Yes, the potential difference (voltage) across the terminals of a battery can be greater than its electromotive force (emf), but this can only occur under certain conditions.
### Key Concepts:
1. **EMF (Electromotive Force):**
- The emf of a battery is the maximum potential difference between the two terminals of the battery when no current is flowing, i.e., when the circuit is open. It represents the energy supplied by the battery per unit charge and is determined by the chemical composition of the battery.
- EMF is the "ideal" voltage of the battery and is constant for a given battery under standard conditions.
2. **Potential Difference (Terminal Voltage):**
- The potential difference across the terminals of a battery, also known as terminal voltage, is the actual voltage you measure across the battery when it is connected in a circuit and supplying current.
- Terminal voltage depends on both the internal resistance of the battery and the current flowing through the circuit.
### When is the Potential Difference Greater than the EMF?
This situation can occur during a process called **charging the battery**. Here’s why:
- **During Discharge (Normal Operation):**
When the battery is supplying power to an external load, the terminal voltage is typically *less* than the emf because of the **internal resistance** of the battery. This internal resistance causes a voltage drop when current flows, and the terminal voltage \( V \) is given by:
\[
V = \text{EMF} - I \cdot r
\]
where:
- \( V \) = terminal voltage
- \( I \) = current drawn by the load
- \( r \) = internal resistance of the battery
- EMF = electromotive force (constant for the battery)
This means that when current is flowing, the terminal voltage drops below the emf due to the loss inside the battery.
- **During Charging:**
When you charge the battery, current is forced into it in the reverse direction. In this case, the potential difference across the terminals can exceed the emf because you are pushing energy into the battery. The terminal voltage in this case is given by:
\[
V = \text{EMF} + I \cdot r
\]
This equation shows that the terminal voltage is greater than the emf during charging because of the internal resistance. The external charger applies a voltage greater than the emf to drive current into the battery, overcoming both the emf and the voltage drop due to internal resistance.
### Summary:
- **Discharging Battery:** Terminal voltage \( V \) is less than the emf due to the internal resistance and voltage drop as current flows.
- **Charging Battery:** Terminal voltage \( V \) can be greater than the emf because external energy is pushing current in, overcoming internal resistance and the battery's emf.
Thus, under charging conditions, the potential difference across the battery terminals can exceed the emf.