Question

The standard electrode potential $\left(E^{\ominus }\right)$ for $OC{l}^{-}|C{l}^{-}$ and $C{l}^{-}|\frac{1}{2}C{l}_2$ respectively are $0.94$ V and $-1.36$ V. The $E^{\ominus }$ value of $OC{l}^{-}|\frac{1}{2}C{l}_2$ will be:

• A. $-0.42$ V
• B. $-2.20$ V
• C. $0.52$ V
• D. $1.04$ V

Answer 1

The correct option is C $0.52$ V

$E^{\circ }$ for $O\overline{Cl}/\overline{Cl}\Rightarrow 0.94v$

$E^{\circ }$ for $\overline{Cl}/1/2C_{12}\Rightarrow -1.36v$

$O\overline{Cl}\to \overline{Cl}---\left(1\right)\to \Delta C{1}^0,n=3$

$\overline{Cl}\to \frac{1}{2}C{l}_2---\left(2\right)\to \Delta C{2}^0$ $n=1$

$\left(1\right)+\left(2\right)$

$O\overline{Cl}\to 1/2C{l}_2--\left(3\right)\to \Delta C^0$ $n=1$

So

$\Delta C^0=\Delta C_1^0+\Delta C_2^0$

$-nF{E}^0=-nF{E}_1^0-nF{E}_2^0$

$-1\times F\times E^0=-3\times F\times 0.94-1.36\times 1\times F$

So $[E^0=0.52v]$