Electrode potentials and electrochemical cells

In a nutshell

An electrochemical cell turns a redox reaction into electricity. Instead of letting electrons pass straight from the reducing agent to the oxidising agent, the cell forces them through an external wire, and the "push" on those electrons is a measurable voltage.

This subtopic is about measuring that push (the electrode potential), using tabulated standard electrode potentials (E^⦵) to calculate a cell's EMF and to predict whether a redox reaction will happen, and how real batteries and fuel cells are built from the same idea.

Assumed knowledge: Oxidation, reduction and redox equations.

Core content

Half-equations: the IUPAC convention

By IUPAC convention, every electrode reaction is written as a reduction, with the electrons on the left:

$\text{oxidised form} + n\,e^- \rightleftharpoons \text{reduced form}$

For example:

• Zn²⁺(aq) + 2e^- ⇌ Zn(s)
• Fe³⁺(aq) + e^- ⇌ Fe²⁺(aq)

Two things follow from writing them this way, and both earn marks:

• The species on the left are the oxidising agents (they gain electrons).
• The species on the right are the reducing agents (they lose electrons).

Use the ⇌ sign: each electrode is a redox equilibrium that can run either way depending on what it is connected to.

Half-cells and the electrochemical cell

A cell is made of two half-cells joined together.

A simple metal half-cell is a strip of metal dipping into a solution of its own ions, for example a copper rod in copper(II) sulfate solution. Each half-cell is sitting at its own redox equilibrium, e.g. Cu²⁺(aq) + 2e^- ⇌ Cu(s).

When the two half-cells are connected, the metal with the greater tendency to be oxidised (here zinc) releases electrons onto its electrode:

• Zn(s) ⟶ Zn²⁺(aq) + 2e^-

Electrons build up on the zinc more than on the copper, so a potential difference forms between the electrodes. The zinc is the negative electrode and the copper is the positive electrode. Electrons flow through the wire from the negative electrode to the positive electrode.

This potential difference is measured with a high-resistance (high-impedance) voltmeter. The high resistance stops any real current from flowing, so the half-cell equilibria stay at standard conditions and you measure the maximum potential difference, the cell's EMF.

Still don't get it? · why there is a voltage at all

Picture two water tanks on shelves at different heights, joined by a pipe. Water runs from the higher tank to the lower one, and the bigger the height difference, the harder it pushes. The water does not care about the absolute height of either shelf, only the gap between them.

Electrons behave the same way. Each half-cell sits at its own "electron height": zinc holds its electrons loosely (it would rather be Zn²⁺), so its electrode is a high shelf; copper holds them more tightly, so its electrode is a low shelf. Join the two and electrons run downhill through the wire, from the zinc (high, negative) to the copper (low, positive).

The voltmeter reading is the size of that gap, not the height of either shelf on its own. That is the key exam idea: you can never measure one electrode's potential by itself, only the difference between two electrodes. So one electrode (the standard hydrogen electrode) is chosen as "ground level" and everything else is quoted relative to it.

The salt bridge

The salt bridge completes the circuit and keeps both solutions electrically neutral as the reaction proceeds.

• It works because it contains free-moving ions that migrate into each solution to balance the charge.
• It is usually filter paper or a gel soaked in a solution of an unreactive salt, typically potassium nitrate (KNO₃).
• The salt must not react with either electrode or electrode solution. For example, potassium chloride is unsuitable with a copper half-cell because chloride ions would form complexes with the copper ions.
• A metal wire cannot be used in place of a salt bridge, because it would set up its own electrode system with the solutions.

The single most common exam error here: the salt bridge allows the movement of ions, never the flow of electrons. Electrons only travel through the external wire.

The standard hydrogen electrode

Because you can only ever measure a difference between two electrodes, one reference electrode is defined as the zero point: the standard hydrogen electrode (SHE), assigned a potential of exactly 0 V by convention.

Its equilibrium is:

$2H^+(aq) + 2e^- \rightleftharpoons H_2(g) \qquad E^{\ominus} = 0.00 \text{ V (by definition)}$

The SHE must have all of these:

1. Hydrogen gas at 100 kPa.
2. A solution of 1.00 mol dm^-3 hydrogen ions (usually 1.00 mol dm^-3 HCl).
3. A temperature of 298 K.
4. A platinum electrode, coated in porous platinum black, which provides an inert conducting surface for the equilibrium (there is no metal in the H⁺/H₂ system to act as an electrode).

To measure a standard electrode potential, the SHE is connected to the half-cell in question, placed on the left, and the voltmeter reading under standard conditions is that half-cell's standard electrode potential.

Standard electrode potential, E^⦵

The standard electrode potential, E^⦵, of a half-cell is the EMF measured when it is connected to a standard hydrogen electrode, under standard conditions:

• all solutions of ions at 1.00 mol dm^-3
• a temperature of 298 K
• any gases at 100 kPa
• no current flowing (a high-resistance voltmeter).

Why the conditions matter

Each half-cell is an equilibrium, so by Le Chatelier's principle its potential shifts if the conditions change. That is why E must be quoted under fixed standard conditions.

For Mⁿ⁺(aq) + ne^- ⇌ M(s): increasing [Mⁿ⁺] pushes the equilibrium to the right, pulling electrons off the electrode and making the electrode potential more positive. Lowering the ion concentration makes it more negative. A temperature change shifts it too. Quote E^⦵ only at the standard values above.

For a half-cell built from two ions in the same solution (for example Fe³⁺/Fe²⁺), both ions must be at 1.00 mol dm^-3, and an inert platinum electrode is used to make electrical contact. Watch the concentration trap: 1.00 mol dm^-3 iron(III) sulfate, Fe₂(SO₄)₃, gives 2.00 mol dm^-3 Fe³⁺, which is not standard.

The electrochemical series

Listing standard electrode potentials in order gives the electrochemical series. A selection:

Reading the series:

• The more positive E^⦵, the more the equilibrium lies to the right, so the left-hand species is a better oxidising agent. The strongest oxidising agents (e.g. F₂) sit at the top left.
• The more negative E^⦵, the more the equilibrium lies to the left, so the right-hand species is a better reducing agent. The strongest reducing agents (e.g. Li) sit at the bottom right.

The conventional representation of a cell

A cell is written as a cell diagram using a fixed set of rules:

• A single vertical line | is a phase boundary (for example solid electrode | aqueous ions).
• A double vertical line || is the salt bridge.
• The more negative electrode is written on the left, the more positive on the right.
• The most oxidised species of each half-cell are placed next to the salt bridge (||); the fully reduced species (metals) go on the outside.
• A comma separates two species that are in the same phase (for example two ions in one solution).
• Include state symbols.

So the zinc-copper cell is:

Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)

For an ion/ion half-cell there is no metal to act as an electrode, so an inert platinum electrode is added on the outside:

Pt(s) | Fe²⁺(aq), Fe³⁺(aq) || ...

The standard hydrogen electrode is written Pt(s) | H₂(g) | H⁺(aq), and goes on the left when it is used to measure a standard electrode potential.

Calculating the EMF of a cell

The EMF (electromotive force) is the difference between the two standard electrode potentials:

$E^{\ominus}_{cell} = E^{\ominus}_{\text{positive electrode}} - E^{\ominus}_{\text{negative electrode}}$

Equivalently, for a cell written the conventional way (negative on the left), $E^{\ominus}_{cell} = E^{\ominus}_{\text{right}} - E^{\ominus}_{\text{left}}$. A real spontaneous cell always gives a positive EMF. The full working is in Worked examples.

Using E^⦵ to predict the direction of redox reactions

Put the two half-equations side by side. Then:

• the more positive half-cell runs forwards (reduction, left to right), so its left-hand species is reduced;
• the more negative half-cell runs backwards (oxidation, right to left), so its right-hand species is oxidised.

The reaction is feasible if the overall EMF is positive:

$E^{\ominus}_{cell} = E^{\ominus}_{\text{(reduced, more positive)}} - E^{\ominus}_{\text{(oxidised, more negative)}} > 0$

Still don't get it? · "more positive reduces, more negative oxidises"

Think of a game of pass-the-parcel where the parcel is a pair of electrons, and each half-cell is a player. A player with a more positive E has a strong grip and pulls the electrons towards itself (it gets reduced). A player with a more negative E has a weak grip and lets the electrons go (it gets oxidised). Electrons always move from the weak grip to the strong grip.

Now the exam trap, which examiner reports flag every year. You must compare the two species that are on the same side of their half-equations. Take chlorine oxidising iron(II):

• Cl₂ + 2e^- ⇌ 2Cl^-, E^⦵ = +1.36 V
• Fe³⁺ + e^- ⇌ Fe²⁺, E^⦵ = +0.77 V

Cl₂ and Fe³⁺ are both on the left (both are the oxidising agents). Cl₂ has the more positive E^⦵, so Cl₂ is a stronger oxidising agent than Fe³⁺ and pulls electrons off Fe²⁺. The mark is lost if you write "Cl₂ is a stronger oxidising agent than Fe²⁺" (wrong species) or just "chlorine is more reactive". Compare oxidising agent with oxidising agent, using the exact E^⦵ notation, e.g. E^⦵ Cl₂/Cl^- > E^⦵ Fe³⁺/Fe²⁺.

A positive EMF tells you a reaction is thermodynamically feasible, but not that it will actually happen at a useful rate. If the activation energy is high the reaction may be very slow or not occur at observable speed. E^⦵ says nothing about rate.

Required practical 8: measuring the EMF of an electrochemical cell

Method:

1. Set up two half-cells. For a metal/metal-ion half-cell, clean the metal electrode with sandpaper to remove the oxide layer and wipe it with propanone to remove grease, then stand it in 1.00 mol dm^-3 solution of its ions.
2. Connect the two solutions with a salt bridge (filter paper soaked in saturated KNO₃), making sure both ends dip into the solutions.
3. Connect the electrodes to a high-resistance voltmeter and record the EMF.

Handling the results:

• Compare the measured EMF with the value calculated from E^⦵ data; small differences arise because conditions are not perfectly standard.
• To investigate conditions, change the concentration (or temperature) of one half-cell and record how the EMF changes, then explain the shift using Le Chatelier's principle.

Commercial cells: non-rechargeable, rechargeable and fuel cells

Electrochemical cells are a portable, commercial source of electrical energy. They fall into three types:

• Non-rechargeable (irreversible) cells: the cell reaction cannot be reversed, so once the reactants are used up the cell is flat and is discarded.
• Rechargeable cells: the cell reaction is reversed by applying an external voltage, so the cell can be used again. This only works if the products stay attached to the electrodes rather than dispersing.
• Fuel cells: reactants (a fuel and oxygen) are supplied continuously, so the cell runs without being recharged (covered below).

To work out what happens in any cell from given electrode data, apply the same rule as for any redox prediction: the more positive half-cell is reduced at the positive electrode; the more negative half-cell is oxidised at the negative electrode. Add the half-equations (scaled so the electrons cancel) and simplify to get the overall discharge reaction.

The lithium cell is the AQA worked example. Its simplified electrode reactions are:

• Positive electrode (reduction): Li⁺ + CoO₂ + e^- ⟶ Li⁺[CoO₂]^-
• Negative electrode (oxidation): Li ⟶ Li⁺ + e^-

The reagents are absorbed onto powdered graphite, which acts as a support medium and lets the ions react without a solvent. Water cannot be used, because lithium reacts with water.

The alkaline hydrogen-oxygen fuel cell

A fuel cell uses the energy from reacting a fuel with oxygen to produce a voltage, with the reactants fed in continuously.

In alkaline conditions the electrode reactions are:

• Negative electrode (oxidation): 2H₂(g) + 4OH^-(aq) ⟶ 4H₂O(l) + 4e^-
• Positive electrode (reduction): O₂(g) + 2H₂O(l) + 4e^- ⟶ 4OH^-(aq)
• Overall: 2H₂(g) + O₂(g) ⟶ 2H₂O(l)

How the electrode reactions generate a current: at the negative electrode, hydrogen is oxidised and releases electrons into the external circuit; these electrons flow through the circuit to the positive electrode, where they reduce oxygen. The flow of electrons through the external wire is the current, and the OH^- ions move through the electrolyte to complete the circuit.

Unlike an ordinary cell, a fuel cell delivers a constant voltage because the reactants are replenished continuously, so their concentrations stay constant and the EMF does not fall over time.

Benefits and risks of fuel cells

Benefits:

• A hydrogen-oxygen fuel cell's only product is water, so there is far less pollution and no CO₂ at the point of use.
• It is more efficient than a petrol or diesel engine: a greater proportion of the chemical energy is converted into useful energy, and less is wasted as heat.

Risks and limitations:

• Hydrogen is flammable and explosive, making it hazardous to store and transport (as a pressurised liquid or adsorbed on a solid).
• Hydrogen is often made by electrolysis or from hydrocarbons; unless the electricity comes from renewable sources, making the hydrogen still releases CO₂, so the cell is not automatically "clean".
• Fuel cells are expensive and have a limited lifetime, and toxic chemicals may be used in their manufacture.

Worked examples

Example 1: Calculate the EMF of a cell and write its conventional representation.

A cell is made from a zinc half-cell and a copper half-cell under standard conditions.

• Zn²⁺(aq) + 2e^- ⇌ Zn(s), E^⦵ = -0.76 V
• Cu²⁺(aq) + 2e^- ⇌ Cu(s), E^⦵ = +0.34 V

Copper has the more positive E^⦵, so it is the positive electrode; zinc is the negative electrode.

$E^{\ominus}_{cell} = E^{\ominus}_{\text{positive}} - E^{\ominus}_{\text{negative}}$

$E^{\ominus}_{cell} = (+0.34) - (-0.76)$

$E^{\ominus}_{cell} = +1.10 \text{ V}$

Conventional representation (negative electrode on the left, most oxidised species next to the salt bridge):

Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)

Where the mark is lost: subtracting the wrong way, or treating the two negatives as a subtraction of magnitudes. Taking +0.34 and -0.76 means +0.34 minus -0.76, which adds to +1.10 V, not +0.42 V.

Example 2: Use E^⦵ data to predict whether a reaction occurs, and write the equation.

Will chlorine oxidise iron(II) ions to iron(III) ions?

• Cl₂(g) + 2e^- ⇌ 2Cl^-(aq), E^⦵ = +1.36 V
• Fe³⁺(aq) + e^- ⇌ Fe²⁺(aq), E^⦵ = +0.77 V

The Cl₂/Cl^- electrode is more positive, so it runs forwards (Cl₂ is reduced) and the Fe³⁺/Fe²⁺ electrode runs backwards (Fe²⁺ is oxidised). Comparing the two oxidising agents (both on the left): E^⦵ Cl₂/Cl^- > E^⦵ Fe³⁺/Fe²⁺, so Cl₂ is the stronger oxidising agent and oxidises Fe²⁺.

$E^{\ominus}_{cell} = (+1.36) - (+0.77) = +0.59 \text{ V}$

The EMF is positive, so the reaction is feasible. Scaling the iron half-equation by 2 so the electrons cancel:

$Cl_2(g) + 2Fe^{2+}(aq) \rightarrow 2Cl^-(aq) + 2Fe^{3+}(aq)$

Where the mark is lost: comparing Cl₂ with Fe²⁺ (wrong species), confusing chlorine with the chloride ion, or quoting a positive EMF without saying which species is oxidised and which is reduced.

Example 3: Deduce the electrode reactions and EMF of a rechargeable cell from given data.

A nickel-cadmium cell uses the following electrode data:

• NiO(OH)(s) + H₂O(l) + e^- ⇌ Ni(OH)₂(s) + OH^-(aq), E^⦵ = +0.52 V
• Cd(OH)₂(s) + 2e^- ⇌ Cd(s) + 2OH^-(aq), E^⦵ = -0.88 V

The nickel electrode is more positive, so on discharge it is reduced at the positive electrode (Ni goes from +3 to +2); the cadmium electrode is more negative, so it is oxidised at the negative electrode (Cd goes from 0 to +2).

Scale the nickel half-equation by 2 so the electrons balance, then add:

• Positive (reduction): 2NiO(OH) + 2H₂O + 2e^- ⟶ 2Ni(OH)₂ + 2OH^-
• Negative (oxidation): Cd + 2OH^- ⟶ Cd(OH)₂ + 2e^-

Cancelling the 2OH^- that appears on both sides gives the overall discharge equation:

$Cd(s) + 2NiO(OH)(s) + 2H_2O(l) \rightarrow Cd(OH)_2(s) + 2Ni(OH)_2(s)$

$E^{\ominus}_{cell} = E^{\ominus}_{\text{positive}} - E^{\ominus}_{\text{negative}} = (+0.52) - (-0.88) = +1.40 \text{ V}$

On recharging, an external voltage drives this reaction in reverse.

Where the mark is lost: giving nickel as +2 in NiO(OH) (it is +3), forgetting to scale the half-equations so the electrons cancel, not cancelling the OH^- ions, or writing the discharge equation when the question asks for recharging.

Common exam mistakes

• Comparing the wrong species when using E^⦵ data. You must compare oxidising agent with oxidising agent (both on the left of their half-equations). Saying "fluorine is a better oxidising agent than water" loses the mark; water is the reducing agent, so you compare F₂ with oxygen (E^⦵ F₂/F^- > E^⦵ O₂/H₂O).
• Saying the salt bridge carries electrons. It allows the movement of ions to complete the circuit; electrons only flow through the external wire.
• Loose E^⦵ notation. Write "the E^⦵ of the Fe²⁺/Fe electrode is the most negative", not "the electrode potential of iron". Quoting it the wrong way round (E^⦵ Au/Au⁺ instead of E^⦵ Au⁺/Au) is also penalised.
• Incomplete standard conditions. For the SHE, give all of 298 K, 100 kPa, 1.00 mol dm^-3 H⁺ and a Pt electrode, and name the acid (HCl). For an Fe³⁺/Fe²⁺ half-cell, state that both ions are at 1.00 mol dm^-3.
• The iron(III) sulfate concentration trap. 1.00 mol dm^-3 Fe₂(SO₄)₃ gives 2.00 mol dm^-3 Fe³⁺, which is not standard.
• Cell-diagram slips. Omitting state symbols, putting a comma instead of a phase line between H₂ and H⁺, adding an extra phase boundary at an electrode end, or leaving out (or wrongly adding) the Pt electrode. Copper and iron electrodes do not need a Pt electrode; ion/ion systems do.
• Not simplifying the overall equation. After adding half-equations, cancel species that appear on both sides (spectator OH^-, or a Li⁺ that appears on both sides of a lithium-cell equation).
• Discharge vs recharge. If the question asks for the recharging reaction, reverse the discharge equation; do not hand back the discharge equation.
• EMF sign errors. E^⦵_cell = E^⦵(positive) - E^⦵(negative); subtracting a negative value adds. A spontaneous cell has a positive EMF.
• Treating "feasible" as "fast". A positive EMF means a reaction is thermodynamically feasible, not that it is quick; a high activation energy can make it slow.
• Fuel-cell confusion. The main advantage over a combustion engine is greater efficiency; burning hydrogen produces water, not CO₂; and the lithium cell contains no carbon electrode, so it produces no CO₂.

Key definitions

• Oxidising agent - an electron acceptor (a species that gains/takes electrons). Not "electron pair acceptor", which is a Lewis acid.
• Reducing agent - an electron donor (a species that loses/gives electrons).
• Standard electrode potential (E^⦵) - the EMF of a half-cell connected to a standard hydrogen electrode, measured under standard conditions of 298 K, 100 kPa and all solutions of ions at 1.00 mol dm^-3, with no current flowing.
• Standard hydrogen electrode - the reference half-cell, assigned a standard electrode potential of exactly 0 V, consisting of hydrogen gas at 100 kPa bubbled over a platinum electrode in 1.00 mol dm^-3 H⁺(aq) at 298 K.
• EMF of a cell (E^⦵_cell) - the difference between the standard electrode potentials of the two half-cells: E^⦵_cell = E^⦵(positive electrode) - E^⦵(negative electrode).
• Salt bridge - a connection containing free-moving ions (e.g. KNO₃) that completes the circuit and keeps the solutions electrically neutral, without allowing electrons to pass.
• Fuel cell - a cell that uses the energy from the reaction of a fuel with oxygen, supplied continuously, to produce a voltage without needing to be recharged.

Specification

• I can write half-equations for electrode reactions using the IUPAC convention (reduction, electrons on the left).
• I can state that the species on the left of a half-equation are oxidising agents and those on the right are reducing agents.
• I can describe the standard hydrogen electrode and state the conditions it needs.
• I can explain that electrode potentials are measured relative to the standard hydrogen electrode, and why standard conditions are required.
• I can state that E^⦵ refers to 298 K, 100 kPa and 1.00 mol dm^-3 solutions of ions.
• I can write and apply the conventional representation of a cell, including a platinum electrode where needed.
• I can use E^⦵ values to predict the direction of simple redox reactions.
• I can calculate the EMF of a cell.
• I can measure the EMF of an electrochemical cell (Required practical 8).
• I can use given electrode data to deduce the reactions in non-rechargeable and rechargeable cells, and deduce the EMF.
• I can give the electrode reactions in an alkaline hydrogen-oxygen fuel cell and explain how they generate a current.
• I can discuss the benefits and risks of using electrochemical cells and fuel cells.

Related notes

• Oxidation, reduction and redox equations
• Transition metals (uses electrode potentials for redox reactions of transition-metal ions)