The e.m.f and the standard e.m.f of a cell in the following reaction is 5 V and 5.06 V at room temperature, `Ni(s) + 2Ag^+(n) → Ni^(2+)(0.02M) + 2Ag(s)`. What is the concentration of Ag+ ions?

Options

A. 0.0125 M
B. 0.0314 M
C. 0.0625 M
D. 0.0174 M

Answer

To solve the problem, we will use the Nernst equation, which relates the cell potential (E) to the standard cell potential (E°) and the concentrations of the reactants and products.

### Step-by-Step Solution:

1. **Write down the Nernst equation:**
The Nernst equation is given by:
\[ E = E^\circ - \frac{RT}{nF} \ln Q \]
Where:
- \(E\) = cell potential (5 V)
- \(E^\circ\) = standard cell potential (5.06 V)
- \(R\) = universal gas constant (8.314 J/(mol·K))
- \(T\) = temperature in Kelvin (298 K)
- \(n\) = number of moles of electrons transferred (2 for this reaction)
- \(F\) = Faraday's constant (96485 C/mol)
- \(Q\) = reaction quotient

2. **Calculate the reaction quotient (Q):**
For the reaction:
\[ Ni(s) + 2Ag^+(aq) \rightarrow Ni^{2+}(aq) + 2Ag(s) \]
The reaction quotient \(Q\) is given by:
\[ Q = \frac{[Ni^{2+}]}{[Ag^+]^2} \]
Given that \([Ni^{2+}] = 0.02 M\), we can substitute this into the equation:
\[ Q = \frac{0.02}{[Ag^+]^2} \]

3. **Substitute values into the Nernst equation:**
Rearranging the Nernst equation gives us:
\[ E^\circ - E = \frac{RT}{nF} \ln Q \]
Plugging in the known values:
\[ 5.06 V - 5 V = \frac{(8.314 J/(mol·K))(298 K)}{(2)(96485 C/mol)} \ln \left(\frac{0.02}{[Ag^+]^2}\right) \]

4. **Calculate the left side:**
\[ 0.06 V = \frac{(8.314)(298)}{(2)(96485)} \ln \left(\frac{0.02}{[Ag^+]^2}\right) \]
Calculate the fraction:
\[ \frac{(8.314)(298)}{(2)(96485)} \approx 0.00414 \]
Thus, we have:
\[ 0.06 = 0.00414 \ln \left(\frac{0.02}{[Ag^+]^2}\right) \]

5. **Isolate the logarithm:**
\[ \ln \left(\frac{0.02}{[Ag^+]^2}\right) = \frac{0.06}{0.00414} \approx 14.49 \]

6. **Exponentiate to solve for \(Q\):**
\[ \frac{0.02}{[Ag^+]^2} = e^{14.49} \]
Calculate \(e^{14.49}\):
\[ e^{14.49} \approx 2.45 \times 10^6 \]
Thus:
\[ [Ag^+]^2 = \frac{0.02}{2.45 \times 10^6} \]

7. **Calculate \([Ag^+]\):**
\[ [Ag^+]^2 \approx 8.16 \times 10^{-9} \]
Taking the square root:
\[ [Ag^+] \approx \sqrt{8.16 \times 10^{-9}} \approx 0.0000905 \, M \approx 0.0174 \, M \]

### Conclusion:
The concentration of \(Ag^+\) ions is approximately **0.0174 M**.