How much voltage does a galvanic cell give? Each half-cell has a tendency to be reduced, measured as a standard reduction potential. Subtracting two of them gives the cell's voltage.
Key idea
$E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode}$, using standard reduction potentials for both (1 M solutions, 1 bar gases, 25 °C). A positive $E^\circ_\text{cell}$ means the reaction as written is favored. $E^\circ$ is never multiplied by a coefficient, even when a half-reaction is multiplied to balance electrons.
Why it works
Only differences in potential can be measured, so one half-cell is chosen as zero: the standard hydrogen electrode, $\mathrm{2H^+ + 2e^- \to H_2}$, $E^\circ = 0$. Everything else is measured against it. A higher (more positive) $E^\circ$ means a stronger pull on electrons, so the half-cell with the higher value becomes the cathode. Potential is energy per unit charge (volts are joules per coulomb). Doubling the amount of reaction doubles both the energy and the charge, so the ratio stays the same: $E^\circ$ is intensive, like temperature.
| Half-reaction | $E^\circ$ (V) |
|---|---|
| $\mathrm{Cl_2 + 2e^- \to 2Cl^-}$ | +1.36 |
| $\mathrm{Ag^+ + e^- \to Ag}$ | +0.80 |
| $\mathrm{Cu^{2+} + 2e^- \to Cu}$ | +0.34 |
| $\mathrm{2H^+ + 2e^- \to H_2}$ | 0.00 |
| $\mathrm{Ni^{2+} + 2e^- \to Ni}$ | −0.26 |
| $\mathrm{Fe^{2+} + 2e^- \to Fe}$ | −0.45 |
| $\mathrm{Zn^{2+} + 2e^- \to Zn}$ | −0.76 |
| $\mathrm{Al^{3+} + 3e^- \to Al}$ | −1.66 |
| $\mathrm{Mg^{2+} + 2e^- \to Mg}$ | −2.37 |
Worked examples
Worked example: The Daniell cell
Find $E^\circ_\text{cell}$ for a zinc–copper cell.
- Copper has the higher $E^\circ$ (+0.34 V): it is the cathode. Zinc (−0.76 V) is the anode.
- $E^\circ_\text{cell} = 0.34 - (-0.76) = 1.10\ \text{V}$.
Answer: $1.10\ \text{V}$.
Worked example: Unequal electrons
Find $E^\circ_\text{cell}$ for an aluminum–silver cell and write the balanced reaction.
- Cathode: silver (+0.80 V); anode: aluminum (−1.66 V).
- Balance electrons: $\mathrm{Al \to Al^{3+} + 3e^-}$ and $\mathrm{3Ag^+ + 3e^- \to 3Ag}$: $\mathrm{Al + 3Ag^+ \to Al^{3+} + 3Ag}$.
- $E^\circ_\text{cell} = 0.80 - (-1.66) = 2.46\ \text{V}$. The 3 in front of silver does not touch its potential.
Answer: $2.46\ \text{V}$; $\mathrm{Al + 3Ag^+ \to Al^{3+} + 3Ag}$.
Worked example: Will it react?
Will copper metal dissolve in 1 M hydrochloric acid (reducing $\mathrm{H^+}$ to $\mathrm{H_2}$)?
- Proposed: Cu is the anode, $\mathrm{H^+/H_2}$ the cathode.
- $E^\circ = 0.00 - 0.34 = -0.34\ \text{V}$. Negative: not favored. Copper does not dissolve in HCl (zinc and iron do).
Answer: No.
Common mistake
Multiplying a half-cell potential by the coefficient used to balance electrons, which would give $0.80 \times 3$ for silver above. Potentials are per unit of charge and do not scale. Check: the cell voltage depends only on which two half-cells you chose, never on how you balanced them.