Question

In acid medium, the standard reduction potential of $NO$ converted to $N_{2}O$ is $1.59V$. Its standard potential in alkaline medium would be:

• A. $-1.59V$
• B. $-0.764V$
• C. $0.764V$
• D. $0.062V$

Answer 1

The correct option is C $0.764V$

$NO⟶N_{2}O$ $E^o=1.59V$

The reaction in acidic medium,

$2NO+2H^{+}⟶N_{2}O+H_{2}O$; $E^o=1.59$

In basic medium,

$2NO+H_{2}O⟶N_{2}O+2O{H}^{-}$; Let $E^o=x$

Now, in alkaline medium,

$H_{2}O\rightleftharpoons H^{+}+O{H}^{-}$

$\Rightarrow K={10}^{-14}$

or, $2H_{2}O\rightleftharpoons 2H^{+}+2O{H}^{-}$, $K={10}^{-28}$

We know, $E^o=\frac{0.059}{n}\log K$

$=\frac{0.059}{2}\log {10}^{-28}=0.826$

For standard potential,

$E_{acidic}^o+E_{basic}^o=E_{cell}^o$

$\Rightarrow 1.59+x=0.826$

$\Rightarrow x=0.764$