Cu_ aq^ + is unstable in solution and undergoes simultaneous oxidation and reduction according to the reaction : 2 Cu^ + (aq) leftharpoons Cu^ 2+ (aq)+Cu(s) choose correct E^ ^ for above reaction if E^ ^ Cu^ 2+ =0.34 V

Correct answer: C. +0.38 V 2 Cu^ + Cu^ +2 +Cu2 ^ - +Cu^ +2 Cu ; E_ 1 ^ ^ =0.34 V ; (i) ^ - +Cu^ +2 Cu^ + ; E_ 2 ^ ^ =0.15 V ; (ii) Cu^ + + ^ - arrow Cu ; E_ 3 ^ ^ =? ...(iii) Now, Δ G_ 1 ^ ^ =-nFE_ 1 ^ ^ =-2 × 0.34 FΔ G_ 2 ^ ^ =-1 × 0.15 F, Δ G_ 3 ^ ^ =-1 × E_ 3 ^ ^ F Again, Δ G_ 1 ^ ^ =Δ G_ 2 ^ ^ +Δ G_ 3 ^ ^ -0.68 F=-0.15 F-E_ 3 F0 E_ 3 ^ ^ =0.68-0.15=0.53 VE_ cell ^ ^ =E_ cathode ^ ^ (Cu^ + / Cu)=0.53-0.15=0.38 V

BITSAT 2014

Exam: BITSAT
Year: 2014
Subject: Chemistry
Chapter: Electrochemistry

BITSAT 2014 · Chemistry · Electrochemistry

Cu_ aq^ + is unstable in solution and undergoes simultaneous...

\mathrm{Cu}_{\mathrm{aq}^{+}

is unstable in solution and undergoes simultaneous oxidation and reduction according to the reaction :

2 \mathrm{Cu}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})

choose correct

\mathrm{E}^{{}^{\circ}}

for above reaction if

\mathrm{E}^{{}^{\circ}} \mathrm{Cu}^{2+}=0.34 \mathrm{ V}

and

\mathrm{E}^{{}^{\circ}} \mathrm{Cu}^{2+}=0.15 \mathrm{ V}

Options

A $-0.38 \mathrm{ V}$
B $+0.49 \mathrm{ V}$
C $+0.38 \mathrm{ V}$
D $-0.19 \mathrm{ V}$

Correct answer

C. $+0.38 \mathrm{ V}$

Step-by-step solution

2 \mathrm{Cu}^{+} \longrightarrow \mathrm{Cu}^{+2}+\mathrm{Cu}2 \mathrme^{-}+\mathrm{Cu}^{+2} \longrightarrow \mathrm{Cu} ; \mathrm{E}_{1}^{{}^{\circ}}=0.34 \mathrm{ V} ; \ldots

(i)

\mathrme^{-}+\mathrm{Cu}^{+2} \longrightarrow \mathrm{Cu}^{+} ; \mathrm{E}_{2}^{{}^{\circ}}=0.15 \mathrm{ V} ; \ldots

(ii)

\mathrm{Cu}^{+}+\mathrme^{-} \rightarrow \mathrm{Cu} ; \mathrm{E}_{3}^{{}^{\circ}}=?

...(iii) Now,

\Delta \mathrm{G}_{1}^{{}^{\circ}}=-\mathrm{nFE}_{1}^{{}^{\circ}}=-2 \times 0.34 \mathrm{ F}\Delta \mathrm{G}_{2}^{{}^{\circ}}=-1 \times 0.15 \mathrm{ F}, \Delta \mathrm{G}_{3}^{{}^{\circ}}=-1 \times \mathrm{E}_{3}^{{}^{\circ}} \mathrm{F}

Again,

\Delta \mathrm{G}_{1}^{{}^{\circ}}=\Delta \mathrm{G}_{2}^{{}^{\circ}}+\Delta \mathrm{G}_{3}^{{}^{\circ}}\Rightarrow-0.68 \mathrm{ F}=-0.15 \mathrm{ F}-\mathrm{E}_{3} \mathrm{ F}0 \Rightarrow \mathrm{E}_{3}^{{}^{\circ}}=0.68-0.15=0.53 \mathrm{ V}\mathrm{E}_{\text{cell }}^{{}^{\circ}}=\mathrm{E}_{\text{cathode }}^{{}^{\circ}}\left(\mathrm{Cu}^{+} / \mathrm{Cu}\right)=0.53-0.15=0.38 \mathrm{ V}

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