Electrochemistry 1 Question 9
9. For the following electrochemical cell at $298 \mathrm{K}$,
$\operatorname{Pt}(s) \mid H _{2}(g, 1$ bar $) \mid H^{+}(a q, 1 \mathrm{M})$ $ | M^{4+}(a q), M^{2+}(a q) \mid \operatorname{Pt}(s) $
$E_{\text {cell }}=0.092 \mathrm{V}$ when $\frac{\left[M^{2+}(a q)\right]}{\left[M^{4+}(a q)\right]}=10^{x}$
Given : $E_{M^{4+} / M^{2+}}^{\circ}=0.151 \mathrm{V} ; 2.303 \frac{R T}{F}=0.059 \mathrm{V}$
The value of $x$ is
(2016 Adv.)
(a) -2
(b) -1
(c) 1
(d) 2
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Answer:
Correct Answer: 9. (d)
Solution:
- Oxidation at anode
$H_2(g) \longrightarrow 2 H^+(a q)+2 e^- ;$
$ \quad E_{\text{SHE}}^{\circ}=0.00 \mathrm{V}$
Reduction at cathode
$ \begin{aligned} & M^{4+}(a q)+2 e^{-} \longrightarrow M^{2+}(a q) ; E_{M^{4+} / M^{2+}}^{\circ}=0.151 \mathrm{V} \\ & \text { Net: } M^{4+}(a q)+H _{2}(g) \longrightarrow M^{2+}(a q)+2 H^{+}(a q) ; \end{aligned} $
$ \begin{aligned} K & =\frac{[M^{2+}][H^{+}]^{2}}{[M^{4+}] p_{H_2}}(E_{\text {cell }}^{\circ}=0.151 \mathrm{V})=\frac{[M^{2+}]}{[M^{4+}]} \\ E_{\text {cell }} & =E_{\text {cell }}^{\circ}-\frac{0.059}{2} \log K \\ \end{aligned} $
$0.092 =0.151-\frac{0.059}{2} \log \frac{\left[M^{2+}\right]}{\left[M^{4+}\right]} $
$\therefore \quad 0.059 =\frac{0.059}{2} \log 10^x $
$\therefore \quad \log 10^x =2 $
$x =2$
BETA
