In a cell that utilizes the reaction: $Zn(s)+2H^{+}(aq)\rightleftharpoons Zn^{2+}(aq)+H_{2}(g)$ addition of $H_{2}SO_{4}$ to cathode compartment will :

Question

Accepted Answer

To understand why Option 1 is the correct answer, we need to analyze the effect of adding $ H_2SO_4 $ to the cathode compartment of the given electrochemical cell reaction:

$$ Zn(s) + 2H^+(aq) \rightleftharpoons Zn^{2+}(aq) + H_2(g) $$

### Explanation for Option 1:

1. **Effect on $ E_{cell} $:**
   - The addition of $ H_2SO_4 $ to the cathode compartment increases the concentration of $ H^+ $ ions. According to the Nernst equation, increasing the concentration of reactants (in this case, $ H^+ $) will increase the cell potential ($ E_{cell} $). This is because the reaction quotient $ Q $ decreases, leading to an increase in $ E_{cell} $.

2. **Shift in Equilibrium:**
   - According to Le Chatelier's principle, increasing the concentration of $ H^+ $ ions will shift the equilibrium to the right to counteract the change. This means more $ Zn^{2+} $ and $ H_2 $ will be produced, favoring the forward reaction.

Thus, the addition of $ H_2SO_4 $ increases the $ E_{cell} $ and shifts the equilibrium to the right, which is why Option 1 is correct.

### Explanation for Why Other Options are Incorrect:

- **Option 2: Lower the $ E_{cell} $ and shift equilibrium to the right.**
  - This option is incorrect because increasing the concentration of $ H^+ $ ions actually increases the $ E_{cell} $, not lowers it. While the equilibrium does shift to the right, the cell potential increases, not decreases.

- **Option 3: Lower the $ E_{cell} $ and shift equilibrium to the left.**
  - This option is incorrect because, as explained, the addition of $ H_2SO_4 $ increases the $ E_{cell} $ and shifts the equilibrium to the right, not to the left.

- **Option 4: Increase the $ E_{cell} $ and shift equilibrium to the left.**
  - This option is incorrect because while the $ E_{cell} $ does increase, the equilibrium shifts to the right, not to the left. The increase in $ H^+ $ concentration drives the reaction forward, not backward.

In conclusion, Option 1 is the correct answer because the addition of $ H_2SO_4 $ increases the $ E_{cell} $ and shifts the equilibrium to the right, consistent with the principles of electrochemistry and Le Chatelier's principle.