formal step. The Reason identifies the two direct consequences that make the Assertion necessary.
118. The balanced reaction
\[
2Fe^{3+}(aq)+Sn^{2+}(aq)\rightarrow2Fe^{2+}(aq)+Sn^{4+}(aq)
\]
has which electron number and reaction quotient?
ⓐ. \(n=1\), \(Q=\frac{[Fe^{2+}][Sn^{4+}]}{[Fe^{3+}][Sn^{2+}]}\)
ⓑ. \(n=4\), \(Q=\frac{[Fe^{3+}]^2[Sn^{2+}]}{[Fe^{2+}]^2[Sn^{4+}]}\)
ⓒ. \(n=2\), \(Q=\frac{[Fe^{3+}]^2[Sn^{2+}]}{[Fe^{2+}]^2[Sn^{4+}]}\)
ⓓ. \(n=2\), \(Q=\frac{[Fe^{2+}]^2[Sn^{4+}]}{[Fe^{3+}]^2[Sn^{2+}]}\)
Correct Answer: \(n=2\), \(Q=\frac{[Fe^{2+}]^2[Sn^{4+}]}{[Fe^{3+}]^2[Sn^{2+}]}\)
Explanation: The iron reduction half-reaction is:
\[
Fe^{3+}+e^-\rightarrow Fe^{2+}
\]
It is multiplied by \(2\).
The tin oxidation half-reaction is:
\[
Sn^{2+}\rightarrow Sn^{4+}+2e^-
\]
Therefore, the balanced reaction transfers:
\[
n=2
\]
Products appear in the numerator of \(Q\), and reactants appear in the denominator.
Stoichiometric coefficients become exponents:
\[
Q=\frac{[Fe^{2+}]^2[Sn^{4+}]}{[Fe^{3+}]^2[Sn^{2+}]}
\]
The electron number is determined from the balanced redox reaction, not by adding ionic charges.
119. At fixed temperature, an increase in \(Q\) for the forward cell reaction generally causes \(E_{\mathrm{cell}}\) to:
ⓐ. increase because the product activities become larger
ⓑ. remain unchanged because only \(E^\circ_{\mathrm{cell}}\) matters
ⓒ. become exactly zero under all conditions
ⓓ. decrease because the Nernst correction is subtracted
Correct Answer: decrease because the Nernst correction is subtracted
Explanation: The Nernst equation is \(E=E^\circ-\frac{RT}{nF}\ln Q\). When \(Q\) increases, \(\ln Q\) also increases. The subtracted correction therefore becomes larger, reducing the cell potential for the forward reaction. A large \(Q\) means the mixture contains relatively more products than reactants. This composition weakens the forward driving tendency and moves the system closer to equilibrium.
120. A reaction quotient equal to unity gives:
ⓐ. \(E=0\) under every condition
ⓑ. \(E=E^\circ\)
ⓒ. \(E=-E^\circ\)
ⓓ. \(E=\frac{RT}{nF}\)
Correct Answer: \(E=E^\circ\)
Explanation: When \(Q=1\), both \(\ln Q\) and \(\log Q\) are zero. The entire concentration-dependent correction term in the Nernst equation therefore vanishes. The actual potential equals the standard potential for that reaction at the specified temperature. This does not necessarily mean that the system is at equilibrium. Equilibrium requires \(Q=K\) and \(E=0\), which occurs only when the equilibrium constant also has the value \(1\).
