Verified Solution • 2025–26 Session moderate

QJEEAdv16Ch9f3b: JEE Advanced 2016 Chemistry - Class 12 JEEAdv16Ch9f3b Solution: For the following electrochemical cell at 298 K Pt(s)

For the following electrochemical cell at 298 K

Pt(s) | H2 (g, 1 bar) | H+ (aq, 1 M) || M4+ (aq), M2+ (aq) | Pt (s)

Ecell = 0.092 V when [M2+(aq)][M4+(aq)]{{\left[ {{M^{2 + }}(aq)} \right]} \over {\left[ {{M^{4 + }}(aq)} \right]}} = 10x

Give, EM4+/M2+oE_{{M^{4+}}/{M^{2 + }}}^o = 0.151 V; 2.303 RT/F = 0.059 V

The value of x is

Multiple Choice Options

  • A -2
  • B -1
  • C 1
  • D 2

Step-by-Step Verified Solution

For the given electrochemical cell, the half-cell reactions are

∙\bullet At anode : H2(g) ⇌\rightleftharpoons 2H+(aq) + 2e −-

∙\bullet At cathode : M4+(aq) + 2e −- ⇌\rightleftharpoons M2+(aq).

The overall cell reaction is

M4+(aq) + H2(g) →\to M2+(aq) + 2H+(aq)

From Nernst equation, we have

Ecell=Ecello−2.303RTnFlog⁡[M2+][H+]2[M4+]pH2{E_{cell}} = E_{cell}^o - {{2.303RT} \over {nF}}\log {{[{M^{2 + }}]{{[{H^ + }]}^2}} \over {[{M^{4 + }}]{p_{{H_2}}}}}

Substituting the given values, we get

Ecell=(EM4+/M2+o−EH+/H2o)0.0592log⁡[M2+][H+]2[M4+]pH2{E_{cell}} = (E_{{M^{4 + }}/{M^{2 + }}}^o - E_{{H^ + }/{H_2}}^o){{0.059} \over 2}\log {{[{M^{2 + }}]{{[{H^ + }]}^2}} \over {[{M^{4 + }}]{p_{{H_2}}}}}

0.092=(0.151−0)−0.0592log⁡10x0.092 = (0.151 - 0) - {{0.059} \over 2}\log {10^x}

0.092=0.151−0.0245log⁡10x0.092 = 0.151 - 0.0245\log {10^x}

−0.059=−0.0245x⇒x=2- 0.059 = - 0.0245x \Rightarrow x = 2

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