Galvanic cells

A galvanic cell has two half-cells — a metal electrode dipped in a solution of its own ions — connected two ways: an external wire carrying electrons, and a salt bridge carrying ions internally.

• The electrode where oxidation occurs is the anode, the negative terminal, since it is the source of the electrons that leave through the wire.
• The electrode where reduction occurs is the cathode, the positive terminal, since electrons arrive there and are consumed.
• Electrons in the external wire travel anode to cathode; conventional current runs the opposite way, cathode to anode, outside the cell.
• The salt bridge (a gel-filled U-tube of an inert electrolyte such as $\text{KCl}$, $\text{KNO}_3$, or $\text{NH}_4\text{NO}_3$) keeps both compartments electrically neutral: anions drift toward the anode to balance the new cations forming there, and cations drift toward the cathode to replace the ions being plated out. Without it, charge would build up on both sides within seconds and the reaction would stall.

The cell's EMF, measured with no current flowing (a potentiometer, not an ordinary ammeter, so the measurement does not itself disturb the cell), is the difference of the two standard reduction potentials:

where: $E_{cathode}$ and $E_{anode}$ are both quoted as reduction potentials (never flip the sign for the anode — subtracting already accounts for it being oxidised), and $E_{cell}$ is positive exactly when the reaction as written is spontaneous. That spontaneity connects to thermodynamics through the charge moved and the driving force behind it:

where: $n$ is the number of moles of electrons transferred per mole of reaction as balanced, $F = 96{,}500\ \text{C/mol}$ is the Faraday constant (the charge on one mole of electrons), and $\Delta G$ is the Gibbs free energy change of the cell reaction. A positive $E_{cell}$ forces $\Delta G$ negative — exactly the condition for a spontaneous process — so the sign of the measured voltage and the sign of the thermodynamic driving force are two views of the same fact.

Problem: A galvanic cell is built from a silver electrode in $1\ M\ \text{AgNO}_3$ and a nickel electrode in $1\ M\ \text{Ni(NO}_3)_2$, with $E^\circ_{\text{Ag}^+/\text{Ag}} = +0.80\ \text{V}$ and $E^\circ_{\text{Ni}^{2+}/\text{Ni}} = -0.25\ \text{V}$. Identify the anode and cathode, write the cell notation, calculate $E^\circ_{cell}$, and find $\Delta G^\circ$ for the reaction as balanced.

Step 1. Compare the two reduction potentials. Silver's is higher ($+0.80\ \text{V}$), so Ag is reduced and acts as the cathode; nickel's is lower ($-0.25\ \text{V}$), so Ni is oxidised and acts as the anode.

Step 2. Write the half-reactions, balancing electrons. Anode: $\text{Ni(s)} \rightarrow \text{Ni}^{2+}(aq) + 2e^-$. Cathode: $2\text{Ag}^+(aq) + 2e^- \rightarrow 2\text{Ag(s)}$, doubled so both half-reactions exchange the same $2$ electrons. Cell notation: $\text{Ni(s)} \mid \text{Ni}^{2+}(aq,\,1M) \parallel \text{Ag}^+(aq,\,1M) \mid \text{Ag(s)}$.

Step 3. Calculate the EMF: $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.80 - (-0.25) = +1.05\ \text{V}$, unchanged by the doubling in Step 2 because electrode potential is intensive.

Step 4. Convert to Gibbs free energy using $\Delta G^\circ = -nFE^\circ_{cell}$ with $n = 2$ (electrons transferred in the balanced equation) and $F = 96{,}500\ \text{C/mol}$: $\Delta G^\circ = -(2)(96{,}500)(1.05) = -202{,}650\ \text{J/mol} \approx -202.7\ \text{kJ/mol}$. The reaction is spontaneous as written: $E^\circ_{cell} = +1.05\ \text{V}$ and $\Delta G^\circ \approx -202.7\ \text{kJ/mol}$.

Competitive stretch

If instead $4$ moles of $\text{Ag}^+$ reacted with $2$ moles of Ni (still the same balanced ratio scaled up), $n$ would be $4$ and $\Delta G^\circ$ would double to about $-405.3\ \text{kJ/mol}$ — but $E^\circ_{cell}$ stays exactly $+1.05\ \text{V}$, because voltage is a per-electron quantity while $\Delta G$ is an extensive, per-mole-of-reaction quantity.

• Treating the salt bridge as a path for electrons — it carries only ions; electrons travel exclusively through the external wire, never through solution.
• Forgetting to double a half-reaction to balance electrons, and then wrongly doubling $E^\circ$ as well — $E^\circ$ is intensive and never changes with the stoichiometric multiplier, even though $n$ (and hence $\Delta G$) does.
• Flipping the sign of the anode's reduction potential before subtracting — the formula $E_{cell} = E_{cathode} - E_{anode}$ already uses both as tabulated reduction potentials; subtracting is what accounts for oxidation at the anode.
• Assuming $E_{cell} > 0$ under standard conditions guarantees spontaneity forever — once concentrations drift from $1\ M$, only the concentration-corrected $E_{cell}$ (via the Nernst equation, covered separately) governs actual spontaneity.

Ask the tutor: want to practice picking anode and cathode for a new pair of electrode potentials, or see how $\Delta G$ changes once the reaction is scaled to a different number of moles?