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 analyze the electrochemical cell and apply the Nernst equation. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the half-reactions
The cell consists of two half-reactions:
- **Anode (oxidation)**: \( H_2(g) \rightarrow 2H^+(aq) + 2e^- \)
- **Cathode (reduction)**: \( Ag^+(aq) + e^- \rightarrow Ag(s) \)

### Step 2: Write the overall cell reaction
To balance the electrons, we multiply the cathode reaction by 2:
- Overall reaction:
\[ H_2(g) + 2Ag^+(aq) \rightarrow 2H^+(aq) + 2Ag(s) \]

### Step 3: Calculate the standard cell potential (\(E^\circ_{cell}\))
The standard reduction potential for the cathode reaction is given as:
- \( E^\circ_{Ag^+/Ag} = +0.799 \, V \)
The standard reduction potential for the anode reaction (hydrogen) is:
- \( E^\circ_{H^+/H_2} = 0 \, V \)

Thus, the standard cell potential is:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.799 \, V - 0 \, V = 0.799 \, V \]

### Step 4: Apply the Nernst equation
The Nernst equation is given by:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log \left( \frac{[products]}{[reactants]} \right) \]
For our reaction, \( n = 2 \) (2 electrons are transferred).

Substituting the values:
\[ E_{cell} = 0.799 \, V - \frac{0.0591}{2} \log \left( \frac{[H^+]^2}{[Ag^+]^2} \right) \]
Given \( E_{cell} = 1.023 \, V \) and \( pH = 5.5 \), we can find the concentration of \( H^+ \):
\[ [H^+] = 10^{-pH} = 10^{-5.5} = 3.16 \times 10^{-6} \, M \]

### Step 5: Substitute into the Nernst 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 6: 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 7: Solve for \( [Ag^+] \)
Multiplying through by \(-\frac{2}{0.0591}\):
\[ \frac{0.224 \times 2}{0.0591} = \log \left( \frac{(3.16 \times 10^{-6})^2}{[Ag^+]^2} \right) \]
Calculating the left side:
\[ \log \left( \frac{(3.16 \times 10^{-6})^2}{[Ag^+]^2} \right) \approx 7.57 \]

Exponentiating gives:
\[ \frac{(3.16 \times 10^{-6})^2}{[Ag^+]^2} = 10^{7.57} \]
Solving for \( [Ag^+] \):
\[ [Ag^+]^2 = \frac{(3.16 \times 10^{-6})^2}{10^{7.57}} \]
\[ [Ag^+] = \sqrt{\frac{(3.16 \times 10^{-6})^2}{10^{7.57}}} \]
Calculating this gives:
\[ [Ag^+] \approx 2 \times 10^{-2} \, M \]

### Final Answer
Thus, the value of \( x \) (the concentration of \( Ag^+ \)) is:
\[ x = 2 \times 10^{-2} \, M \]

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