Question

What is the potential of an electrode which originally contained $0.1 M NO _{3}^{-}$and $0.4$ $MH ^{+}$and which has been treated by $60 \%$ of the cadmium necessary to reduce all the $NO _{3}^{-}$to $NO ( g )$ at $1\, atm$.
Given, $NO _{3}^{-}+4 H ^{+}+3 e ^{-} \rightarrow NO +2 H _{2} O , E^{\circ}=0.95\, V$ 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 $Cd ^{2+}$
$NO _{3}^{-}( aq )+4 H ^{+}( aq )+3 e ^{-} \rightarrow NO ( g )+2 H _{2} O (l)$
$X=0.06$
$0.1- x 0.4-4 x$
$\left[ NO _{3}^{-}\right]$ remaining $=0.1-0.06 \approx 0.04\, M$
$\left[ H ^{+}\right]$ remaining $=0.4-4 \times 0.06=0.4-0.24=0.16\, M$
$E _{ NO _{3}^{-} / NO }= E _{ NO _{3}^{-} / NO }^{ o }-\frac{0.591}{3} \log \frac{1}{\left[ NO _{3}^{-}\right]\left[ H ^{+}\right]^{4}}$
$=0.95-\frac{0.0591}{3} \log \frac{1}{(0.04)(0.16)^{4}}=0.95$
$-\frac{0.0591}{3} \log \frac{1}{4 \times 10^{-2} \times 10^{-8} \times 16^{4}}$
$=0.95-\frac{0.0591}{3} \log \frac{1}{4 \times 10^{-2} \times 10^{-8} \times 2^{+16}}$
$=0.95-\frac{0.0591}{3} \log \frac{10^{10}}{2^{2} \times 2^{16}}$
$=0.95-\frac{0.0591}{3}[10-18 \times 0.3010]$
$=0.95-\frac{0.0591}{3} \times 4.582$
$=0.95-0.09=0.86\, V$