At `25^@C` potential of the cell, `Pt, H_2(g), HCl (aq)"//"AgCl(s), Ag(s)` is 0.22 V. If `E^@` of silver electrode is 0.8 V , calculate the solubility of AgCI in water .

Answer

To solve the problem, we need to calculate the solubility of AgCl in water based on the given cell potential and the standard electrode potential of the silver electrode. Let's break it down step by step.

### Step 1: Write the half-reactions
At the anode (oxidation):
\[ \text{H}_2(g) \rightarrow 2\text{H}^+(aq) + 2e^- \]

At the cathode (reduction):
\[ \text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s) \]

### Step 2: Write the cell potential equation
The overall cell potential \( E^\circ_{\text{cell}} \) can be expressed as:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{Ag}^+/Ag} - E^\circ_{\text{H}^+/H_2} \]

Given:
- \( E^\circ_{\text{Ag}^+/Ag} = 0.8 \, \text{V} \)
- \( E^\circ_{\text{H}^+/H_2} = 0 \, \text{V} \)

Thus,
\[ E^\circ_{\text{cell}} = 0.8 \, \text{V} - 0 \, \text{V} = 0.8 \, \text{V} \]

### Step 3: Use the Nernst equation
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log Q \]
Where:
- \( n \) is the number of moles of electrons transferred (which is 2 in this case).
- \( Q \) is the reaction quotient.

For the reaction:
\[ Q = \frac{[\text{H}^+]^2}{[\text{Ag}^+]^2} \]

### Step 4: Substitute known values into the Nernst equation
We know:
- \( E_{\text{cell}} = 0.22 \, \text{V} \)
- \( E^\circ_{\text{cell}} = 0.8 \, \text{V} \)

Substituting into the Nernst equation:
\[ 0.22 = 0.8 - \frac{0.059}{2} \log \left( \frac{[\text{H}^+]^2}{[\text{Ag}^+]^2} \right) \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ 0.22 - 0.8 = -\frac{0.059}{2} \log \left( \frac{[\text{H}^+]^2}{[\text{Ag}^+]^2} \right) \]
\[ -0.58 = -\frac{0.059}{2} \log \left( \frac{[\text{H}^+]^2}{[\text{Ag}^+]^2} \right) \]

### Step 6: Solve for the log term
Multiply both sides by -2:
\[ 1.16 = 0.059 \log \left( \frac{[\text{H}^+]^2}{[\text{Ag}^+]^2} \right) \]

### Step 7: Isolate the log term
\[ \log \left( \frac{[\text{H}^+]^2}{[\text{Ag}^+]^2} \right) = \frac{1.16}{0.059} \]
Calculating gives:
\[ \log \left( \frac{[\text{H}^+]^2}{[\text{Ag}^+]^2} \right) \approx 19.66 \]

### Step 8: Solve for the concentrations
Taking the antilog:
\[ \frac{[\text{H}^+]^2}{[\text{Ag}^+]^2} = 10^{19.66} \]

Since at standard conditions, \( [\text{H}^+] = 1 \, \text{M} \):
\[ \frac{1^2}{[\text{Ag}^+]^2} = 10^{19.66} \]
Thus:
\[ [\text{Ag}^+]^2 = 10^{-19.66} \]
\[ [\text{Ag}^+] = 10^{-9.83} \]

### Step 9: Calculate the solubility of AgCl
Since the solubility of AgCl in water is equal to the concentration of Ag+ ions:
\[ \text{Solubility of AgCl} = 10^{-9.83} \, \text{mol/L} \]

### Final Answer
The solubility of AgCl in water is approximately \( 1.47 \times 10^{-10} \, \text{mol/L} \).

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