For the following electrochemical cell at 298 K Pt(s) | H_2 (g, 1 bar) | H^+ (aq, 1 M) || M^4+ (aq), M^2+ (aq) | Pt (s) Ecell = 0.092 V when \ left[ M^2 + (aq) ] \over [ M^4 + (aq) ] = 10x Give, E_M^4+/M^2 + ^o = 0.151 V; 2.303 RT/F = 0.059 V The value of x is

Electrochemistry PYQ — JEE Advanced 2016 · Shift 2 · Q2 with Solution

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Correct answer: D

The cell is

\text{Pt}(s)\,|\,H_2(g,1\,\text{bar})\,|\,H^+(aq,1\,M)\;||\;M^{4+}(aq),M^{2+}(aq)\,|\,\text{Pt}(s)

The left electrode is the standard hydrogen electrode (SHE), so

E_{\text{left}}=0\,\text{V}

Hence,

E_{\text{cell}}=E_{\text{right}}-E_{\text{left}}=E_{\text{right}}

So,

E_{M^{4+}/M^{2+}}=0.092\,\text{V}

M^{4+}+2e^- \rightarrow M^{2+}

Given standard reduction potential:

E^\circ_{M^{4+}/M^{2+}}=0.151\,\text{V}

For the reduction

M^{4+}+2e^- \rightarrow M^{2+}

Reaction quotient is

Q=\frac{[M^{2+}]}{[M^{4+}]}

Thus,

E=E^\circ-\frac{0.059}{2}\log\frac{[M^{2+}]}{[M^{4+}]}

Substitute the values:

0.092=0.151-\frac{0.059}{2}\log\frac{[M^{2+}]}{[M^{4+}]}

0.151-0.092=\frac{0.059}{2}\log\frac{[M^{2+}]}{[M^{4+}]}

0.059=\frac{0.059}{2}\log\frac{[M^{2+}]}{[M^{4+}]}

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

\log\frac{[M^{2+}]}{[M^{4+}]}=2

Therefore,

\frac{[M^{2+}]}{[M^{4+}]}=10^2

Given that

\frac{[M^{2+}]}{[M^{4+}]}=10^x

we get

x=2

So the correct option is D.