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
Answer & explanation
Correct answer: option 2
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
$E_\text{cell} = E^\circ_\text{cell} - \frac{0.0591}{2} \log(1)$
$\log(1) = 0$
$E_\text{cell} = 1.25 - 0 = 1.25\,\text{V}$