Question

For the following electrochemical cell at $298 {K}$,

${Pt}_{(s)} \mid {H}_{2(g)},(1$ bar $)\left|{H}_{(a q)}^{+},(1 {M})\right|\left|M_{(a q)}^{4+}, M_{(a q)}^{2+}\right| {Pt}_{(s)}$

$E_{\text {cell }}=0.092 {~V}$ when $\frac{\left[M_{(a q)}^{2+}\right]}{\left[M_{(a q)}^{4+}\right]}=10^x$

Given : $E_{M^{4+} / M^{2+}}^{\circ}=0.151 {~V} ; 2.303 \frac{R T}{F}=0.059 {~V}$ The value of $x$ is

(a) $-2$

(b) $-1$

(c) 1

(d) 2

Answer 1

Correct Option : (D)

Explanation —

At anode: ${H}_{2({g})} \rightarrow 2 {H}^{+}\left({aq)}+2 e^{-}\right.$

$\frac{\text { At cathode }: M^{4+}{ }_{(a q)}+2 e^{-} \rightarrow M^{2+}{ }_{(a q)}}{{H}_{2(g)}+M^{4+}{ }_{(a q)} \rightarrow M^{2+}{ }_{(a q)}+2 {H}^{+}{ }_{(a q)}}$

$E_{\text {cell }}=E^{\circ}{ }_{\text {cell }}-\frac{0.059}{2} \log \frac{\left[M^{2+}\right]\left[{H}^{+}\right]^2}{\left[M^{4+}\right]}$

$0.092=\left(E_{M^{4+} / M^{2+}}^{\circ}-E_{{H}^{+} / {H}_2}^{\circ}\right)-\frac{0.059}{2} \log \left(10^x\left[{H}^{+}\right]^2\right)$

$0.092=(0.151-0)-\frac{0.059}{2} \log \left(10^x \times 1^2\right)$

$0.092=0.151-0.0295 \log 10^x$

$0.0295 \log 10^x=0.151-0.092$

$\log 10^x=\frac{0.059}{0.0295}=2$

$10^x=\text { Antilog } 2=10^2 \therefore x=2$