Question

Find the emf of the cell

Ni(s) + 2 Ag⁺ (0.001 M) → Ni²⁺ (0.001 M) + 2Ag(s)

(Given that E°cell = 1.05 V, \(\frac{2.303RT}{F} = 0.059\) at 298 K)

• A. 1.0385 V
• B. 1.385 V
• C. 0.9615 V
• D. 1.05 V

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The electromotive force (emf) of a non-standard cell is calculated using the Nernst equation, which relates the cell potential to the standard cell potential and the reaction quotient.
Key Formula or Approach:
The Nernst equation is:
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.059}{n} \log Q \]
For the reaction: \(Ni(s) + 2Ag^{+}(aq) \rightarrow Ni^{2+}(aq) + 2Ag(s)\)
The number of electrons transferred, \(n = 2\).
The reaction quotient, \(Q = \frac{[Ni^{2+}]}{[Ag^{+}]^{2}}\).
Step 2: Detailed Explanation:
Substitute the given values into the equation:
\(E^{\circ}_{cell} = 10.5 V\)
\([Ni^{2+}] = 0.001 M = 10^{-3} M\)
\([Ag^{+}] = 0.001 M = 10^{-3} M\)
\[ E_{cell} = 10.5 - \frac{0.059}{2} \log \left( \frac{10^{-3}}{(10^{-3})^{2}} \right) \]
\[ E_{cell} = 10.5 - 0.0295 \log \left( \frac{10^{-3}}{10^{-6}} \right) \]
\[ E_{cell} = 10.5 - 0.0295 \log(10^{3}) \]
\[ E_{cell} = 10.5 - 0.0295 \times 3 \]
\[ E_{cell} = 10.5 - 0.0885 = 10.4115 V \]
Step 3: Final Answer:
Based on the provided \(E^{\circ}_{cell} = 10.5 V\), the calculated emf is \(10.4115 V\), which is not in the options.
However, if there was a typo and \(E^{\circ}_{cell}\) was intended to be \(1.05 V\):
\(E_{cell} = 1.05 - 0.0885 = 0.9615 V\), which matches option (3).
Given the PDF states (NA), it signifies a technical error in the question's values.