Question

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

• A. -2
• B. -1
• C. 1
• D. 2

Answer 1

Answer: (D)

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