Question

The emf of the cell \(\mathrm{Ni}_{(s)} \left| \mathrm{Ni}^{2+}(0.16 \, \mathrm{M}) \right| \mathrm{Ag}^{+}(0.002 \, \mathrm{M}) \left| \mathrm{Ag}_{(s)} \right.\) is (\(E^0_{\text{cell}} = 1.05 \, \mathrm{V}\)):

• A. \(-0.91 \, \mathrm{V}\)
• B. \(+0.46 \, \mathrm{V}\)
• C. \(+0.91 \, \mathrm{V}\)
• D. \(-0.75 \, \mathrm{V}\)

Hint:

For cell emf calculations:
- Write the cell reaction and determine \(n\).
- Use the Nernst equation with concentrations for \(Q\).
- Ensure units and logarithms are handled correctly.

Answer 1

The Correct Option is C

The electromotive force (emf) of an electrochemical cell is calculated using the Nernst equation: \[ E_{\text{cell}} = E^0_{\text{cell}} - \frac{0.0592}{n} \log Q \] where \(E^0_{\text{cell}}\) is the standard cell potential, \(n\) is the number of electrons transferred, and \(Q\) is the reaction quotient.
Step 1: Identify the cell reaction The cell notation is \(\mathrm{Ni}_{(s)} \left| \mathrm{Ni}^{2+}(0.16 \, \mathrm{M}) \right| \mathrm{Ag}^{+}(0.002 \, \mathrm{M}) \left| \mathrm{Ag}_{(s)} \right.\). - Anode (left, oxidation): \(\mathrm{Ni}_{(s)} \rightarrow \mathrm{Ni}^{2+} + 2e^-\) - Cathode (right, reduction): \(\mathrm{Ag}^{+} + e^- \rightarrow \mathrm{Ag}_{(s)}\) Balance the electrons by multiplying the cathode reaction by 2: \[ 2\mathrm{Ag}^{+} + 2e^- \rightarrow 2\mathrm{Ag}_{(s)} \] Overall cell reaction: \[ \mathrm{Ni}_{(s)} + 2\mathrm{Ag}^{+} \rightarrow \mathrm{Ni}^{2+} + 2\mathrm{Ag}_{(s)} \] Thus, \(n = 2\) (2 electrons transferred).
Step 2: Write the reaction quotient \(Q\) For the reaction \(\mathrm{Ni} + 2\mathrm{Ag}^{+} \rightarrow \mathrm{Ni}^{2+} + 2\mathrm{Ag}\), the reaction quotient is: \[ Q = \frac{[\mathrm{Ni}^{2+}]}{[\mathrm{Ag}^{+}]^2} \] (Solids have activity = 1.) Given: - \([\mathrm{Ni}^{2+}] = 0.16 \, \mathrm{M}\) - \([\mathrm{Ag}^{+}] = 0.002 \, \mathrm{M}\) \[ Q = \frac{0.16}{(0.002)^2} = \frac{0.16}{0.000004} = 40000 \]
Step 3: Apply the Nernst equation Given \(E^0_{\text{cell}} = 1.05 \, \mathrm{V}\), \(n = 2\), and at 298 K (\(0.0592 \, \mathrm{V}\)): \[ E_{\text{cell}} = 1.05 - \frac{0.0592}{2} \log (40000) \] Calculate \(\log (40000) = \log (4 \times 10^4) = \log 4 + 4 \approx 0.602 + 4 = 4.602\). Then: \[ \frac{0.0592}{2} \times 4.602 \approx 0.0296 \times 4.602 \approx 0.1362 \] \[ E_{\text{cell}} = 1.05 - 0.1362 \approx 0.9138 \, \mathrm{V} \] Rounding to two decimal places, \(E_{\text{cell}} \approx 0.91 \, \mathrm{V}\).
Step 4: Match with options The closest option is (C) \(+0.91 \, \mathrm{V}\).