Question

For the following cell having a hydrogen electrode and a normal calomel electrode, the electromotive force (EMF) is $0.634$ V at $298$ K.
$Pt\left[H_2\right(1atm\left)|H^{+}\right(pH=x)\parallel KCl(1N\left)|H{g}_{2}C{l}_2\right(s\left)|Hg\right(l)$
If $E^o$ of $C{l}^{-}|H{g}_{2}C{l}_2|Hg$ is $0.28$ V, then select the incorrect statements.

• A. The cell reaction is $H_2\left(s\right)+2Hg\left(l\right)+2C{l}^{-}\left(g\right)\to 2H^{+}\left(l\right)+H{g}_{2}C{l}_2\left(s\right)$.
• B. The value of $x$ is $5$.
• C. The value of $x$ is $6$.
• D. If semi-normal calomel cathode is used in place of normal cathode cell, then the emf will decrease.

Answer 1

Answer: (A), (B), (D)

The cell reaction is $H_2\left(s\right)+2Hg\left(l\right)+2C{l}^{-}\left(g\right)\to 2H^{+}\left(l\right)+H{g}_{2}C{l}_2\left(s\right)$.
The value of $x$ is $5$.
If semi-normal calomel cathode is used in place of normal cathode cell, then the emf will decrease.

(A) The correct cell reaction is $H_2\left(g\right)+H{g}_{2}C{l}_2\left(s\right)\to 2H^{+}\left(aq\right)+2Hg\left(l\right)+2C{l}^{-}\left(aq\right)$

(B) and (C) $E_{cell}^0=E_{calomel}^0-E_{SHE}^0=0.28V-0.0V=0.28V$

$E_{cell}=E_{cell}^0-\frac{0.0592}{n}log\frac{\left[C{l}^{-}{]}^2\right[H^{+}{]}^2}{P_{H_2}}$

$0.634V=0.28V-\frac{0.0592}{2}log\frac{1^2\times ({10}^{-x}{)}^2}{1}$

$-11.95=log({10}^{-x}{)}^2$
$1.097\times {10}^{-12}=({10}^{-x}{)}^2$
$\left({10}^{-}x\right)={10}^{-6}$
$x=6$
Hence, the value of x is 6.

(D) If semi-normal calomel cathode is used in place of normal cathode cell, then the emf will increase.
With decrease in chloride ion concentration, the term $-log\frac{\left[C{l}^{-}{]}^2\right[H^{+}{]}^2}{P_{H_2}}$will increase. Due to this, the cell potential will increase.

Hence, the options A, B and D are incorrect.