For the following electrochemical cell at 298 K, $$\begin{aligned} & \operatorname{Pt}(s) \mid \mathrm{H}_{2}(g, 1 \text { bar })\left|\mathrm{H}^{+}(a q, 1 \mathrm{M}) \| \mathrm{M}^{4+}(a q), \mathrm{M}^{2+}(a q)\right| \mathrm{Pt}(s) \\ & E_{\text {cell }}=0.092 \mathrm{~V} \text { when } \frac{\left[\mathrm{M}^{2+}(a q)\right]}{\left[\mathrm{M}^{4+}(a q)\right]}=10^{x} . \end{aligned}$$ Given : $E_{\mathrm{M}^{4+} / \mathrm{M}^{2+}}^{0}=0.151 \mathrm{~V} ; 2.303 \frac{R T}{F}=0.059 \mathrm{~V}$ The value of $x$ is

JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Redox Reactions and Electrochemistry

For the following electrochemical cell at 298 K,

Pt (s) ∣ H_{2} (g, 1 bar) H^{+} (a q, 1 M) ∥ M^{4 +} (a q), M^{2 +} (a q) Pt (s) E_{cell} = 0.092 V when \frac{[ M ^{2 +} ( a q ) ]}{[ M ^{4 +} ( a q ) ]} = 1 0^{x} .

Given :

E_{M^{4 +} / M^{2 +}}^{0} = 0.151 V; 2.303 \frac{R T}{F} = 0.059 V

The value of

x

Answer: (D)

The cell reaction is:

H_{2} + M^{4 +} ⇌ 2 H^{+} + M^{2 +}

with

n = 2

electrons transferred. Using the Nernst equation:

E_{cell} = E_{cell}^{0} - \frac{0.059}{2} lo g \frac{[ H ^{+} ] ^{2} [ M ^{2 +} ]}{[ M ^{4 +} ] \times P _{H_{2}}}

With

[H^{+}] = 1 M

P_{H_{2}} = 1 bar

, and

\frac{[ M ^{2 +} ]}{[ M ^{4 +} ]} = 1 0^{x}

0.092 = 0.151 - \frac{0.059}{2} x

\frac{0.059 x}{2} = 0.151 - 0.092 = 0.059

x = 2

JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Single correct option