A Daniell cell $Zn ∣ Zn X^{2 +} ∣∣ Cu X^{2 +} ∣ Cu$ is working at 298 K. Which of the following changes will INCREASE the cell EMF?

Answer: A, B

E = E^{\circ} - \frac{0.0591}{2} lo g \frac{[ Zn X ^{2 +} ]}{[ Cu X ^{2 +} ]}

. Lowering

[Zn X^{2 +}]

or raising

[Cu X^{2 +}]

makes

Q

smaller and

E

larger, so (A) and (B) increase EMF. Doubling both leaves the ratio unchanged (C). Ammonia complexes

Cu X^{2 +}

[Cu (NH X_{3}) X_{4}] X^{2 +}

, lowering free

[Cu X^{2 +}]

and decreasing EMF (D).

For the cell Al | Al^3+ (0.01 M) || Sn^2+ (0.1 M) | Sn at 298 K, E deg(Al^3+/Al) = -1.66 V and E deg(Sn^2+/Sn) = -0.14 V. The cell EMF in mV... JEE Main level
A Daniell cell Zn | Zn^2+ (1.0 M) || Cu^2+ (0.02 M) | Cu is operated at 350 K. Assume E deg_cell = 1.10 V does not change with temperature. ... JEE Advanced level
E deg(Cl2/Cl-) = 1.36 V. The reduction potential of a chlorine electrode at 298 K with p_Cl2 = 0.1 bar and [Cl-] = 0.2 M is (take 2.303RT/F ... JEE Main level
The cell Pt | H2 (1 bar) | H+ (1 M) || Cu^2+ (x M) | Cu shows an EMF of 0.290 V at 298 K. Given E deg(Cu^2+/Cu) = 0.34 V and 2.303RT/F = 0.0... JEE Advanced level
For two divalent metals M and N, the cell M | M^2+ (0.01 M) || N^2+ (0.1 M) | N has EMF 0.443 V at 298 K. The value of log_10 K for the reac... JEE Advanced level
The reduction potential of the hydrogen electrode Pt | H2 (4 bar) | H+ in a solution of pH 4 at 298 K is (take 2.303RT/F = 0.0591 V, log 2 =... JEE Main level