Question

$\text{Pt}(s)H_2(g)(1bar)\left|H^{+}(aq)(1M)\right|\left|M^{3+}(aq),M^{+}(aq)\right|\text{Pt}(s)$
The $E_{\text{cell }}$ for the given cell is $0.1115V$ at $298K$ when $\frac{\left[M^{+}(aq)\right]}{\left[M^{3+}(aq)\right]}=10^a$
The value of $a$ is ____
Given : $E_{M^{3+}/M^{+}}^{\theta }=0.2V$
$\frac{2.303RT}{F}=0.059V$

Answer 1

Answer: 3

Overall reaction :-
$H_{2(g)}+M_{\text{(aq) }}^{3+}>M_{\text{(aq) }}^{+}+2H_{\text{(aq) }}^{+}$
$E_{\text{Cell }}=E_{\text{Cathode }}^o-E_{\text{anode }}^{\circ }-\frac{0.059}{2}\log \frac{\left[M^{+}\right]\times 1^2}{\left[M^{+3}\right]1}$
$0.1115=0.2-\frac{0.059}{2}\log \frac{\left[M^{+}\right]}{\left[M^{+3}\right]}$
$3=\log \frac{\left[M^{+}\right]}{\left[M^{+3}\right]}$
$\therefore a=3$