Electrochemistry - Result Question 60
9. For the following electrochemical cell at $298 K$,
$\operatorname{Pt}(s) \mid H _2(g, 1$ bar $) \mid H^{+}(a q, 1 M)$
$$ | M^{4+}(a q), M^{2+}(a q) \mid \operatorname{Pt}(s) $$
$E _{\text {cell }}=0.092 V$ when $\frac{\left[M^{2+}(a q)\right]}{\left[M^{4+}(a q)\right]}=10^{x}$ Given : $E _{M^{4+} / M^{2+}}^{\circ}=0.151 V ; 2.303 \frac{R T}{F}=0.059 V$
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
$H _2(g) \longrightarrow 2 H^{+}(a q)+2 e^{-} ; \quad E _{SHE}^{\circ}=0.00 V$
Reduction at cathode
$$ \begin{aligned} & M^{4+}(a q)+2 e^{-} \longrightarrow M^{2+}(a q) ; E _{M^{4+} / M^{2+}}^{\circ}=0.151 V \ & \text { Net: } M^{4+}(a q)+H _2(g) \longrightarrow M^{2+}(a q)+2 H^{+}(a q) ; \end{aligned} $$
$$ \begin{aligned} K & =\frac{\left[M^{2+}\right]\left[H^{+}\right]^{2}}{\left[M^{4+}\right] p _{H _2}}\left(E _{\text {cell }}^{\circ}=0.151 V\right)=\frac{\left[M^{2+}\right]}{\left[M^{4+}\right]} \ E _{\text {cell }} & =E _{\text {cell }}^{\circ}-\frac{0.059}{2} \log K \
BETA
