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: \[ \log \left( \frac{1.5}{0.2} \right) = \log (7.5) \approx 0.875 \] \[ E_{\text{cell}} = 0.46 \, \text{V} - 0.0591 \times 0.875 \] \[ E_{\text{cell}} = 0.46 \, \text{V} - 0.0518 \, \text{V} \] \[ E_{\text{cell}} = 0.4082 \, \text{V} \approx 0.40 \, \text{V} \]

Answer 2

Step 1:
To calculate the emf of the cell, we need to use the Nernst equation:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q \] where:
- \( E_{\text{cell}} \) is the emf of the cell.
- \( E^\circ_{\text{cell}} \) is the standard emf of the cell.
- \( n \) is the number of electrons involved in the reaction.
- \( Q \) is the reaction quotient.
For the given cell, the standard emf of the cell \( E^\circ_{\text{cell}} \) is the difference between the standard reduction potentials of the two half-cells: \[ E^\circ_{\text{cell}} = E^\circ_{\text{Ag}^+/ \text{Ag}} - E^\circ_{\text{Cu}^{2+}/\text{Cu}} = 0.80 \, \text{V} - 0.34 \, \text{V} = 0.46 \, \text{V} \]

Step 2:
Now, calculate the reaction quotient \( Q \). Since 1 Faraday of electricity is passed through Cu²⁺ (1.5 M) and 0.1 Faraday through Ag⁺ (0.2 M), the concentrations of Cu²⁺ and Ag⁺ will change according to the stoichiometry of the half-reactions. However, based on the provided data, the change in concentrations will be considered for the final concentrations.
So, we can use the given formula to calculate the emf, assuming no significant changes in concentration:
\[ E_{\text{cell}} = 0.46 \, \text{V} - 0.06 \, \text{V} = 0.40 \, \text{V} \]

Final Answer:
The emf of the cell formed at 298 K is 0.4 V.