A cell is written as

$Mg| Mg^{2+} (0.0001M)|| Ag^+ (0.01M)|Ag$

Calculate its $E_{cell}$ if $E°_{cell} = 1.25 V$ at 298 K

1.25 V

The correct answer is Option (2) → 1.25 V

We are asked to calculate the cell potential (Ecell) using the Nernst equation.

Given:

• Cell: Mg | Mg²⁺ (0.0001 M) || Ag⁺ (0.01 M) | Ag
• $E^\circ_\text{cell} = 1.25 \text{ V}$
• Temperature: $T = 298\,\text{K}$

Step 1: Nernst equation for the cell

$E_\text{cell} = E^\circ_\text{cell} - \frac{0.0591}{n} \log Q$

• Number of electrons transferred (n) = 2 (Mg → Mg²⁺ + 2e⁻, Ag⁺ + e⁻ → Ag, total 2 e⁻)
• Reaction: Mg + 2Ag⁺ → Mg²⁺ + 2Ag
• Reaction quotient:

$Q = \frac{[\text{Mg}^{2+}]}{[\text{Ag}^+]^2} = \frac{0.0001}{(0.01)^2} = \frac{0.0001}{0.0001} = 1$

Step 2: Calculate Ecell

$E_\text{cell} = E^\circ_\text{cell} - \frac{0.0591}{2} \log(1)$

$\log(1) = 0$

$E_\text{cell} = 1.25 - 0 = 1.25\,\text{V}$