**Determining the amount of NaOH required to make the pH of the cathodic compartment equal to 7.0**

To determine the amount of NaOH required to make the pH of the cathodic compartment equal to 7.0, we need to consider the reactions occurring in the cell and the relationship between pH and concentration of H+ ions.

1. **Cell Reaction:**
The cell reaction in the given cell is:
Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

2. **Standard Reduction Potential:**
The standard reduction potential for the Zn2+/Zn half-reaction is given as -0.76 V.

3. **Nernst Equation:**
The Nernst equation relates the cell potential (E) to the concentrations of the reactants and products involved in the cell reaction:
E = E° - (0.0592/n) log(Q)
where E is the cell potential, E° is the standard reduction potential, n is the number of electrons transferred, and Q is the reaction quotient.

4. **pH and Concentration of H+ ions:**
pH is a measure of the concentration of H+ ions in a solution. The relationship between pH and the concentration of H+ ions is given by the equation:
pH = -log[H+]

5. **Determining the Concentration of H+ ions:**
To determine the concentration of H+ ions in the cathodic compartment, we need to calculate the cell potential (E) using the Nernst equation. Then, we can use the pH equation to find the concentration of H+ ions.

6. **Calculations:**
Given:
E° = 0.701 V
E°Zn2+/Zn = -0.76 V
pH = 7.0

From the Nernst equation:
E = E° - (0.0592/n) log(Q)

Rearranging the equation:
log(Q) = (E° - E)/(0.0592/n)

Substituting the given values:
log(Q) = (-0.76 - 0.701)/(0.0592/2)
log(Q) = -24.831

From the pH equation:
[H+] = 10^(-pH)

Substituting the given pH value:
[H+] = 10^(-7)

Since the concentration of H+ ions in the cathodic compartment is equal to the concentration of H+ ions in the NaOH solution, we can equate the two concentrations:
[H+] = [OH-]

Therefore, the concentration of OH- ions is also 10^(-7) M.

7. **Calculating the Amount of NaOH:**
The concentration of NaOH can be calculated using the formula:
Concentration (M) = Number of Moles / Volume (L)

The equivalent weight of NaOH is given as 40 g/equivalent.

Therefore, the moles of NaOH required can be calculated as:
Moles of NaOH = Concentration (M) * Volume (L)

Substituting the given values:
Moles of NaOH = (10^(-7) M) * (1.0 L