Chemistry - Electrochemistry Question with Solution | TestHub

After addition of Cd it oxidises into Cd²⁺

$\begin{array}{ccccccccc}{\text{NO}}_3^{-}\left(\text{aq}\right)& +& 4{\text{H}}^{+}\left(\text{aq}\right)& +& 3{\text{e}}^{-}& \to & \text{NO}\left(\text{g}\right)& +& 2{\text{H}}_2\text{O}\left(\mathit{\text{l}}\right)\\ \text{0.1}-\text{x}& & \text{0.4}-4\text{x}& & & & & & \end{array}$

X = 0.06

$\left[{\text{NO}}_3^{-}\right]remaining=\text{0.1}-\text{0.06}\approx \text{0.04 M}$

$\left[{\text{H}}^{+}\right]remaining=\text{0.4}-4\times \text{0.06}=\text{0.4}-\text{0.24}=\text{0.16 M}$

${\text{E}}_{{\text{NO}}_3^{-}/\text{NO}}={\text{E}}_{{\text{NO}}_3^{-}/\text{NO}}^{\circ }-\frac{\text{0.591}}{3}\text{log}\frac{1}{\left[{\text{NO}}_3^{-}\right]{\left[{\text{H}}^{+}\right]}^4}$

$=\text{0.95}-\frac{\text{0.0591}}{3}\text{log}\frac{1}{\left(\text{0.04}\right){\left(\text{0.16}\right)}^4}=\text{0.95}-\frac{\text{0.0591}}{3}\text{log}\frac{1}{4\times 10^{-2}\times 10^{-8}\times 16^4}$

$=\text{0.95}-\frac{\text{0.0591}}{3}\text{log}\frac{1}{4\times 10^{-2}\times 10^{-8}{\times 2}^{+16}}$

$=\text{0.95}-\frac{\text{0.0591}}{3}\text{log}\frac{10^{10}}{2^2\times 2^{16}}$

$=\text{0.95}-\frac{\text{0.0591}}{3}\left[10-18\times \text{0.3010}\right]=\text{0.95}-\frac{\text{0.0591}}{3}\times \text{4.582}$

$=\text{0.95}-\text{0.09}=\text{0.86 V}$