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
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → decreases by a small fraction
The cell potential under non-standard conditions is given by the Nernst equation. For this cell two electrons are transferred (Mg → \( \ce{Mg^{2+}} \) + 2e−; \( \ce{2Ag^{+}} \) + 2e− → 2Ag), so n = 2, and the solids do not appear in the concentration term:
\( \displaystyle E_{cell} = E_{cell}^{0} - \frac{0.059}{2}\log\frac{[\ce{Mg^2+}]}{[\ce{Ag+}]^{2}} \)
\( \ce{Mg^{2+}} \) is a product, so it sits in the numerator. Doubling it makes the log term larger and Ecell smaller. The change is
\( \displaystyle \Delta E = -\frac{0.059}{2}\log 2 = -0.0295 \times 0.301 = -0.0089\ \mathrm{V} \)
Starting from 1 M solutions, \( E_{cell} \) falls from 3.17 V to about 3.161 V, a drop of less than 0.3%. Because the concentration enters through a logarithm, doubling it changes the potential only slightly. So \( E_{cell} \) decreases by a small fraction.
Why the other options are incorrect
Option (1) doubled — Incorrect. The cell potential is not proportional to concentration; a twofold change in concentration shifts it by only (0.059/2) log 2, about 0.009 V.
Option (2) halved — Incorrect. For the same reason the potential cannot fall to half; the change is a few millivolts.
Option (3) increases but less than double — Incorrect. The potential decreases, not increases: adding more of a product pushes the reaction backwards, so the driving force falls.