Electrochemistry 1 Question 9
9. For the following electrochemical cell at $298 \mathrm{~K}$,
$\operatorname{Pt}(s) \mid \mathrm{H}_{2}(g, 1$ bar $) \mid \mathrm{H}^{+}(a q, 1 \mathrm{M})$
$$ | M^{4+}(a q), M^{2+}(a q) \mid \operatorname{Pt}(s) $$
$E_{\text {cell }}=0.092 \mathrm{~V}$ 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
Electrochemistry
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Answer:
Correct Answer: 9. (a)
Solution:
- Oxidation at anode
$\mathrm{H}{2}(g) \longrightarrow 2 \mathrm{H}^{+}(a q)+2 e^{-} ; \quad E{\mathrm{SHE}}^{\circ}=0.00 \mathrm{~V}$
Reduction at cathode
$$ \begin{aligned} & M^{4+}(a q)+2 e^{-} \longrightarrow M^{2+}(a q) ; E_{M^{4+} / M^{2+}}^{\circ}=0.151 \mathrm{~V} \ & \text { Net: } M^{4+}(a q)+\mathrm{H}_{2}(g) \longrightarrow M^{2+}(a q)+2 \mathrm{H}^{+}(a q) ; \end{aligned} $$
$$ \begin{aligned} K & =\frac{\left[M^{2+}\right]\left[\mathrm{H}^{+}\right]^{2}}{\left[M^{4+}\right] p_{\mathrm{H}{2}}}\left(E{\text {cell }}^{\circ}=0.151 \mathrm{~V}\right)=\frac{\left[M^{2+}\right]}{\left[M^{4+}\right]} \ E_{\text {cell }} & =E_{\text {cell }}^{\circ}-\frac{0.059}{2} \log K \
