Question

In the electrochemical cell: Zn|ZnSO₄ (0.01 M)||CuSO₄ (1.0 M)|Cu, the emf of this Daniel cell is E₁. When the concentration of ZnSO₄ is changed to 1.0 M and that CuSO₄ changed to 0.01 M, the emf changes to E₂. From the above, which one is the relationship between E₁ and E₂?

Options

• E₁ < E₂

• E₁ > E₂

• E₂ ≥ E₁

• E₁ = E₂

Answer 1

E₁ > E₂

Explanation:

`"E"_"cell" = "E"_"cell"^0 - 0.0591/2 log  (["Zn"^(2+)])/(["Cu"^(2+)])`

E₁ = `"E"_"cell"^0 - 0.0591/2 log  10^-2/1`

E₁ = `"E"_"cell"^0 + 0.0591` .....................(1)

\[\ce{Zn_{(s)} -> Zn^{2+}_{( aq)} + 2e^-}\]

\[\ce{Cu^{2+}_{( aq)} + 2e^- -> Cu_{(s)}}\]

\[\ce{Zn_{(s)} + Cu^{2+}_{( aq)} -> Zn^{2+}_{( aq)} + Cu_{(s)}}\]

E₂ = `"E"_"cell"^0 - 0.0591/2 log  1/10^-2`

E₂ = `"E"_"cell"^0 - 0.0591` .....................(2)

∴ E₁ > E₂