Consider the following Galvanic cell (Image) By what value the ce…

MCQ

Consider the following Galvanic cell (Image) By what value the cell voltage change when concentration of ions in anodic and cathodic compartments both increased by factor of $10$ at $298\, K.$

A
$+0.0591$
B
$-0.0591$
✓
$-0.1182$
D
$0$

✓

Answer

Correct option: C.

$-0.1182$

c
The cell reaction is

$H_{2}+C l_{2} \rightarrow 2 H^{+}+2 C l^{-}$

The expression for the cell potential at $298 K$ is $E_{c e l l}=E_{c e l l}^{0}-\frac{0.0591}{2} \log \frac{\left|H^{+}\right|^{2}|C l^-|^{2}}{P_{H_{2}} P_{C l_{2}}}$

When, concentration of ions in anodic and cathodic

compartment both increased by factor of $10,$ the expression for the cell potential becomes,

$E_{\text {cell }}=E_{\text {cell }}^{0}-\frac{0.0591}{2} \log \frac{\left(10 \times\left[H^{+}\right]\right)^{2} \times\left(10 \times[C l^-)^{2}\right.}{P_{H_{2}} P_{C l_{2}}}$

$E_{\text {cell }}=E_{\text {cell }}^{0}-\frac{0.0591}{2} \log \left(10000 \times \frac{\left.\left[H^{+}\right]^{2} \mid C l-\right]^{2}}{P_{H_{2}} P_{C l_{2}}}\right)$

$E_{ cell }=E_{ cell }^{0}-\frac{0.0591}{2} \log \frac{\left.\mid H^{+}\right]^{2}[C l-]^{2}}{P_{H_{2}} P_{C l_{2}}}-\frac{0.0591}{2} \times 4$

$E_{ cell }=E_{ cell }^{0}-\frac{0.0591}{2} \log \frac{\left.\mid H^{+}\right]^{2}\left[C l-\left.\right|^{2}\right.}{P_{H_{2}} P_{C l_{2}}}-0.1182 \ldots$

The increase in the cell potential can be obtained by subtracting above equations. The increase in the cell potential is

$E_{\alpha e l l}^{0}-\frac{0.0591}{2} \log \frac{\left[H^{+}\right]^{2}\left[C l^{-}\right]^{2}}{P_{H_{2}} P_{C l_{2}}}-0.1182-\left[E_{\text {cell }}^{0}-\frac{0.0591}{2} \log \frac{\left[H^{+}\right]^{2}\left[C l^{-}\right]^{2}}{P_{H_{2}} P_{C l_{2}}}\right]$

Hence, the increase in the cell potential is $-0.1182 V$.