Question

1 Faraday electricity was passed through Cu$^{2+}$ (1.5 M, 1 L)/Cu and 0.1 Faraday was passed through Ag$^+$ (0.2 M, 1 L) electrolytic cells. After this, the two cells were connected as shown below to make an electrochemical cell. The emf of the cell thus formed at 298 K is:

Given: $ E^\circ_{\text{Cu}^{2+}/\text{Cu}} = 0.34 \, \text{V} $ $ E^\circ_{\text{Ag}^+/ \text{Ag}} = 0.8 \, \text{V} $ $ \frac{2.303RT}{F} = 0.06 \, \text{V} $

Hint:

The Nernst equation helps to adjust the standard electrode potential by accounting for the concentration of the ions involved in the reaction. For each 10-fold change in concentration, the potential changes by approximately 0.0591 V at 298 K.

Answer 1

Correct Answer: 0.4

Step 1: Standard Electrode Potentials
The given standard electrode potentials are: \[ E^\circ_{\text{Cu}^{2+}/\text{Cu}} = 0.34 \, \text{V} \] \[ E^\circ_{\text{Ag}^+/\text{Ag}} = 0.80 \, \text{V} \] 
Step 2: Applying the Nernst Equation
The Nernst equation gives the relationship between the emf of the electrochemical cell, the standard electrode potentials, and the concentrations of the ions involved: \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q \] Where:
\(E^\circ_{\text{cell}}\) is the standard cell potential, \(n\) is the number of electrons involved, \(Q\) is the reaction quotient. 
Step 3: Determining the Standard Cell Potential
The standard cell potential is the difference between the two half-cell potentials: \[ E^\circ_{\text{cell}} = E^\circ_{\text{Ag}^+/\text{Ag}} - E^\circ_{\text{Cu}^{2+}/\text{Cu}} = 0.80 \, \text{V} - 0.34 \, \text{V} = 0.46 \, \text{V} \] 
Step 4: Reaction Quotient (Q)
The reaction quotient \(Q\) is given by the ratio of the concentrations of the products over the reactants. The concentrations of Cu\(^{2+}\) and Ag\(^+\) ions change due to the passage of electricity: After 1 Faraday is passed through the Cu\(^{2+}\) cell, the concentration of Cu\(^{2+}\) is reduced by a factor based on the number of moles reduced. 
Since 1 Faraday corresponds to 1 mole of electrons, the concentration of Cu\(^{2+}\) decreases. 
Similarly, after 0.1 Faraday is passed through the Ag\(^+\) cell, the concentration of Ag\(^+\) decreases. 
Now, using the changes in concentration:
Initially, the concentration of Cu\(^{2+}\) is 1.5 M, and after 1 Faraday, the concentration of Cu\(^{2+}\) decreases as per the stoichiometry. Initially, the concentration of Ag\(^+\) is 0.2 M, and after 0.1 Faraday, the concentration of Ag\(^+\) decreases accordingly. \[ Q = \frac{[\text{Cu}^{2+}]}{[\text{Ag}^+]} \] 
Step 5: Substituting Values into the Nernst Equation
Substitute the concentrations and other known values into the Nernst equation: \[ E_{\text{cell}} = 0.46 \, \text{V} - \frac{0.0591}{1} \log \left( \frac{1.5 \, \text{M}}{0.2 \, \text{M}} \right) \] Now, calculate the logarithmic term: \[ \log \left( \frac{1.5}{0.2} \right) = \log (7.5) \approx 0.875 \] \[ E_{\text{cell}} = 0.46 \, \text{V} - 0.0591 \times 0.875 \] \[ E_{\text{cell}} = 0.46 \, \text{V} - 0.0518 \, \text{V} \] \[ E_{\text{cell}} = 0.4082 \, \text{V} \approx 0.40 \, \text{V} \]