In the Cell reaction
$Mg(s) + 2Ag^+(aq) → Mg^{2+}(aq) + 2Ag(s)$
$E_{cell}^0= 3.17V$. By doubling the concentration of $Mg^{2+}, E_{cell}$ is

decreases by a small fraction

The correct answer is Option (4) → decreases by a small fraction

The effect of changing ion concentration on cell potential.

This is handled by the Nernst equation:

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

Where:

• $Q = \frac{[\text{Mg}^{2+}]}{[\text{Ag}^+]^2}$
• $n = 2$ electrons transferred

Initial cell potential:

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

Effect of doubling [Mg²⁺]:

$Q→2⋅Q$

$\Delta E = - \frac{0.0591}{2} \log 2 \approx -0.0183\, \text{V}$

• Ecell decreases slightly
• It does not double or halve

Correct answer: decreases by a small fraction