Question

The standard oxidation potential for the half cell
$N{O}_2^{-}\text{(g)}+H_{2}O\to N{O}_3^{-}\text{(aq)}+2H^{+}\text{(aq)}+2e^{-}$ is $–0.78\text{V}$.
Calculate the reduction potential in $9$ molar $H^{+}$ assuming all other species at unit concentration. What will be the reduction potential in neutral medium?
(Given, $\log 9=0.95$

• A. $0.837\text{V},0.367\text{V}$
• B. $0.749\text{V},0.947\text{V}$
• C. $1.896\text{V},0.801\text{V}$
• D. $0.942\text{V},0.078\text{V}$

Answer 1

The correct option is A $0.837\text{V},0.367\text{V}$

$E_{\text{oxi}}=E_{\text{oxi}}^{\circ }-\frac{0.059}{n}\log [H^{+}{]}^2$
$E_{\text{oxi}}=-0.78-\frac{0.059}{2}\log [9{]}^2$
$=-0.78-0.03\times 2\log 9$
$=-0.78-0.03\times 2\times 0.95$
$=-0.78-0.057$
$E_{\text{oxi.}}=-0.837\text{V}$
$E_{\text{reduction}}=-E_{\text{oxidation}}=0.837\text{V}$
In neutral medium; $\left[H^{+}\right]=\left[O{H}^{-}\right]={10}^{-7}$
$E_{\text{oxi}}=-0.78-\frac{0.059}{2}\log [{10}^{-7}{]}^2$
$=-0.78-\frac{0.059}{2}\times 2\log {10}^{-7}$
$=-0.78+0.059\times 7$
$E_{\text{oxi.}}=-0.367\text{V}$
Answer: $0.873\text{V},0.367$