Question

Calculate the EMF of the Galvanic cell: $ \text{Zn} | \text{Zn}^{2+}(1.0 M) \parallel \text{Cu}^{2+}(0.5 M) | \text{Cu} $ Given: $ E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.763 \, \text{V} $ and $ E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.350 \, \text{V} $

• A. 1.40 V
• B. 1.10 V
• C. 0.40 V
• D. 0.92 V

Hint:

To calculate the EMF of a galvanic cell, use the Nernst equation by considering the standard reduction potentials and the concentrations of the ions involved.

Answer 1

The Correct Option is D

To calculate the EMF of the Galvanic cell, we will use the Nernst equation: \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592}{n} \log Q \] Where: \( E^\circ_{\text{cell}} \) is the standard EMF of the cell, \( n \) is the number of electrons transferred, \( Q \) is the reaction quotient. 
Step 1: Calculate the standard EMF of the cell (\( E^\circ_{\text{cell}} \))
The standard EMF of the cell is the difference between the standard electrode potentials of the cathode and anode: \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] The cathode is Cu (\( E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.350 \, \text{V} \)), The anode is Zn (\( E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.763 \, \text{V} \)). Thus: \[ E^\circ_{\text{cell}} = 0.350 \, \text{V} - (-0.763 \, \text{V}) = 0.350 \, \text{V} + 0.763 \, \text{V} = 1.113 \, \text{V} \] 
Step 2: Calculate the reaction quotient (\( Q \))
The reaction quotient \( Q \) is given by:
\[ Q = \frac{[\text{Cu}^{2+}]}{[\text{Zn}^{2+}]} \] Substitute the concentrations of the ions: \[ Q = \frac{0.5}{1.0} = 0.5 \] 
Step 3: Apply the Nernst equation
The number of electrons transferred in the reaction is 2, as the half-reaction for both zinc and copper involves 2 electrons. Substitute the values into the Nernst equation: \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592}{n} \log Q \] \[ E_{\text{cell}} = 1.113 \, \text{V} - \frac{0.0592}{2} \log 0.5 \] We know that \( \log 0.5 = -0.3010 \), so: \[ E_{\text{cell}} = 1.113 \, \text{V} - \frac{0.0592}{2} \times (-0.3010) \] \[ E_{\text{cell}} = 1.113 \, \text{V} + 0.0089 \, \text{V} \] \[ E_{\text{cell}} = 1.1219 \, \text{V} \] Thus, the EMF of the Galvanic cell is approximately 1.12 V. 
Step 4: Final Answer
The closest answer to our calculated value is: \[ 1.10 \, \text{V} \quad \text{(Option B)} \] Thus, the correct answer is (B) 1.10 V.