Question

The standard reduction -potential for the half-cell:
$N{O}_3^{-}\left(aq\right)+2H^{+}+e^{-}\to N{O}_2\left(g\right)+H_{2}O$ is 0.78V.

$[\text{Take}\frac{2.303RT}{F}=0.06,log2=0.3]$
Select correct options.

• A. The reduction potential of a half-cell in 8 M H${}^{+}$ (considering other species to be at unit concentration) is 0.834 volt.
• B. The reduction potential of a half-cell in 8 M H${}^{+}$ ( (considering other species to be at unit concentration) is 0.672 volt.
• C. The reduction potential of a half-cell in a neutral solution (considering other species to be at unit concentration) is 1.62 volt.
• D. The reduction potential of a half-cell in a neutral solution (considering other species to be at unit concentration) is 0.36 volt.

Answer 1

Answer: (A), (D)

The correct options are
A The reduction potential of a half-cell in 8 M H${}^{+}$ (considering other species to be at unit concentration) is 0.834 volt.
D The reduction potential of a half-cell in a neutral solution (considering other species to be at unit concentration) is 0.36 volt.

For (A) and (B) option:

$E_{cell}=E_{cell}^0-\frac{0.06}{n}log\frac{1}{[H^{+}{]}^2}=0.78V-\frac{0.06}{1}log\frac{1}{8}=0.78V-(-0.054V)=0.834V$.

Thus, the statement (A) is correct and the statement (B) is incorrect.

For (C) and (D) option:

$E_{cell}=E_{cell}^0-\frac{0.06}{n}log\frac{1}{[H^{+}{]}^2}=0.78V-\frac{0.06}{1}log\frac{1}{(1\times {10}^{-7})}=0.78V-\left(0.42V\right)=-0.36V$.

Thus, the statement (D) is correct and the statement (C) is incorrect.