- In order to explain this, we need to remind ourselves of the Nernst equation
\(E({\rm{ cell }}) = E({\rm{ cell }},{\rm{ standard }}) - \frac{{0.0592V}}{n} \cdot \log Q\)
- When the battery runs out, it means that its voltage becomes\(\;0\).
- From the Nernst equation, we can conclude that this happens when
\(E({\rm{ cell }},{\rm{ standard }}) = \frac{{0.0592V}}{n} \cdot \log Q\)
\(E({\rm{ cell }},{\rm{ standard }}) = \frac{{0.0592V}}{n} \cdot \log K\)
- We can see that the cell voltage will become zero when the chemical reaction occurring in the battery reaches equilibrium.
- That means that when we turn the battery on, a chemical reaction starts happening and it goes on, like any chemical reaction, until it reaches equilibrium.
- And when it reaches equilibrium, we see from the Nernst equation that the \(E(cell)\) becomes zero, which means that the battery dies when the chemical reaction occurring inside reaches equilibrium.