Question

The e.m.f. of a Daniell cell at 298 K is E₁.
ZnSO₄ CuSO₄
Zn Cu
(0.01 M) (1.0 M)
When the concentration of ZnSO₄ is 1.0 M and that of CuSO₄
is 0.01 M, the e.m.f. changed to E₂. What is the relationship
between E₁ and E2?
(a) E₂ = 0 ≠ E₁ (b) E₁ > E₂
(c) E₁ < E₂ (d) E₁ = E₂

Answer 1

(B)E1>E2

• For the Deniell cell, the cell reaction is

• Zn(s)+Cu2+(aq.)⇔Cu(s)+Zn2+(aq.)

• Applying Nernst equation, we have (at 2981∘C)

• Ecell=E∘cell−0.0591Vn10gCZn2+CZn2+

• Since E∘cell is positive, Ecell will decrease if we increase the magnitude of concentration ratio, i.e., CZn2+/CCu2+.
• Since  0.01M1.0M<1.0M0.01M , we have E1>E2.