Question

The emf of Daniell cell at $298 K$ is $E_{1}$ $Zn \left| ZnSO _{4}(0.01 M )\right|\left| CuSO _{4}(1.0 M )\right| Cu$ When the concentration of $ZnSO _{4}$ is $1.0 M$ and that of $CuSO _{4}$ is $0.01 M$, the emf changed to $E_{2}$. What is the relation between $E_{1}$ and $E_{2}$ ?

• A. $E_1 = E_2$
• B. $E_2 = 0 \ne E_2$
• C. $E_1 > E_2$
• D. $E_1 < E_2$

Answer 1

The Correct Option is C

In a Daniel cell, 

At cathode: Cu²⁺(aq) + 2e⁻ → Cu(s)
At anode: Zn(s) → Zn²⁺(aq) + 2e⁻

E_cell = E°_cell - \(\frac{0.0591}{n}\) log \(\frac{anode}{cathode}\)

Zn| ZnSO₄ (0.01 M) 

CuSO₄ | Cu (1.0 M) 

In the equation, anode = Zn²⁺ . Cathode = Cu²⁺ 

Substituting the value: 

E₁ = E° - \(\frac{0.0591}{n}\)log\(\frac{0.01}{1.0}\)

= E° - \(\frac{0.0591}{2} log 10^{-2}\)

E° + 0.0591 V

Now, for E₂ : Zn (1.0 M) Cu (0.01) 

E₂ = E° - \(\frac{0.0591}{2} \)\(log \frac{1}{0.01}\)

E°- \(\frac{0.0591}{2} log^2\)

E° - 0.0591 V

E₁ > E₂