Question

In a cell that utilizes the reaction, what will happen if we add $H_{2}SO_{4}$ to cathode compartment?
$\left(\text{Zn}\right)_{\left(\text{s}\right)} + 2 \text{H}_{\left(\text{aq}\right)}^{+} \rightarrow \text{Zn}_{\left(\text{aq}\right)}^{2 +} + \left(\text{H}\right)_{2 \left(\text{g}\right)}$

• 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

Answer: (C)

$\left(\text{Zn}\right)_{\left(\text{s}\right)} + 2 \text{H}_{\left(\text{aq}\right)}^{+} ⇌ \, \text{Zn}_{\left(\text{aq}\right)}^{2 +} + \left(\text{H}\right)_{2 \left(\text{g}\right)}$
$\text{E}_{\text{ cell}} = \text{E}_{\text{ cell}}^{\text{o}} - \frac{\text{0.059}}{2} \textit{ \, log \, } \frac{\left[\text{Zn}^{2 +}\right] \times \textit{p}_{\text{H}_{2}}}{\left[\text{H}^{+}\right]^{2}}$
On adding $\text{H}_{2} \text{SO}_{4}$ the $\left[\text{H}^{+}\right]$ will increase therefore $\text{E}_{\text{ cell}}$ will also increase and the equilibrium will shift towards the right.