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 $E_{2}$ ?

Updated On: Aug 21, 2023
  • $E_1 = E_2$
  • $E_2 = 0 \ne E_2$
  • $E_1 > E_2$
  • $E_1 < E_2$
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The Correct Option is C

Solution and Explanation

In a Daniel cell, 

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

Ecell = 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
 

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Concepts Used:

Nernst Equation

This equation relates the equilibrium cell potential (also called the Nernst potential) to its concentration gradient across a membrane. If there is a concentration gradient for the ion across the membrane, an electric potential will form, and if selective ion channels exist the ion can cross the membrane.