Question

In a cell that utilises the reaction, $Zn\left(s\right)+2H^{+}\left(aq\right)\rightleftharpoons {Zn}^{2+}\left(aq\right)+H_2\left(g\right)$, addition of $H_2{SO}_4$ to cathode compartment will:

• A. lower the $E_{cell}$ and shift equilibrium to the left
• B. lower the $E_{cell}$ and shift the equilibrium to the right
• C. increase the $E_{cell}$ and shift the equilibrium to the right
• D. increase the $E_{cell}$ and shift the equilibrium to the left

Answer 1

Answer: (C)

increase the $E_{cell}$ and shift the equilibrium to the right

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

$Q=\frac{\left[{Zn}^{2+}\right]}{{\left[H^{+}\right]}^2}$

Corresponding cell is:

$Zn\left|{Zn}^{2+}\right(C_1\left)\right|\left|H^{+}\right(aq\left)\right|Pt\left(H_2\right)$

and

$E_{cell}=E_{cell}^o-\frac{0.0591}{2}\log K=E_{cell}^o-\frac{0.0591}{2}\log \frac{\left[{Zn}^{2+}\right]}{{\left[H^{+}\right]}^2}$

If $H_2{SO}_4$ is added to the cathodic compartment, (towards reactant side), then $Q$ decreases (due to increase in $H^{+}$)

Hence equilibrium is displaced towards the right and $E_{cell}$ increases.