Question

In a cell that utilises the reactions,

$Zn\left(s\right)+2H^{+}\left(aq\right)\to Z{n}^{2+}\left(aq\right)+H_2\left(g\right)$ addition of $H_{2}S{O}_4$ to cathode compartment will:

• A. increase the E and shift equilibrium to the left
• B. lower the E and shift equilibrium to the right
• C. increase the E and shift equilibrium to the right
• D. lower the E and shift equilibrium to the left

Answer 1

The correct option is C increase the E and shift equilibrium to the right

$Zn\left(s\right)+2H^{+}\left(aq\right)\to Z{n}^{2+}\left(aq\right)+H_2\left(g\right)$ addition of $H_{2}S{O}_4$

As we add $H_{2}S{O}_4$ in solution, conc. of hydrogen ion increases and according to the principal, reaction moves forward.

According to nernst equation,

$E=E^o-\left(0.0591/n\right)logQ$

$E_{cell}=E_{cell}^o+\frac{0.059}{2}log\frac{[H^{+}{]}^2}{P_{H_2}\times \left[Z{n}^{2+}\right]}$

Addition of $H_{2}S{O}_4$ will increase $\left[H^{+}\right]$ and thus $E_{cell}$ will be more and equilibrium will be shifted to right.

So, the correct option is $C$