Electrode Potentials and Cell Voltage

Introduction

A torch battery, a car battery and the corrosion of an iron gate all depend on the same idea: different redox couples have different tendencies to gain electrons. Tables of standard electrode potentials put a number on that tendency. The single most useful calculation you can do with such a table is to predict the voltage of a cell and the direction in which it will run. The rule is short — cathode minus anode — but most exam errors come from applying it carelessly, so this page builds a reliable routine.

Core explanation

Reduction potentials only. Every entry in a standard table is written as a reduction half-equation, for example Cu²⁺(aq) + 2e⁻ → Cu(s), E° = +0.34 V. The more positive E° is, the more strongly the oxidised form (here Cu²⁺) pulls electrons towards itself. The zero of the scale is the standard hydrogen electrode, 2H⁺(aq) + 2e⁻ → H₂(g), defined as 0.00 V.

Selected standard reduction potentials at 298 K:

Half-equation E° / V --- --- Mg²⁺ + 2e⁻ → Mg −2.37 Al³⁺ + 3e⁻ → Al −1.66 Zn²⁺ + 2e⁻ → Zn −0.76 Fe²⁺ + 2e⁻ → Fe −0.44 Ni²⁺ + 2e⁻ → Ni −0.25 2H⁺ + 2e⁻ → H₂ 0.00 Cu²⁺ + 2e⁻ → Cu +0.34 I₂ + 2e⁻ → 2I⁻ +0.54 Fe³⁺ + e⁻ → Fe²⁺ +0.77 Ag⁺ + e⁻ → Ag +0.80 Br₂ + 2e⁻ → 2Br⁻ +1.07 Cl₂ + 2e⁻ → 2Cl⁻ +1.36

Which electrode is which? When two half-cells are connected, the couple with the more positive E° undergoes reduction: it is the cathode . The couple with the less positive E° is forced to run backwards, as an oxidation: it is the anode . Electrons flow through the external wire from anode to cathode.

The cell potential. The standard cell potential is

E°cell = E°(cathode) − E°(anode)

using both values exactly as they appear in the reduction table. You never reverse the sign of the anode value yourself; the subtraction already accounts for the reversal. If you choose the electrodes correctly, E°cell is positive, which corresponds to a spontaneous reaction under standard conditions.

No multiplication by coefficients. To write the overall equation you may multiply a half-equation so that electrons cancel — for example doubling Ag⁺ + e⁻ → Ag when it is paired with Cu. The potential stays at +0.80 V. Potential is energy per unit charge, an intensive quantity: twice as much silver deposited involves twice the charge and twice the energy, but the same energy per coulomb.

Checking the direction of a proposed reaction. If a question asks "does reaction X occur?", assign cathode and anode according to the reaction as written, not according to the table. A negative E°cell means the reaction as written is not spontaneous under standard conditions; the reverse reaction is.

Step-by-step reasoning

1. Write both couples as reduction half-equations with their E° values. 2. Choose the more positive E° as the cathode (for a galvanic cell) or, for a stated reaction, identify which species is reduced. 3. Calculate E°cell = E°(cathode) − E°(anode). 4. Balance electrons by multiplying half-equations, but leave the potentials unchanged. 5. Add the half-equations and check that atoms and charge balance.

Visual explanation

Imagine the potentials as rungs on a vertical ladder with positive values at the top. Draw both couples on the ladder. The upper couple is the cathode, the lower one is the anode, and the vertical distance between the rungs is the cell potential. Electrons "fall" from the lower rung up to the higher-potential couple through the wire.

Real-world analogy

Think of two water tanks at different heights joined by a pipe. Water flows from the higher-energy tank to the lower one, and the pressure available depends only on the height difference, not on how much water is in each tank. The cell potential is that height difference.

Real-world example

A zinc–copper (Daniell) cell gives about 1.10 V. Galvanised steel is protected because zinc (−0.76 V) is oxidised in preference to iron (−0.44 V): in any local cell formed by moisture, zinc is the anode and corrodes sacrificially while the iron remains intact.

Why?

Why subtract the anode potential rather than add its negative? They are numerically the same, but the subtraction keeps every value in its reduction form, so there is only one table convention to remember and no need to flip signs. Fewer manipulations mean fewer sign errors in multi-step problems.

Common misconception

"Because two silver ions are needed per copper atom, the silver potential should be doubled." Potentials are intensive and are never scaled by coefficients. Doubling a half-equation doubles the charge transferred, not the voltage.

Worked example

Question: A cell is made from a Ni²⁺/Ni half-cell and an Ag⁺/Ag half-cell under standard conditions. Identify the electrodes, calculate E°cell and write the cell reaction.

Reasoning: Ag⁺/Ag (+0.80 V) is more positive than Ni²⁺/Ni (−0.25 V), so silver is the cathode and nickel the anode. E°cell = +0.80 − (−0.25) = +1.05 V. Electrons are balanced by doubling the silver half-equation (potential unchanged): 2Ag⁺ + Ni → 2Ag + Ni²⁺.

Answer: Cathode Ag, anode Ni, E°cell = +1.05 V, reaction 2Ag⁺(aq) + Ni(s) → 2Ag(s) + Ni²⁺(aq).

Quick check

1. Will bromine oxidise iodide ions under standard conditions? Give the cell potential for the reaction Br₂ + 2I⁻ → 2Br⁻ + I₂. Answer: Yes. Br₂ is reduced (cathode, +1.07 V) and I⁻ is oxidised (anode, +0.54 V), so E°cell = +1.07 − 0.54 = +0.53 V, which is positive.

Exam focus

Always state which half-cell is the cathode before substituting numbers, and write E°cell = E°(cathode) − E°(anode) explicitly. Examiners award marks for the positive sign, for units (V) and for a balanced overall equation with electrons cancelled. Never scale a potential with coefficients.

Advanced insight

A positive E°cell predicts thermodynamic feasibility only under standard conditions. Real concentrations shift potentials (the Nernst equation), and a feasible reaction can still be too slow to observe: the reduction of water by magnesium at room temperature is thermodynamically very favourable but kinetically sluggish because of an oxide film and a large activation barrier.

Summary

Standard electrode potentials are tabulated as reductions relative to the hydrogen electrode. In a galvanic cell the more positive couple is the cathode and the less positive couple is the anode. E°cell = E°(cathode) − E°(anode), using table values unchanged. Coefficients used to balance electrons never multiply the potential, and a positive E°cell indicates a spontaneous reaction under standard conditions.

Practice questions

1. Calculate E°cell for a cell made from Mg²⁺/Mg and Cu²⁺/Cu half-cells. Answer: Copper is the cathode: E°cell = +0.34 − (−2.37) = +2.71 V. 2. Does Fe³⁺ oxidise I⁻ to I₂ under standard conditions? Justify with a calculation. Answer: Yes. Fe³⁺ is reduced (+0.77 V) and I⁻ is oxidised (+0.54 V), giving E°cell = +0.23 V, which is positive. 3. For the proposed reaction Zn²⁺ + Cu → Zn + Cu²⁺, calculate E° and comment. Answer: Zn²⁺ would be the species reduced, so E° = −0.76 − (+0.34) = −1.10 V; negative, so the reaction is not spontaneous and the reverse reaction occurs instead. 4. A student writes E°cell for the Al/Ag cell as 3(+0.80) − (−1.66). Identify the error and give the correct value. Answer: The silver potential must not be multiplied by the coefficient; E°cell = +0.80 − (−1.66) = +2.46 V.