Question

A galvanic cell consists of the following: \[ \text{Zn(s)} | \text{Zn}^{2+}(0.01M) || \text{Cu}^{2+}(0.1M) | \text{Cu(s)} \] The standard reduction potentials of the two electrodes are given as \( E^0(\text{Zn}^{2+}/\text{Zn}) = -0.763 \, \text{V} \) and \( E^0(\text{Cu}^{2+}/\text{Cu}) = 0.337 \, \text{V} \). The emf of the above cell will be:

• A. 1.13 V
• B. 1.50 V
• C. 0.455 V
• D. 1.10 V

Hint:

In a galvanic cell, the emf is the difference between the reduction potentials of the cathode and the anode.

Answer 1

The Correct Option is A

Step 1: Understanding the formula for emf.
The emf of the galvanic cell is given by: \[ E_{\text{cell}} = E^0(\text{cathode}) - E^0(\text{anode}) \] Here, the cathode is copper (\( \text{Cu}^{2+}/\text{Cu} \)) and the anode is zinc (\( \text{Zn}^{2+}/\text{Zn} \)). Step 2: Substituting the given values.
\[ E_{\text{cell}} = 0.337 - (-0.763) = 0.337 + 0.763 = 1.13 \, \text{V} \] Step 3: Conclusion.
The emf of the galvanic cell is \( \boxed{1.13} \, \text{V} \). The correct answer is (1) 1.13 V.