At equimolar concentration of `Fe^(2+)` and `Fe^(3+)`, what must `[Ag^(+)]` be so that the voltage of the galvanic cell made from the `(Ag^(+) | Ag)` and `(Fe^(3+)|Fe^(2+)` electrodes equals zero?
`Fe^(2+) + Ag^(+) rightarrow Fe^(3+) + Ag`
`E_(Ag^(+), Ag)^(@)`= 0.7991, `E_(Fe^(3+)//Fe^(2+))^(@) = 0.771`

Options

A. 0.34
B. 0.44
C. 0.47
D. 0.61

Explanation

To solve the problem, we need to find the concentration of \( [Ag^+] \) such that the voltage of the galvanic cell made from the \( (Ag^+ | Ag) \) and \( (Fe^{3+} | Fe^{2+}) \) electrodes equals zero.

### Step-by-Step Solution:

1. **Identify the Standard Reduction Potentials:**
- Given:
- \( E^\circ (Ag^+ | Ag) = 0.7991 \, V \)
- \( E^\circ (Fe^{3+} | Fe^{2+}) = 0.771 \, V \)

2. **Calculate the Standard Cell Potential:**
- The overall cell reaction can be represented as:
\[ Fe^{2+} + Ag^+ \rightarrow Fe^{3+} + Ag \]
- The standard cell potential \( E^\circ_{cell} \) can be calculated using:
\[ E^\circ_{cell} = E^\circ (cathode) - E^\circ (anode) \]
- Here, \( Ag^+ \) is reduced (cathode) and \( Fe^{3+} \) is oxidized (anode):
\[ E^\circ_{cell} = E^\circ (Ag^+ | Ag) - E^\circ (Fe^{3+} | Fe^{2+}) = 0.7991 - 0.771 = 0.028 \, V \]

3. **Set Up the Nernst Equation:**
- At equilibrium, the cell potential \( E_{cell} = 0 \):
\[ 0 = E^\circ_{cell} - \frac{0.0591}{n} \log \left( \frac{[Fe^{3+}]}{[Fe^{2+}][Ag^+]} \right) \]
- Here, \( n = 1 \) (one electron transfer).

4. **Substituting Known Values:**
- Since \( [Fe^{3+}] = [Fe^{2+}] \) (equimolar concentrations), we can denote both concentrations as \( C \):
\[ 0 = 0.028 - 0.0591 \log \left( \frac{C}{C \cdot [Ag^+]} \right) \]
- This simplifies to:
\[ 0 = 0.028 - 0.0591 \log \left( \frac{1}{[Ag^+]} \right) \]

5. **Rearranging the Equation:**
- Rearranging gives:
\[ 0.0591 \log \left( \frac{1}{[Ag^+]} \right) = 0.028 \]
- Dividing both sides by \( 0.0591 \):
\[ \log \left( \frac{1}{[Ag^+]} \right) = \frac{0.028}{0.0591} \approx 0.473 \]

6. **Solving for \( [Ag^+] \):**
- Taking the antilogarithm:
\[ \frac{1}{[Ag^+]} = 10^{0.473} \]
- Therefore:
\[ [Ag^+] = \frac{1}{10^{0.473}} \approx 0.344 \, M \]

### Final Answer:
The concentration of \( [Ag^+] \) must be approximately \( 0.344 \, M \).

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