Question

The e.m.f. of a Daniell cell at $298\, K$ is $E _{1}$. $Zn / SO _{4}(0.01 M ) \| CuSO _{4}(1.0 M ) / Cu$ When the concentration of $ZnSO _{4}$ is $1.0\, M$ and that of $CuSO _{4}$ is $0.01\, M$, the e.m.f. is changed to $E _{2}$. What is the relationship between $E _{1}$ and $E _{2}$ :

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

Answer 1

The Correct Option is B

To understand the relationship between the e.m.f. values of the Daniell cell under different concentration conditions, we use the Nernst equation. The Daniell cell can be represented as:

Zn | Zn^{2+} (c_1) \| Cu^{2+} (c_2) | Cu

where c_1 and c_2 are the concentrations of Zn^{2+} and Cu^{2+}, respectively.

The cell reaction is:

Zn(s) + Cu^{2+} (aq) \rightarrow Zn^{2+} (aq) + Cu(s)

The Nernst equation for the cell is given by:

E = E^{\circ} - \frac{RT}{nF} \ln \frac{[Zn^{2+}]}{[Cu^{2+}]}

where:

• E is the e.m.f. under non-standard conditions.
• E^{\circ} is the standard e.m.f. of the cell.
• R is the universal gas constant (8.314 J/(mol·K)).
• T is the temperature in Kelvin.
• n is the number of moles of electrons transferred in the balanced equation (which is 2 here).
• F is Faraday's constant (96485 C/mol).

Initially, the concentrations are [Zn^{2+}] = 0.01\, M and [Cu^{2+}] = 1.0\, M. For the first case (E_1):

E_1 = E^{\circ} - \frac{RT}{2F} \ln \frac{0.01}{1.0}

For the second case (E_2) where [Zn^{2+}] = 1.0\, M and [Cu^{2+}] = 0.01\, M:

E_2 = E^{\circ} - \frac{RT}{2F} \ln \frac{1.0}{0.01}

By solving these equations, we can see:

• E_1 involves taking the logarithm of a number less than 1, which will give a negative value and effectively increase the term E^{\circ} - \text{term} relative to E^{\circ}.
• E_2 involves taking the logarithm of a number greater than 1, which makes the term more positive, decreasing the effective value relative to E^{\circ}.

From this, we conclude:

E_1 > E_2

This confirms that the e.m.f. E_1 is greater than E_2 when the concentration conditions are adjusted as described.