Question

In a cell that utilizes the reaction $Zn\left(s\right)$ $+2H^{+}$ (aq) $\to Z{n}^{2+}\left(aq\right)+H_2\left(g\right)$ addition of $H_{2}S{O}_4$ to cathode compartment, will:

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

Answer 1

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

$Zn\left(s\right)$ $+2H^{+}$ (aq) $\to Z{n}^{2+}\left(aq\right)+H_2\left(g\right)$

$E=E^0-\frac{0.059}{2}log\frac{\left[Z{n}^{+2}\right]}{[H^{+}{]}^2}$

So as we add sulphuric acid, conc. of hydrogen ion increases and equilibrium shifts to right.

Also according to the equation emf of the cell will be increased by adding acid.