A cell is set up at 298 K as Zn(s) |…

A cell is set up at 298 K as Zn(s) | Zn^2+(1.0 M) \| Cu^2+ | Cu(s). The copper half-cell contains 0.10 M total copper(II), present almost entirely as…

One or more options correct

A cell is set up at

298\ \mathrm{K}

\mathrm{Zn(s)\,|\,Zn^{2+}(1.0\ M)\,\|\,Cu^{2+}\,|\,Cu(s)}

. The copper half-cell contains

0.10\ \mathrm{M}

total copper(II), present almost entirely as

\mathrm{[Cu(NH_3)_4]^{2+}}

with a free

\mathrm{NH_3}

concentration of

1.0\ \mathrm{M}

. Given:

E^\circ(\mathrm{Zn^{2+}/Zn})=-0.76\ \mathrm{V}

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

, formation constant

K_f=1.0\times10^{13}

for

\mathrm{[Cu(NH_3)_4]^{2+}}

, and

\dfrac{2.303RT}{F}=0.0592\ \mathrm{V}

. The correct statement(s) is/are

A. The emf of the cell is about $0.69\ \mathrm{V}$ .
B. The electrode potential of the copper half-cell is higher than $+0.34\ \mathrm{V}$ because the complex is more stable than $\mathrm{Cu^{2+}(aq)}$ .
C. The cell reaction is non-spontaneous in the forward direction, because complex formation makes $E_{cell}$ negative.
D. The standard reduction potential of the $\mathrm{[Cu(NH_3)_4]^{2+}/Cu}$ couple is about $-0.43\ \mathrm{V}$ .

Show answer and solution

Answer: A

Free

\mathrm{[Cu^{2+}]}=\dfrac{[\mathrm{Cu(NH_3)_4^{2+}}]}{K_f[\mathrm{NH_3}]^4}=\dfrac{0.10}{10^{13}\times1^4}=1.0\times10^{-14}\ \mathrm{M}

.
Cell reaction

\mathrm{Zn+Cu^{2+}\to Zn^{2+}+Cu}

with

E^\circ_{cell}=1.10\ \mathrm{V}

E_{cell}=1.10-\dfrac{0.0592}{2}\log\dfrac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]}=1.10-0.0296\times14=0.686\ \mathrm{V}\approx0.69\ \mathrm{V}

. So (A) is correct.
(B) Complexation lowers free

\mathrm{[Cu^{2+}]}

, so the copper electrode potential falls to

0.34+0.0296\log(10^{-14})=-0.07\ \mathrm{V}

and (B) is false.
(C)

E_{cell}

is still positive (

+0.69\ \mathrm{V}

), so the reaction remains spontaneous; (C) is false.
(D) The standard potential of the complex couple is

E^\circ=0.34-\dfrac{0.0592}{2}\log K_f=0.34-0.385=-0.045\ \mathrm{V}

, not

-0.43\ \mathrm{V}

(that value,

0.34-0.0592\times13

, arises from forgetting the factor

\tfrac12

for the two-electron couple). (D) is false.

Report a mistake

← Previous Electrochemistry question Next Electrochemistry question →

Original question written in the JEE Advanced 2023 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake