Question

Consider the following Galvanic cell. By what value the cell voltage change when concentration of ions in anodic and cathodic compartment both increased by factor of $10$ at $298K?$

• A. $+0.0591$
• B. $-0.0591$
• C. $-0.1182$
• D. $0$

Answer 1

Answer: (C)

$-0.1182$

The cell reaction is
$H_2+C{l}_2\to 2H^{+}+2C{l}^{-}$
The expression for the cell potential at 298 K is
$E_{cell}=E_{cell}^0-\frac{0.0591}{2}log\frac{\left[H^{+}{]}^2\right[C{l}^{-}{]}^2}{P_{H_2}{P}_{C{l}_2}}$......(1)
When, concentration of ions in anodic and cathodic compartment both increased by factor of 10, the expression for the cell potential becomes,
$E_{cell}=E_{cell}^0-\frac{0.0591}{2}log\frac{(10\times [H^{+}]{)}^2\times (10\times \left[C{l}^{-}\right]{)}^2}{P_{H_2}{P}_{C{l}_2}}$
$E_{cell}=E_{cell}^0-\frac{0.0591}{2}log(10000\times \frac{\left[H^{+}{]}^2\right[C{l}^{-}{]}^2}{P_{H_2}{P}_{C{l}_2}})$
$E_{cell}=E_{cell}^0-\frac{0.0591}{2}log\frac{\left[H^{+}{]}^2\right[C{l}^{-}{]}^2}{P_{H_2}{P}_{C{l}_2}}-\frac{0.0591}{2}\times 4$
$E_{cell}=E_{cell}^0-\frac{0.0591}{2}log\frac{\left[H^{+}{]}^2\right[C{l}^{-}{]}^2}{P_{H_2}{P}_{C{l}_2}}-0.1182$......(2)
The increase in the cell potential can be obtained by subtracting equation (1) from (2)
The increase in the cell potential is
$E_{cell}^0-\frac{0.0591}{2}log\frac{\left[H^{+}{]}^2\right[C{l}^{-}{]}^2}{P_{H_2}{P}_{C{l}_2}}-0.1182-[E_{cell}^0-\frac{0.0591}{2}log\frac{\left[H^{+}{]}^2\right[C{l}^{-}{]}^2}{P_{H_2}{P}_{C{l}_2}}]$
Hence, the increase in the cell potential is - 0.1182 V.
Thus option C is correct answer.