The emf of the cell, `Pt|H_(2)(1atm)|H^(+) (0.1M, 30mL)||Ag^(+) (0.8M)Ag` is `0.9V`. Calculate the emf when `40mL` of `0.05M NaOH` is added.

Explanation

To solve the problem step by step, we will follow the process outlined in the video transcript while providing detailed explanations for each step.

### Step 1: Identify the Cell Reactions
The cell is represented as:
\[ \text{Pt} | \text{H}_2(1 \text{ atm}) | \text{H}^+(0.1 \text{ M}, 30 \text{ mL}) || \text{Ag}^+(0.8 \text{ M}) | \text{Ag} \]

In this cell:
- The anode reaction (oxidation) is:
\[ \text{H}_2 \rightarrow 2\text{H}^+ + 2e^- \]
- The cathode reaction (reduction) is:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \]

To balance the number of electrons transferred, we multiply the cathode reaction by 2:
\[ 2\text{Ag}^+ + 2e^- \rightarrow 2\text{Ag} \]

### Step 2: Apply the Nernst Equation
The Nernst equation is given by:
\[ E = E^0 - \frac{0.0591}{n} \log \frac{[\text{H}^+]^2}{[\text{Ag}^+]^2} \]
where:
- \( E \) is the cell potential (emf),
- \( E^0 \) is the standard cell potential,
- \( n \) is the number of electrons transferred (which is 2 in this case),
- \( [\text{H}^+] \) is the concentration of hydrogen ions,
- \( [\text{Ag}^+] \) is the concentration of silver ions.

Given that the emf of the cell is 0.9 V, we can rearrange the equation to find \( E^0 \):
\[ 0.9 = E^0 - \frac{0.0591}{2} \log \frac{(0.1)^2}{(0.8)^2} \]

### Step 3: Calculate \( E^0 \)
First, calculate the logarithmic term:
\[ \log \frac{(0.1)^2}{(0.8)^2} = \log \frac{0.01}{0.64} = \log(0.015625) \approx -1.806 \]

Now substitute this value into the Nernst equation:
\[ 0.9 = E^0 - \frac{0.0591}{2} \times (-1.806) \]
\[ 0.9 = E^0 + 0.0534 \]
\[ E^0 = 0.9 - 0.0534 = 0.8466 \text{ V} \approx 0.85 \text{ V} \]

### Step 4: Calculate New Concentration of \( \text{H}^+ \) After Adding NaOH
When 40 mL of 0.05 M NaOH is added, the concentration of \( \text{H}^+ \) will change due to the neutralization reaction. The total volume becomes \( 30 \text{ mL} + 40 \text{ mL} = 70 \text{ mL} \).

Calculate the moles of \( \text{H}^+ \) and \( \text{OH}^- \):
- Moles of \( \text{H}^+ \) before adding NaOH:
\[ \text{Moles of } \text{H}^+ = 0.1 \text{ M} \times 0.030 \text{ L} = 0.003 \text{ moles} \]
- Moles of \( \text{OH}^- \) from NaOH:
\[ \text{Moles of } \text{OH}^- = 0.05 \text{ M} \times 0.040 \text{ L} = 0.002 \text{ moles} \]

Now, calculate the new concentration of \( \text{H}^+ \):
\[ \text{Moles of } \text{H}^+ \text{ after reaction} = 0.003 - 0.002 = 0.001 \text{ moles} \]
\[ \text{New concentration of } \text{H}^+ = \frac{0.001 \text{ moles}}{0.070 \text{ L}} = \frac{1}{70} \text{ M} \]

### Step 5: Calculate the New EMF Using Nernst Equation
Now we can use the Nernst equation again with the new concentration of \( \text{H}^+ \):
\[ E = E^0 - \frac{0.0591}{2} \log \frac{(1/70)^2}{(0.8)^2} \]

Calculate the logarithmic term:
\[ \log \frac{(1/70)^2}{(0.8)^2} = \log \frac{1/4900}{0.64} = \log(0.0002040816) \approx -3.690 \]

Now substitute this into the Nernst equation:
\[ E = 0.85 - \frac{0.0591}{2} \times (-3.690) \]
\[ E = 0.85 + 0.1087 \]
\[ E \approx 0.9587 \text{ V} \approx 0.90 \text{ V} \]

### Final Answer
The new emf of the cell after adding 40 mL of 0.05 M NaOH is approximately **0.90 V**.

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