Two concentration cells are set up at 298 K: cell I is Zn | Zn^2+ (0.1 M) || Zn^2+ (0.4 M) | Zn and cell II is Ag | Ag+ (0.1 M) || Ag+ (0.4 M) | Ag. The ratio E_II/E_I is

JEE Main levelSingle correct optionDifficulty 2 of 3ec092

Two concentration cells are set up at 298 K: cell I is $Zn ∣ Zn X^{2 +} (0.1 M) ∣∣ Zn X^{2 +} (0.4 M) ∣ Zn$ and cell II is $Ag ∣ Ag X^{+} (0.1 M) ∣∣ Ag X^{+} (0.4 M) ∣ Ag$ . The ratio $E_{II} / E_{I}$ is

Both cells have the same concentration ratio, so

E = \frac{0.0591}{n} lo g 4

in each case. For zinc

n = 2

, for silver

n = 1

. Hence

E_{II} / E_{I} = \frac{0.0591 lo g 4}{( 0.0591/2 ) lo g 4} = 2

. The actual values are 0.0356 V and 0.0178 V.