Electrochemical Cells

Two half-cells, one circuit. A galvanic cell has an electrode dipped in electrolyte on each side. The anode is where oxidation happens — electrons are released there, so it is the negative terminal. The cathode is where reduction happens — electrons are consumed, so it is the positive terminal. In the classic Daniell cell, a zinc rod sits in $\text{ZnSO}_4(aq)$ and a copper rod sits in $\text{CuSO}_4(aq)$:

where: the forward arrow denotes overall spontaneous reaction, $\text{Zn}(s)$ is oxidised at the anode ($\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$) and $\text{Cu}^{2+}(aq)$ is reduced at the cathode ($\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$). Electrons leave the zinc rod, travel through the external wire, and enter the copper rod — that flow of electrons through the wire is the electric current the cell delivers.

The salt bridge. As Zn²⁺ builds up in the anode compartment and Cu²⁺ is consumed in the cathode compartment, each half-cell would otherwise become charge-imbalanced and the reaction would stall. A salt bridge (an inert electrolyte such as KCl in a gel) lets ions migrate between the compartments to keep both sides electrically neutral, completing the internal part of the circuit. It never carries electrons — that is a common confusion addressed below.

Cell notation. By convention the anode (oxidation) is written on the left and the cathode (reduction) on the right; a single vertical line marks a phase boundary and a double vertical line marks the salt bridge:

Standard electrode potential and the zero reference. Every electrode has an intrinsic tendency to gain or lose electrons, but potential can only be measured as a difference, so chemists fix one electrode as a universal zero: the standard hydrogen electrode (SHE), $\text{Pt} \mid \text{H}_2(g,\ 1\ \text{bar}) \mid \text{H}^+(aq,\ 1\ \text{M})$ at $298\ \text{K}$, is assigned $E^\circ = 0.00\ \text{V}$ by definition. Coupling any electrode to the SHE and measuring the EMF gives that electrode's standard reduction potential, $E^\circ$, and these values are tabulated once and reused for any pair of electrodes.

The goal of this derivation is to show, from the physical meaning of reduction potential, why the standard EMF of any galvanic cell equals the cathode's standard reduction potential minus the anode's, and to apply it to recover the Daniell cell's measured $1.10\ \text{V}$.

Step 1. Assign a standard reduction potential to each half-cell relative to the SHE zero. A more positive $E^\circ$ means that species has a stronger pull on electrons — it is more easily reduced. For the Daniell cell, $E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34\ \text{V}$ and $E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76\ \text{V}$.

Step 2. Decide which electrode actually gets reduced. When the two half-cells are connected, the one with the higher (more positive) reduction potential wins the competition for electrons and is reduced — it becomes the cathode. The other is forced into oxidation and becomes the anode. Here $\text{Cu}^{2+}/\text{Cu}$ ($+0.34\ \text{V}$) is the cathode and $\text{Zn}^{2+}/\text{Zn}$ ($-0.76\ \text{V}$) is the anode.

Step 3. Build the cell EMF as a potential drop. The overall EMF is the difference between the electrode that "pulls" electrons in (cathode) and the electrode that "pushes" electrons out (anode), both measured on the same reduction-potential scale, so

Substituting the Daniell values gives $E^\circ_{cell} = 0.34 - (-0.76) = 1.10\ \text{V}$, matching the measured EMF of a real Daniell cell.

Step 4. Note what the formula does not do: it never multiplies $E^\circ$ by the number of electrons transferred, because electrode potential is an intensive quantity — a per-electron property, not a per-mole-of-substance one — unlike $\Delta G$, which does scale with moles reacted.

Problem. For the Daniell cell, $E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34\ \text{V}$ and $E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76\ \text{V}$. Find $E^\circ_{cell}$, and then use it to calculate the standard Gibbs energy change $\Delta G^\circ$ for the reaction $\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$. Take Faraday's constant $F = 96{,}500\ \text{C/mol}$.

Step 1. Assign cathode and anode. Cu²⁺/Cu has the higher reduction potential, so it is the cathode; Zn²⁺/Zn is the anode.

Step 2. Calculate $E^\circ_{cell}$. $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34 - (-0.76) = 1.10\ \text{V}$.

Step 3. Count electrons transferred. The balanced equation transfers $n = 2$ electrons (Zn loses 2, Cu²⁺ gains 2).

Step 4. Apply $\Delta G^\circ = -nFE^\circ_{cell}$: $\Delta G^\circ = -(2)(96{,}500)(1.10) = -212{,}300\ \text{J/mol} = -212.3\ \text{kJ/mol}$.

Step 5. Interpret the sign. A negative $\Delta G^\circ$ confirms the reaction is spontaneous under standard conditions, consistent with the positive $E^\circ_{cell}$ found in Step 2 — the two criteria for spontaneity, a positive EMF and a negative Gibbs energy, always agree. $E^\circ_{cell} = 1.10\ \text{V}$ and $\Delta G^\circ = -212.3\ \text{kJ/mol}$.

Competitive stretch

For the silver–nickel cell (Ag⁺/Ag, $E^\circ = +0.80\ \text{V}$; Ni²⁺/Ni, $E^\circ = -0.25\ \text{V}$), the overall reaction is $\text{Ni}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Ni}^{2+}(aq) + 2\text{Ag}(s)$ with $n = 2$ electrons and $E^\circ_{cell} = 0.80 - (-0.25) = 1.05\ \text{V}$. Even though the silver half-reaction was doubled to balance electrons, $E^\circ_{\text{Ag}^+/\text{Ag}}$ stays at $+0.80\ \text{V}$ (intensive property), giving $\Delta G^\circ = -(2)(96{,}500)(1.05) = -202{,}650\ \text{J/mol} \approx -202.65\ \text{kJ/mol}$.

• Multiplying $E^\circ$ by a stoichiometric coefficient. Electrode potential is intensive (per electron), never scaled when a half-reaction is doubled to balance electrons — unlike $\Delta G$, which does scale with $n$.
• Confusing the salt bridge's job with electron flow. The salt bridge carries ions to keep each half-cell neutral and complete the circuit; electrons travel only through the external wire, never through the salt bridge.
• Mixing up terminal polarity across cell types. In a galvanic cell the anode is negative and the cathode is positive; this reverses in an electrolytic cell, even though "anode = oxidation, cathode = reduction" never changes.
• Getting the sign of $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$ backwards, which turns a genuinely spontaneous cell into an apparently negative EMF — always identify which electrode is actually reduced (higher $E^\circ$) before subtracting.

Ask the tutor: Want to practise picking the anode and cathode from a table of standard reduction potentials, or work through a few more E°cell-to-ΔG° conversions together?