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The emf of Daniell cell at 298 K is E1 Zn | ZnSO4 (0.01 M) | | CuSO4 (1.0 M) | Cu When the concentration of ZnSO4 is 1.0 M and that of CuSO4 is 0.01 M, the emf changed to E2 What is the relation between E1 and E2?
  • a)
    E1 = E2
  • b)
    E2 = 0 ≠ E2
  • c)
    E1 >E2
  • d)
    E1 < />2
Correct answer is option 'C'. Can you explain this answer?
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The emf of Daniell cell at 298 K is E1 Zn | ZnSO4 (0.01 M) | | CuSO4 ...
Relation between E1 and E2 in Daniell cell:

The Daniell cell consists of two half-cells, one with a zinc electrode and the other with a copper electrode. The half-cell reactions are as follows:

Zn → Zn2+ + 2e- (oxidation)
Cu2+ + 2e- → Cu (reduction)

The emf of the Daniell cell is given by the equation:
Ecell = Ered (cathode) - Ered (anode)

Where Ered (cathode) is the reduction potential of the cathode (Cu2+ + 2e- → Cu) and Ered (anode) is the reduction potential of the anode (Zn2+ + 2e- → Zn).

Effect of concentration on the emf of the Daniell cell:

The concentration of the electrolyte solutions in the half-cells can affect the emf of the Daniell cell. According to the Nernst equation, the emf of a cell is given by the equation:

Ecell = E°cell - (RT/nF) * ln(Q)

Where E°cell is the standard emf of the cell, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred in the balanced half-cell reactions, F is the Faraday constant, and Q is the reaction quotient.

Comparison of E1 and E2:

In the given scenario, the emf of the Daniell cell at 298 K is E1. When the concentration of ZnSO4 is increased to 1.0 M and that of CuSO4 is decreased to 0.01 M, the emf changes to E2.

Since the concentration of ZnSO4 is increased and that of CuSO4 is decreased, the reaction quotient Q will change. As a result, the term (RT/nF) * ln(Q) in the Nernst equation will change, leading to a change in the overall emf of the cell.

Conclusion:

Based on the given information, we can conclude that the emf of the Daniell cell will decrease when the concentration of ZnSO4 is increased and that of CuSO4 is decreased. Therefore, the correct answer is option C) E1 > E2.
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Community Answer
The emf of Daniell cell at 298 K is E1 Zn | ZnSO4 (0.01 M) | | CuSO4 ...
Using the relation
=
Substituting the given values in two cases.
Thus, E1 > E2
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The emf of Daniell cell at 298 K is E1 Zn | ZnSO4 (0.01 M) | | CuSO4 (1.0 M) | Cu When the concentration of ZnSO4 is 1.0 M and that of CuSO4 is 0.01 M, the emf changed to E2 What is the relation between E1 and E2?a)E1 = E2b)E2 = 0 ≠ E2c)E1 >E2d)E1 2Correct answer is option 'C'. Can you explain this answer?
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