Question

Find the emf of the following cell. (Take $\frac{RTln10}{F}=0.6$
$\begin{array}{c}Pt,H_2\left(g\right)/H^{+}\left(aq\right)//H^{+}\left(aq\right)/H_2\left(g\right),Pt\\ 10atm{10}^{-9}M{10}^{-3}M0.1atm\end{array}$

• A. 0.24 V
• B. 0.48 V
• C. 0.42 V
• D. 0.84 V

Answer 1

The correct option is C 0.42 V

Cell is equivalent to $\begin{array}{c}Pt,H_2\left(g\right)/H^{+}\left(aq\right)//H^{+}\left(aq\right)/H_2\left(g\right),Pt\\ 10atm{10}^{-9}M{10}^{-3}M0.1atm\end{array}$
oxidation half cell:
$\underset{10atm}{H_2\left(g\right)}\to \underset{{10}^{-9}M}{2H^{+}}+2e^{-}$
reduction half cell:
$\underset{{10}^{-3}M}{2H^{+}(aq.)}+2e^{-}\to \underset{0.1atm}{H_2\left(g\right)}$
overall reaction:
$\underset{{10}^{-3}M}{2H^{+}(aq.)}+\underset{10atm}{H_2\left(g\right)}\to \underset{0.1atm}{H_2\left(g\right)}+\underset{{10}^{-9}M}{2H^{+}}$
apply nearst equation for $n=2$
$E_{cell}^0=0-\frac{0.06}{2}log\frac{({10}^{-9}{)}^2\times 0.1}{10\times ({10}^{-3}{)}^2}=-\frac{0.06}{2}log{10}^{-14}=0.42V$