The electromotive force (EMF) of a cell is calculated as the difference between the half-cell potentials of the cathode (reduction half-cell) and the anode (oxidation half-cell).
Here's the formula:
EMFcell = Ecathode - Eanode
Where:
- EMFcell represents the electromotive force of the cell (measured in volts).
- Ecathode represents the reduction potential of the cathode (positive electrode). This is where reduction occurs.
- Eanode represents the reduction potential of the anode (negative electrode). This is where oxidation occurs. Note that since standard reduction potentials are typically given, you use the reduction potential of the anode in this formula.
Explanation:
The EMF represents the driving force that pushes electrons through the external circuit from the anode to the cathode. A positive EMF value indicates that the reaction is spontaneous under standard conditions.
Example:
Consider a galvanic cell with the following half-reactions:
- Cathode (Reduction): Cu2+(aq) + 2e- → Cu(s) E° = +0.34 V
- Anode (Oxidation): Zn(s) → Zn2+(aq) + 2e- E° = -0.76 V
The EMF of the cell would be:
EMFcell = Ecathode - Eanode = 0.34 V - (-0.76 V) = 1.10 V
Therefore, the EMF of this cell is 1.10 V.
