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

To find the emf of the cell, we use the Nernst equation. The Nernst equation in its logarithmic form is given by: 

\[E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log \frac{[C]^c[D]^d}{[A]^a[B]^b}\]

where:

• E_cell is the cell potential under non-standard conditions.
• E°_cell is the standard cell potential.
• n is the number of moles of electrons exchanged.
• [C], [D], [A], [B] are the concentrations of products and reactants respectively.
• a, b, c, d are the stoichiometric coefficients of the reactants and products from the balanced equation.

For the given cell reaction: 
Ni(s) + 2 Ag⁺ (0.001 M) → Ni²⁺ (0.001 M) + 2Ag(s)
We have:

• E°cell = 1.05 V
• n = 2 (as there are 2 electrons exchanged per reaction)
• Concentration of Ag⁺ = 0.001 M
• Concentration of Ni²⁺ = 0.001 M

Therefore, the reaction quotient Q expression is: 

\[Q = \frac{[\text{Ni}^{2+}]}{[\text{Ag}^+]^2}\]

Substitute the concentrations:

\[Q = \frac{0.001}{(0.001)^2} = 1000\]

Now, apply these values in the Nernst equation:

\[E_{\text{cell}} = 1.05 - \frac{0.059}{2} \log(1000)\]

Calculate:

\[E_{\text{cell}} = 1.05 - 0.0295 \cdot 3\]

(since log(1000) = 3)

\[E_{\text{cell}} = 1.05 - 0.0885 = 0.9615 \, \text{V}\]

Therefore, the emf of the cell is 0.9615 V.