Question

What is the EMF of a galvanic cell if $ E^\circ_{\text{cathode}} = 0.80 \, \text{volts} $ and $ E^\circ_{\text{anode}} = -0.76 \, \text{volts}?$

• A. 1.56 volts
• B. 0.04 volts
• C. -1.56 volts
• D. -0.04 volts

Hint:

To calculate the EMF of a galvanic cell, subtract the standard electrode potential of the anode from that of the cathode.

Answer 1

The Correct Option is A

Step 1: Understanding the EMF of a Galvanic Cell
The EMF of a galvanic cell is calculated using the formula: \[ E_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Step 2: Substituting the Given Values
Given:
\( E^\circ_{\text{cathode}} = 0.80 \, \text{V} \),
\( E^\circ_{\text{anode}} = -0.76 \, \text{V} \).
The EMF of the cell is: \[ E_{\text{cell}} = 0.80 - (-0.76) = 0.80 + 0.76 = 1.56 \, \text{V} \]
Step 3: Conclusion
Thus, the EMF of the galvanic cell is 1.56 volts.