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JEE Main levelSingle correct optionDifficulty 1 of 3ec29

For $Zn ∣ Zn X^{2 +} (0.1 M) ∣∣ Cu X^{2 +} (0.01 M) ∣ Cu$ with $E_{cell}^{\circ} = 1.10 V$ , the EMF at 298 K is closest to

A $1.07 V$
B $1.13 V$
C $1.10 V$
D $1.04 V$

Answer: A

E = 1.10 - \frac{0.0591}{2} lo g \frac{[ Zn X ^{2 +} ]}{[ Cu X ^{2 +} ]} = 1.10 - 0.02955 lo g 10 = 1.0705 V

. Product more concentrated than reactant, so

E < E^{\circ}

.

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