You are given the followin cell at `298 K, Zn|(Zn^(+ +) ._((aq.))),(0.01M)||(HCl_((aq.))),(1.0lit)||(H_(2)(g)),(1.0atm)|Pt` with `E_(cell)=0.701` and `E_(Zn^(2+)//Zn)^(0)=-0.76V`. Which of the following amounts of `NaOH (` equivalent weight `=40 )` will just make the `pH` of cathodic compartment to be equal to `7.0 :`

Options

A. `0.4 g`
B. `4g`
C. `10g`
D. `2g`

Explanation

To solve the problem step by step, we will follow the outlined procedure based on the information provided in the question and the video transcript.

### Step 1: Identify the reactions at the anode and cathode
- At the anode, zinc (Zn) is oxidized to zinc ions (Zn²⁺):
\[ \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \]
- At the cathode, hydrogen ions (H⁺) are reduced to hydrogen gas (H₂):
\[ 2\text{H}^+ + 2e^- \rightarrow \text{H}_2 \]

### Step 2: Write the net reaction
Combining the two half-reactions gives us the overall cell reaction:
\[ \text{Zn} + 2\text{H}^+ \rightarrow \text{Zn}^{2+} + \text{H}_2 \]

### Step 3: Use the Nernst equation to find the concentration of H⁺
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log \left( \frac{[\text{Zn}^{2+}]}{[\text{H}^+]^2} \cdot P_{\text{H}_2} \right) \]
Where:
- \( E_{\text{cell}} = 0.701 \, \text{V} \)
- \( E^\circ_{\text{cell}} = 0.76 \, \text{V} \) (since Zn is oxidized)
- \( n = 2 \) (number of electrons transferred)
- \( [\text{Zn}^{2+}] = 0.01 \, \text{M} \)
- \( P_{\text{H}_2} = 1 \, \text{atm} \)

Substituting the values into the Nernst equation:
\[ 0.701 = 0.76 - \frac{0.059}{2} \log \left( \frac{0.01}{[\text{H}^+]^2} \cdot 1 \right) \]

### Step 4: Rearranging to solve for [H⁺]
Rearranging the equation gives:
\[ 0.701 - 0.76 = -0.0295 \log \left( \frac{0.01}{[\text{H}^+]^2} \right) \]
\[ -0.059 = -0.0295 \log \left( \frac{0.01}{[\text{H}^+]^2} \right) \]
\[ \log \left( \frac{0.01}{[\text{H}^+]^2} \right) = 2 \]
Taking the antilog:
\[ \frac{0.01}{[\text{H}^+]^2} = 10^2 = 100 \]
Thus:
\[ [\text{H}^+]^2 = \frac{0.01}{100} = 0.0001 \]
Taking the square root:
\[ [\text{H}^+] = 10^{-2} \, \text{M} \]

### Step 5: Calculate the number of moles of H⁺ in 1 liter
Given the concentration of H⁺:
\[ \text{Number of moles of H}^+ = [\text{H}^+] \times \text{Volume} = 10^{-2} \, \text{mol/L} \times 1 \, \text{L} = 0.01 \, \text{mol} \]

### Step 6: Determine the amount of NaOH required to neutralize H⁺
To achieve a pH of 7, we need to neutralize the H⁺ ions:
\[ \text{Moles of NaOH required} = \text{Moles of H}^+ = 0.01 \, \text{mol} \]
The equivalent weight of NaOH is 40 g/mol, so:
\[ \text{Weight of NaOH} = \text{Moles} \times \text{Equivalent weight} = 0.01 \, \text{mol} \times 40 \, \text{g/mol} = 0.4 \, \text{g} \]

### Final Answer
The amount of NaOH required to make the pH of the cathodic compartment equal to 7.0 is **0.4 grams**.

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