Cell Potential Under Nonstandard Conditions
Introduced| English |
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| Nernst equation/nɜːnst ɪˈkweɪʒn/ |
A voltage that fades with use
- A fresh battery reads full voltage; a drained one reads low.
- Concentrations shift as the reaction runs, and so does the push.
- The standard value is only a starting point.
- An equation corrects it for the actual mixture.
Voltage depends on concentration
- The actual voltage differs from $E°$ when concentrations are not standard.
- Using up reactants lowers the voltage.
- Piling up products lowers it too.
The actual cell voltage differs from $E°$ when concentrations are not standard.
Nonstandard concentrations shift the voltage away from $E°$.
Building up products lowers the cell voltage.
More product raises $Q$, which lowers $E$.
The Nernst equation
- The Nernst equation 能斯特方程 corrects for concentration:
- Here $Q$ is the reaction quotient of the current mixture.
The Nernst equation uses which quantity for the current mixture?
$Q$ describes the present concentrations.
The equation that corrects cell potential for concentration is the ____ equation.
The Nernst equation adjusts $E°$ using $Q$.
Zero at equilibrium
- As the reaction proceeds, $Q$ rises and $E$ falls.
- At equilibrium, $E = 0$ -- a dead battery.
- There is no more push once $Q = K$.
As a battery discharges, its voltage...
$Q$ rises as it runs, so $E$ falls to zero at equilibrium.
A dead battery corresponds to a cell potential of...
At equilibrium ($Q = K$), $E = 0$.
As a battery discharges, what happens to its voltage?
- Products build up and reactants deplete, so $Q$ rises.
- The Nernst equation shows $E$ falling toward zero.
Nonstandard cell potential
Using the Nernst equation, sort each change by its effect on the cell potential.
A battery's voltage is not fixed -- it drops as it discharges ($Q$ rises toward $K$). At equilibrium $E = 0$, which is a dead battery, not "no reaction possible." And use $Q$ (the current mixture), not $K$, in the Nernst equation.
Cell voltage changes with concentration: the Nernst equation $E = E° - \frac{RT}{nF}\ln Q$ corrects $E°$ for the current mixture. As the reaction runs, $Q$ rises and $E$ falls, reaching $E = 0$ at equilibrium -- a dead battery.