Question

For the following electrochemical cell at $298K$,
$Pt(s)\mid H_2(g,1$ bar $)\left|H^{+}(aq,1M)\right|\left|M^{4+}(aq),M^{2+}(aq)\right|Pt(s)$
$E_{\text{cell }}=0.092V$ when $\frac{\left[M^{2+}(aq)\right]}{\left[M^{4+}(aq)\right]}=10^x$
Given: $E_{M^{4+}/M^{2+}}^0=0.151V;2.303\frac{RT}{F}=0.059V$
The value of $x$ is

• A. -2
• B. -1
• C. 1
• D. 2

Answer 1

Answer: (D)

$H_2\rightleftharpoons 2H^{+}+2e^{-}$
$M^{+4}+2e^{-}\rightleftharpoons M^{+2}$
__________________________
$H_2+M^{+4}\rightleftharpoons 2H^{+}+M^{+2}$
$E_{cell}=\left(E_{M^{+4}/M^{+2}}-E_{H^{+}/H_2}\right)-\frac{0.0591}{2}\log \frac{\left[M^{+2}\right]{\left[H^{+}\right]}^2}{\left[M^{+4}\right]P_{H_2}}$
$0.092=0.151-\frac{0.0591}{2}\log \frac{\left[M^{+2}\right]}{\left[M^{+4}\right]},\frac{(1{)}^2}{(1)}$
$-0.059=-\frac{0.0591}{2}\log \frac{\left[M^{+2}\right]}{\left[M^{+4}\right]}$
$\log \frac{\left[M^{+2}\right]}{\left[M^{+4}\right]}=2$
$\frac{\left[M^{+2}\right]}{\left[M^{+4}\right]}=10^{2}x=2$