Question

Copper reduces $N{O}_3^{-}$ into $NO$ and $N{O}_2$ depending upon concentration of $HN{O}_3$ in solution. Assuming $\left[C{u}^{2+}\right]=0.1M$, and $P_{NO}=P_{N{O}_2}={10}^{-3}$ bar. At which concentration of $HN{O}_3$, thermodynamic tendency for reduction of $N{O}_3^{-}$ into $NO$ and $N{O}_2$ by copper is same?
[Given: $E_{C{u}^{2+}|Cu}^{\circ }=+0.34V,E_{N{O}_3^{-}|NO}^{\circ }=+0.96V,E_{N{O}_3^{-}|N{O}_2}^{\circ }=+.79V$].

• A. ${10}^{1.23}M$
• B. ${10}^{0.56}M$
• C. ${10}^{0.66}M$
• D. ${10}^{0.12}M$

Answer 1

The correct option is C ${10}^{0.66}M$

$3Cu\left(s\right)+8H^{+}+2N{O}_3^{-}\to 3C{u}^{2+}+2NO+4H_{2}O$
$Cu\left(s\right)+4H^{+}+2N{O}_3^{-}\to C{u}^{2+}+2N{O}_2+2H_{2}O$
Let concentration of $HN{O}_3$ is $x$ so $\left[H^{+}\right]=x$ and
$\left[N{O}_3^{-}\right]=x$
$E_{N{O}_3^{-}|NO}-E_{C{u}^{2+}|Cu}=E_{N{O}_3^{-}|N{O}_2}-E_{C{u}^{2+}|Cu}$
or $E_{N{O}_3^{-}|NO}=E_{N{O}_3^{-}|N{O}_2}$
$0.96-\frac{0.0591}{3}log\frac{{10}^{-3}}{x^5}=0.79-\frac{0.0591}{1}log\left(\frac{{10}^{-3}}{x^3}\right)$
$\Rightarrow$ $0.62=\frac{0.0591}{6}log\frac{{10}^{-9}}{x^{10}}=0.45-\frac{0.0591}{2}log\frac{{10}^{-9}}{x^6}$

$logx=0.657\approx 0.66$ or $x={10}^{+0.66}$

Hence option (C) is correct.