JEE Chemistry Electrochemistry MCQs Set 06

Answer: (a) \( \ce{Cu^2+} \) ions can be reduced by \( \ce{H2}(g) \)

Explanation:
1. Hydrogen has \( E^\circ = 0 \), lower than 0.337 V for Cu, so H2 can reduce \( \ce{Cu^2+} \) to Cu.
2. Cu cannot be oxidised by H+; \( \ce{Sn^2+} \) (-0.136 V) cannot be reduced by H2; Cu cannot reduce \( \ce{Sn^2+} \).

Teacher's Note:
1. Compare with 0 V.
2. Higher E is reduced.

Answer: (c) M > B > A

Explanation:
1. Carbon removes oxygen from AO and BO but not from MO, so M is the most reactive (most stable oxide).
2. B removes oxygen from AO, so B is more active than A. The order is M > B > A.

Teacher's Note:
1. Reduction by a more active metal.
2. Carbon cannot reduce MO.

Answer: (b) \( \dfrac{RT}{2F}\ln\dfrac{p_1}{p_2} \)

Explanation:
1. Left electrode: \( \ce{H2}(p_1) \rightarrow 2\ce{H+} + 2e^- \); right electrode: \( 2\ce{H+} + 2e^- \rightarrow \ce{H2}(p_2) \). Net: \( \ce{H2}(p_1) \rightarrow \ce{H2}(p_2) \), n = 2.
2. \( E = -\tfrac{RT}{2F}\ln\tfrac{p_2}{p_1} = \tfrac{RT}{2F}\ln\tfrac{p_1}{p_2} \).

Teacher's Note:
1. Concentration cell.
2. Two electrons per H2.

Answer: (b) \( \ce{Ce^4+ > MnO4^- > Cr2O7^2- > Sn^4+} \)

Explanation:
1. A higher reduction potential means a stronger oxidising agent.
2. Order: Ce4+ (1.61) > MnO4- (1.51) > Cr2O7 2- (1.33) > Sn4+ (0.15).

Teacher's Note:
1. Arrange potentials.
2. Highest E is best oxidiser.

Answer: (b) 0.35 V

Explanation:
1. \( Q = \dfrac{[\ce{Cl-}]^2[\ce{Br2}]}{[\ce{Br-}]^2 p_{\ce{Cl2}}} = \dfrac{(0.01)^2(0.01)}{(0.01)^2(1)} = 0.01 \).
2. \( E = 0.29 - \tfrac{0.059}{2}\log 0.01 = 0.29 + 0.059 = 0.35 \) V.

Teacher's Note:
1. n = 2.
2. log 0.01 = -2.

Answer: (a) 128 S cm\(^2\) mol\(^{-1}\)

Explanation:
1. By Kohlrausch's law: \( \Lambda^\circ(\ce{NaBr}) = \Lambda^\circ(\ce{NaCl}) + \Lambda^\circ(\ce{KBr}) - \Lambda^\circ(\ce{KCl}) \).
2. \( = 126 + 152 - 150 = 128 \) S cm2 mol-1.

Teacher's Note:
1. Add and subtract ionic conductivities.
2. K+ and Cl- cancel.

Answer: (b) \( E_1 < E_2 \)

Explanation:
1. \( E = E^\circ - \tfrac{RT}{nF}\ln[\ce{M^n+}] \). At 298 K and 2 M, \( E_1 = E^\circ - \tfrac{R(298)}{nF}\ln 2 \).
2. At twice the temperature and 1 M, \( \ln 1 = 0 \), so \( E_2 = E^\circ \) (taking \( E^\circ \) as unchanged). Hence \( E_1 < E_2 \).

Teacher's Note:
1. Concentration halves to 1 M.
2. Log term vanishes.

Answer: (c) 67.2

Explanation:
1. Electrolysis of water: \( 2\ce{H2O} \rightarrow 2\ce{H2} + \ce{O2} \) uses 4 mol of electrons.
2. Gases: 2 mol H2 + 1 mol O2 = 3 mol, which is \( 3\times 22.4 = 67.2 \) L at STP.

Teacher's Note:
1. 4 e- per O2.
2. Total moles times 22.4.

Answer: (b) \( \ce{Zn + 2Ag+ -> 2Ag + Zn^2+} \)

Explanation:
1. Cell voltage is the sum of the oxidation potential of the anode and the reduction potential of the cathode.
2. Zn + 2Ag+: \( 0.76 + 0.80 = 1.56 \) V, the largest of the four (Zn + Cu2+ = 1.10 V).

Teacher's Note:
1. Compare totals.
2. Ag+ is the best oxidiser here.

Answer: (c) Increase in the concentration of \( \ce{Ag+} \) ions

Explanation:
1. \( E = E^\circ - \tfrac{0.059}{2}\log\dfrac{[\ce{Sn^2+}]}{[\ce{Ag+}]^2} \).
2. Raising \( [\ce{Ag+}] \) lowers the log term, so E increases. Rod size has no effect and more \( \ce{Sn^2+} \) lowers E.

Teacher's Note:
1. Nernst equation.
2. Reactant concentration helps.

Answer: (c) II and IV

Explanation:
1. \( E^\circ \) of Fe is lowest, then Cu, then Ag. A more active metal displaces a less active one from solution.
2. Fe displaces Cu (II) and Ag (IV). Cu cannot displace Fe (I) and Ag cannot displace Cu (III).

Teacher's Note:
1. Reactivity series.
2. Check each statement.

Answer: (c) 1.212 V

Explanation:
1. Fe is oxidised (anode) and \( \ce{Fe^3+} \) is reduced (cathode).
2. \( E^\circ = 0.771 - (-0.441) = 1.212 \) V.

Teacher's Note:
1. Cathode minus anode.
2. The two half cells.

Answer: (b) \( -0.177 \) V

Explanation:
1. \( 2\ce{H+} + 2e^- \rightarrow \ce{H2} \), \( E = -0.059\,\text{pH} \) at 25 degrees Celsius.
2. \( E = -0.059\times 3 = -0.177 \) V.

Teacher's Note:
1. E degrees is zero for hydrogen.
2. pH 3.

Answer: (a) Cl will oxidise Br\(^-\) and not F\(^-\)

Explanation:
1. The order given is F > Cl > Br for reduction potential, so Cl2 is a weaker oxidiser than F2 but stronger than Br2.
2. Cl2 oxidises \( \ce{Br-} \) to Br2 but cannot oxidise \( \ce{F-} \).

Teacher's Note:
1. Higher E oxidises lower.
2. Cl2 gets reduced.

Answer: (d) decrease in A and increase in B

Explanation:
1. Electrolysis of \( \ce{AgNO3} \) with Pt electrodes gives Ag and O2 and leaves \( \ce{HNO3} \), so pH decreases in A.
2. Electrolysis of LiCl solution gives H2 and Cl2 and leaves LiOH, so pH increases in B.

Teacher's Note:
1. Which ions remain in solution?
2. Acid formed in A, base formed in B.