The EMF of the following cell is 0.86 volts `Ag|AgNO_(3)(0.0093M)||AgNO_(3)(xM)|Ag`. The value of x will be

Options

A. 82.8M
B. 2.28M
C. 0.228M
D. 1.14M

Explanation

To find the value of \( x \) in the electrochemical cell given by the notation \( \text{Ag} | \text{AgNO}_3(0.0093M) || \text{AgNO}_3(xM) | \text{Ag} \) with an EMF of 0.86 volts, we can follow these steps:

### Step 1: Identify the half-reactions
In the given cell notation, we have:
- Anode (oxidation): \( \text{Ag} \rightarrow \text{Ag}^+ + e^- \)
- Cathode (reduction): \( \text{Ag}^+ + e^- \rightarrow \text{Ag} \)

### Step 2: Write the Nernst equation
The Nernst equation is given by:
\[ E = E^\circ - \frac{0.0591}{n} \log Q \]
where:
- \( E \) is the cell potential (0.86 V)
- \( E^\circ \) is the standard cell potential
- \( n \) is the number of moles of electrons transferred (1 in this case)
- \( Q \) is the reaction quotient

### Step 3: Determine the standard cell potential \( E^\circ \)
Since both half-reactions involve silver, the standard reduction potentials for both reactions are equal in magnitude but opposite in sign. Therefore, the standard cell potential \( E^\circ \) is:
\[ E^\circ = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0 - 0 = 0 \text{ V} \]

### Step 4: Calculate the reaction quotient \( Q \)
The reaction quotient \( Q \) is defined as:
\[ Q = \frac{[\text{Ag}^+]_{\text{anode}}}{[\text{Ag}^+]_{\text{cathode}}} = \frac{0.0093}{x} \]

### Step 5: Substitute values into the Nernst equation
Substituting the known values into the Nernst equation:
\[ 0.86 = 0 - \frac{0.0591}{1} \log \left(\frac{0.0093}{x}\right) \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ 0.86 = -0.0591 \log \left(\frac{0.0093}{x}\right) \]
\[ \log \left(\frac{0.0093}{x}\right) = -\frac{0.86}{0.0591} \]

### Step 7: Calculate the logarithm
Calculating the right side:
\[ \log \left(\frac{0.0093}{x}\right) = -14.53 \quad (\text{approx.}) \]

### Step 8: Solve for \( \frac{0.0093}{x} \)
Taking the antilogarithm:
\[ \frac{0.0093}{x} = 10^{-14.53} \]
\[ \frac{0.0093}{x} \approx 3.10 \times 10^{-15} \]

### Step 9: Solve for \( x \)
Rearranging gives:
\[ x \approx \frac{0.0093}{3.10 \times 10^{-15}} \approx 2.99 \times 10^{13} \text{ M} \]

### Conclusion
Thus, the value of \( x \) is approximately \( 2.99 \times 10^{13} \text{ M} \).

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