Consider the cell `|{:(H_(2)(Pt)),(1atm):} :|: {:(H_(3)O^(+)(aq)),(pH =5.5):}:||{:(Ag^(+)),(xM):}:|Ag` . The measured `EMF` of the cell is `1.023 V`. What is the value of `x`?
`E_(Ag^(+),Ag)^(@) +0.799 V.[T = 25^(@)C]`

Options

A. `2 xx 10^(-2)M`
B. `2 xx 10^(-3)M`
C. `1.5 xx 10^(-3)M`
D. `1.5 xx 10^(-2)M`

Explanation

To solve the problem, we need to determine the concentration of Ag⁺ ions (denoted as x) in the given electrochemical cell. Here are the steps to arrive at the solution:

### Step 1: Write the half-reactions
The cell consists of two half-reactions: one at the anode (oxidation) and one at the cathode (reduction).

- **Anode Reaction (Oxidation)**:
\[ H_2(g) \rightarrow 2H^+(aq) + 2e^- \]

- **Cathode Reaction (Reduction)**:
\[ Ag^+(aq) + e^- \rightarrow Ag(s) \]

### Step 2: Combine the half-reactions
To balance the number of electrons transferred, we multiply the cathode reaction by 2:

\[ 2Ag^+(aq) + 2e^- \rightarrow 2Ag(s) \]

Now, the overall balanced reaction is:
\[ H_2(g) + 2Ag^+(aq) \rightarrow 2H^+(aq) + 2Ag(s) \]

### Step 3: Use the Nernst Equation
The Nernst equation relates the cell potential (E) to the standard electrode potentials and the concentrations of the reactants and products:

\[ E = E^\circ - \frac{0.0591}{n} \log Q \]

Where:
- \(E\) = cell potential (1.023 V)
- \(E^\circ\) = standard cell potential
- \(n\) = number of moles of electrons transferred (2 in this case)
- \(Q\) = reaction quotient

### Step 4: Calculate the standard cell potential
The standard cell potential \(E^\circ\) can be calculated as follows:
\[ E^\circ = E^\circ_{cathode} - E^\circ_{anode} \]
Given:
- \(E^\circ_{Ag^+/Ag} = 0.799 \, V\)
- \(E^\circ_{H^+/H_2} = 0 \, V\) (standard hydrogen electrode)

Thus,
\[ E^\circ = 0.799 \, V - 0 \, V = 0.799 \, V \]

### Step 5: Substitute values into the Nernst equation
Now we can substitute the values into the Nernst equation:
\[ 1.023 = 0.799 - \frac{0.0591}{2} \log \left( \frac{[H^+]^2}{[Ag^+]^2} \right) \]

### Step 6: Calculate [H⁺] from pH
Given that pH = 5.5, we can calculate the concentration of \(H^+\):
\[ [H^+] = 10^{-pH} = 10^{-5.5} = 3.16 \times 10^{-6} \, M \]

### Step 7: Substitute [H⁺] into the equation
Substituting \([H^+]\) into the Nernst equation:
\[ 1.023 = 0.799 - \frac{0.0591}{2} \log \left( \frac{(3.16 \times 10^{-6})^2}{[Ag^+]^2} \right) \]

### Step 8: Simplify the equation
Rearranging gives:
\[ 1.023 - 0.799 = - \frac{0.0591}{2} \log \left( \frac{(3.16 \times 10^{-6})^2}{[Ag^+]^2} \right) \]
\[ 0.224 = - \frac{0.0591}{2} \log \left( \frac{(3.16 \times 10^{-6})^2}{[Ag^+]^2} \right) \]

### Step 9: Solve for [Ag⁺]
Now we can solve for \([Ag^+]\):
\[ \log \left( \frac{(3.16 \times 10^{-6})^2}{[Ag^+]^2} \right) = -\frac{0.224 \times 2}{0.0591} \]
Calculating the right-hand side gives:
\[ \log \left( \frac{(3.16 \times 10^{-6})^2}{[Ag^+]^2} \right) = -7.57 \]
Taking the antilog:
\[ \frac{(3.16 \times 10^{-6})^2}{[Ag^+]^2} = 10^{-7.57} \]
\[ [Ag^+]^2 = \frac{(3.16 \times 10^{-6})^2}{10^{-7.57}} \]
Calculating this gives:
\[ [Ag^+]^2 = 9.98 \times 10^{-8} \]
\[ [Ag^+] = \sqrt{9.98 \times 10^{-8}} \approx 3.16 \times 10^{-4} \, M \]

### Conclusion
Thus, the value of \(x\) (the concentration of \(Ag^+\)) is approximately \(0.0316 \, M\).

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