1. Identify the cell reaction

The cell reaction is $Zn(s) + 2H^+(aq) \rightarrow Zn^{2+}(aq) + H_2(g)$

• Oxidation at anode: $Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-$
• Reduction at cathode: $2H^+(aq) + 2e^- \rightarrow H_2(g)$

So, the cathode compartment contains $H^+$ ions.

2. Write the reaction quotient

For the overall reaction, $Q = \frac{[Zn^{2+}]\,P_{H_2}}{[H^+]^2}$

since solid $Zn$ is omitted.

3. Use the Nernst equation

At a given temperature, $E = E^\circ - \frac{0.0591}{2}\log Q$

Substituting $Q$: $E = E^\circ - \frac{0.0591}{2}\log\left(\frac{[Zn^{2+}]P_{H_2}}{[H^+]^2}\right)$

4. Effect of adding $H_2SO_4$ to cathode compartment

$H_2SO_4$ is a strong acid, so adding it increases $[H^+]$.

From $Q = \frac{[Zn^{2+}]P_{H_2}}{[H^+]^2}$

if $[H^+]$ increases, then $Q$ decreases.

Now from Nernst equation:

• If $Q$ decreases, then $\log Q$ decreases.
• Therefore, the term $-\frac{0.0591}{2}\log Q$ increases.
• Hence, cell emf $E$ increases.

5. Effect on equilibrium

The reaction consumes $H^+$: $Zn(s) + 2H^+(aq) \rightarrow Zn^{2+}(aq) + H_2(g)$

By Le Chatelier’s principle, increasing concentration of a reactant ($H^+$) shifts the equilibrium to the right to consume added $H^+$.

6. Check options

• A: lower $E$, shift left — incorrect
• B: increase $E$, shift left — incorrect
• C: increase $E$, shift right — correct
• D: lower $E$, shift right — incorrect

7. Final answer

The addition of $H_2SO_4$ to the cathode compartment will:

• increase the emf $E$
• shift equilibrium to the right

So the correct option is C.