Electrochemistry: Cell Potentials, Free Energy and the Nernst Equation

Inside a galvanic cell: anode, cathode and salt bridge

A galvanic (voltaic) cell turns a spontaneous redox reaction into electrical work. In the classic zinc–copper cell, a zinc strip sits in Zn²⁺ solution and a copper strip sits in Cu²⁺ solution. Zinc is oxidized, Zn → Zn²⁺ + 2e⁻, and the electrons travel through the wire to the copper strip, where Cu²⁺ + 2e⁻ → Cu.

Oxidation always happens at the anode and reduction at the cathode. In a galvanic cell the anode is the negative electrode, because it releases electrons into the wire. A salt bridge, often filled with KNO₃ solution, lets ions move between the two beakers so that neither builds up a net charge. Without it, the current stops almost at once.

Cell notation lists the anode on the left and the cathode on the right: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s). A single bar marks a phase boundary and the double bar marks the salt bridge. If a half-reaction has no solid metal, an inert electrode such as platinum is written at the outer end.

Reading a table of standard reduction potentials

Each half-reaction gets a standard reduction potential, E°, measured against the standard hydrogen electrode, which is assigned exactly 0 V. Standard conditions are 1 M for dissolved species, 1 bar for gases and, in most tables, 25 °C. Every entry is written as a reduction.

The more positive E° is, the more strongly the oxidized form pulls in electrons. F₂ at +2.866 V is the strongest common oxidizing agent in Appendix L. Li⁺ at −3.04 V is the hardest ion to reduce, so lithium metal is a very strong reducing agent. Values used below from Appendix L: Ag⁺/Ag +0.7996 V, Cu²⁺/Cu +0.337 V, Ni²⁺/Ni −0.257 V and Zn²⁺/Zn −0.7618 V.

Worked example: E°cell for the zinc–copper cell

Use E°cell = E°cathode − E°anode, with both values taken as written in the reduction table. Copper has the more positive E°, so it is reduced and is the cathode. E°cell = +0.337 V − (−0.7618 V) = +1.099 V, reported to three decimal places because 0.337 has three.

A positive E°cell means the reaction is spontaneous as written under standard conditions. If you had chosen the electrodes the wrong way round, you would get −1.099 V. That tells you the reverse reaction is the spontaneous one.

The same test works for any redox pair. Will Fe³⁺ oxidize I⁻ to I₂? With Fe³⁺/Fe²⁺ at +0.771 V as the cathode and I₂/I⁻ at +0.5355 V as the anode, E°cell = +0.236 V, so yes. Will Zn²⁺ oxidize copper metal? E°cell = −0.7618 − 0.337 = −1.099 V, so no.

Worked example: why E° is never multiplied by a coefficient

Consider Ni(s) + 2 Ag⁺(aq) → Ni²⁺(aq) + 2 Ag(s). To balance electrons, the silver half-reaction must be doubled. It is tempting to double its E° too, but that would be wrong. E° is an intensive property: it measures energy per unit of charge, and doubling the amount of reaction doubles both the energy and the charge.

So E°cell = +0.7996 V − (−0.257 V) = +1.057 V, with no factor of 2. The amount of reaction enters elsewhere, through n, the number of moles of electrons transferred, which here is 2.

Linking E°cell to ΔG° and K

The maximum electrical work a cell can do equals −ΔG, which gives ΔG° = −nFE°cell. Here F is the Faraday constant, 96,485 C per mole of electrons, and one volt times one coulomb is one joule. A positive E° therefore means a negative ΔG°, and both mean the reaction is spontaneous under standard conditions.

Because ΔG° = −RT ln K, the cell potential also fixes the equilibrium constant. At 25 °C this becomes log K = nE°cell / 0.0592 V, where 0.0592 V is 2.303RT/F at 298.15 K rounded to three significant figures, as OpenStax uses.

Worked example. For the nickel–silver reaction above, n = 2 and E°cell = 1.057 V. ΔG° = −2 × 96,485 C/mol × 1.057 V = −2.04 × 10⁵ J, or −204 kJ per mole of reaction as written. log K = 2 × 1.057 / 0.0592 = 35.7, so K is about 5 × 10³⁵. The reaction goes essentially to completion. Treat K to one significant figure: a 0.001 V change in E° moves log K by about 0.03.

The Nernst equation for nonstandard concentrations

Real cells rarely run at 1 M. The Nernst equation corrects the potential: E = E° − (0.0592 V / n) log Q at 25 °C, where Q is the reaction quotient. Solids are left out of Q, just as they are from K.

Worked example. A zinc–copper cell has [Zn²⁺] = 1.00 M and [Cu²⁺] = 1.0 × 10⁻³ M. For Zn + Cu²⁺ → Zn²⁺ + Cu, Q = [Zn²⁺]/[Cu²⁺] = 1.0 × 10³ and n = 2. E = 1.099 − (0.0592/2) × 3.00 = 1.099 − 0.089 = 1.010 V. Starving the cathode of Cu²⁺ lowers the voltage, as Le Chatelier's principle predicts.

A concentration cell has the same half-reaction on both sides, so E° = 0, and it runs only because the concentrations differ. For copper electrodes in 0.0010 M and 1.0 M Cu²⁺, E = −(0.0592/2) log(0.0010/1.0) = +0.089 V. The dilute side is the anode, because oxidation there makes more Cu²⁺ and evens out the difference. When E reaches zero, Q equals K and the cell is dead.

Electrolysis and Faraday's law

An electrolytic cell uses an external power supply to drive a nonspontaneous reaction, such as plating a metal or splitting water. The anode is still where oxidation happens, but it is now the positive electrode. The amount of product depends only on the charge that passes: charge Q = current I × time t, and moles of electrons = Q / F.

Worked example. A current of 2.50 A flows for 30.0 min through CuSO₄ solution. Q = 2.50 A × 1800 s = 4500 C. Moles of electrons = 4500 / 96,485 = 0.0466 mol. Each Cu²⁺ needs 2 electrons, so 0.0233 mol Cu is deposited, which at 63.55 g/mol is 1.48 g.

Run the calculation backward to find time. Plating 1.00 g of silver (Ag⁺ + e⁻ → Ag) needs 1.00 / 107.87 = 0.00927 mol electrons, which is 894 C. At 0.500 A that takes 1789 s, about 29.8 minutes.

Sign and bookkeeping errors in cell problems

Most mistakes in electrochemistry are sign errors. Write each half-reaction, label it oxidation or reduction, and only then do the arithmetic.

• Flipping the sign of the anode potential and then also subtracting it, which counts the reversal twice.
• Multiplying E° by the coefficient used to balance electrons.
• Mixing up which electrode is positive. The cathode is positive in a galvanic cell, but the anode is positive in an electrolytic cell.
• Using the wrong n in ΔG° = −nFE° or in the Nernst equation. n is the moles of electrons transferred in the balanced overall reaction.
• Leaving ΔG° in joules when the answer is wanted in kilojoules, or the reverse.
• Writing Q upside down in the Nernst equation. Products go on top, as in any reaction quotient.
• Using minutes instead of seconds in Q = It.