Question

What is the potential of an electrode which originally contained $0.1MN{O}_3^{-}$and $0.4$ $M{H}^{+}$and which has been treated by $60\%$ of the cadmium necessary to reduce all the $N{O}_3^{-}$to $NO(g)$ at $1atm$.
Given, $N{O}_3^{-}+4H^{+}+3e^{-}\to NO+2H_{2}O,E^{\circ }=0.95V$ and $\log 2=0.3010$

• A. 0.52 V
• B. 0.44 V
• C. 0.86 V
• D. 0.78 V

Answer 1

Answer: (C)

After addition of Cd it oxidises into $C{d}^{2+}$
$N{O}_3^{-}(aq)+4H^{+}(aq)+3e^{-}\to NO(g)+2H_{2}O(l)$
$X=0.06$
$0.1-x0.4-4x$
$\left[N{O}_3^{-}\right]$ remaining $=0.1-0.06\approx 0.04M$
$\left[H^{+}\right]$ remaining $=0.4-4\times 0.06=0.4-0.24=0.16M$
$E_{N{O}_3^{-}/NO}=E_{N{O}_3^{-}/NO}^o-\frac{0.591}{3}\log \frac{1}{\left[N{O}_3^{-}\right]{\left[H^{+}\right]}^4}$
$=0.95-\frac{0.0591}{3}\log \frac{1}{(0.04)(0.16{)}^4}=0.95$
$-\frac{0.0591}{3}\log \frac{1}{4\times 10^{-2}\times 10^{-8}\times 16^4}$
$=0.95-\frac{0.0591}{3}\log \frac{1}{4\times 10^{-2}\times 10^{-8}\times 2^{+16}}$
$=0.95-\frac{0.0591}{3}\log \frac{10^{10}}{2^2\times 2^{16}}$
$=0.95-\frac{0.0591}{3}[10-18\times 0.3010]$
$=0.95-\frac{0.0591}{3}\times 4.582$
$=0.95-0.09=0.86V$