Grade by Grade
F. Electrochemistry · Electrochemistry

Calculating Cell Potentials and Predicting Spontaneity

F2.6 F2.5 F3.3About 70 minutes
Learning goals
  • I can calculate a standard cell potential from standard reduction potentials.
  • I can identify the anode and cathode of a galvanic cell from E° values.
  • I can use the sign of E°cell to predict whether a redox reaction is spontaneous.
  • I can describe, qualitatively, how the cell potential changes as a cell runs down.

Words to know

standard cell potential (E°cell)
The potential difference of a cell when every species is at standard conditions: E°cell = E°cathode − E°anode.
spontaneous (redox)
A redox reaction with a positive E°cell: it can proceed on its own and could power a galvanic cell.
non-spontaneous (redox)
A redox reaction with a negative E°cell: it will not proceed on its own, but it can be forced to happen with an outside power supply (electrolysis).
internal resistance
Resistance to current inside the cell itself (in the electrolyte, salt bridge and electrodes), which lowers the voltage measured while current flows.

From two numbers to a voltage

In the last lesson you met standard reduction potentials. Each one tells you how strongly a half-cell pulls on electrons compared with the hydrogen electrode. Put two half-cells together and the electrons are pushed from the weaker puller to the stronger one. The voltage of the cell is simply the difference between the two pulls:

E°cell = E°cathode − E°anode

Use both values exactly as they are written in the table (as reduction potentials). Do not change any signs, and do not multiply by coefficients.

Which half-cell is the cathode?

In a galvanic cell, the half-cell with the higher (more positive) E° is the cathode. It has the stronger oxidizing agent, so it is the one that gets reduced. The half-cell with the lower E° is the anode. Picking them this way always gives a positive E°cell for a working galvanic cell.

On a vertical potential scale, E°cell is just the distance between the two half-cells.

+1.0 V0.0 V−1.0 VAg⁺/Ag +0.80Cu²⁺/Cu +0.34Zn²⁺/Zn −0.76Al³⁺/Al −1.66Zn–Cu cell0.34 − (−0.76) = 1.10 VAl–Ag cell0.80 − (−1.66) = 2.46 Vcathode = higher E°anode = lower E°E°cell = the gap
E°cell is the gap between the two half-cells on the potential scale. The higher half-cell is the cathode. The wider the gap, the larger the voltage: aluminium and silver are much farther apart than zinc and copper.
Worked example

Example 1: The zinc–copper cell

Calculate E°cell for Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s).

  1. From the table: E°(Cu²⁺/Cu) = +0.34 V; E°(Zn²⁺/Zn) = −0.76 V.
  2. Copper has the higher E°, so it is the cathode; zinc is the anode.
  3. E°cell = E°cathode − E°anode = (+0.34 V) − (−0.76 V) = +1.10 V.

E°cell = +1.10 V. The positive sign confirms a spontaneous reaction: Zn + Cu²⁺ → Zn²⁺ + Cu.

Worked example

Example 2: Analyse an aluminium–silver cell completely

An aluminium electrode in 1.0 mol/L Al(NO₃)₃ is connected to a silver electrode in 1.0 mol/L AgNO₃ with a KNO₃ salt bridge. Find the cathode, the anode, E°cell and the overall equation.

  1. E°(Ag⁺/Ag) = +0.80 V; E°(Al³⁺/Al) = −1.66 V. Silver is higher, so silver is the cathode and aluminium is the anode.
  2. E°cell = 0.80 − (−1.66) = +2.46 V.
  3. Cathode: Ag⁺ + e⁻ → Ag (×3). Anode: Al → Al³⁺ + 3e⁻ (the table row reversed).
  4. Overall: Al(s) + 3Ag⁺(aq) → Al³⁺(aq) + 3Ag(s). (The 3 multiplies the silver half-reaction, but the E° of silver stays +0.80 V.)
  5. Electrons flow from Al to Ag in the wire; NO₃⁻ moves toward the aluminium, K⁺ toward the silver.

Cathode Ag, anode Al, E°cell = +2.46 V, Al(s) + 3Ag⁺(aq) → Al³⁺(aq) + 3Ag(s). Notation: Al(s) | Al³⁺(aq) || Ag⁺(aq) | Ag(s).

salt bridge: KNO₃(aq)NO₃⁻K⁺AlAgV2.46 V (standard)e⁻e⁻Al³⁺(aq)NO₃⁻(aq)Ag⁺(aq)NO₃⁻(aq)anode (−)cathode (+)Al(s) → Al³⁺(aq) + 3e⁻E° = −1.66 V (as a reduction)Ag⁺(aq) + e⁻ → Ag(s)E° = +0.80 V
The aluminium–silver cell from Example 2. Aluminium has the lower E°, so it is the anode; silver has the higher E°, so it is the cathode. Under standard conditions the voltmeter would read about 2.46 V.

Predicting spontaneity from E°cell

You can test any redox equation, not just a cell, with the same formula:

  1. Decide which species is reduced (its half-reaction is the "cathode") and which is oxidized (the "anode") in the equation as written.
  2. Calculate E°cell = E°(reduced species' half-reaction) − E°(oxidized species' half-reaction).
  3. Positive: spontaneous as written. Negative: non-spontaneous as written (the reverse reaction is spontaneous). Zero: a mixture at standard conditions is already at equilibrium (no net reaction).

This gives the same predictions as the "oxidizing agent above the reducing agent" rule from Predicting Redox Reactions and Investigating Them in the Lab, but now with a number that tells you how strongly the reaction is driven.

CalculateE°cell = E°cathode − E°anodeE°cell > 0spontaneous as written;can run a galvanic cellE°cell = 0at equilibrium understandard conditionsE°cell < 0non-spontaneous; theREVERSE is spontaneous
Reading the sign of E°cell. A positive value means the reaction is spontaneous as written; a negative value means it needs an outside energy source (or that the reverse reaction is the spontaneous one).
Worked example

Example 3: Copper in two acids

Copper does not react with 1.0 mol/L hydrochloric acid, but it does react with 1.0 mol/L nitric acid. Use E° values to explain why.

  1. In HCl the only oxidizing agent that could react is H⁺: Cu + 2H⁺ → Cu²⁺ + H₂. Reduced: H⁺ (0.00 V). Oxidized: Cu (+0.34 V). E°cell = 0.00 − 0.34 = −0.34 V: non-spontaneous.
  2. In HNO₃, nitrate in acid is a much stronger oxidizing agent: NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O, E° = +0.96 V. E°cell = 0.96 − 0.34 = +0.62 V: spontaneous.
  3. Balanced: 3Cu(s) + 2NO₃⁻(aq) + 8H⁺(aq) → 3Cu²⁺(aq) + 2NO(g) + 4H₂O(l) (from the lesson on balancing in acid).

Copper is attacked by nitrate, not by H⁺. The acid in HNO₃ supplies the H⁺ the nitrate half-reaction needs.

Worked example

Example 4: Iron(III) and iodide

Is 2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(s) spontaneous under standard conditions?

  1. Reduced: Fe³⁺ → Fe²⁺, E° = +0.77 V. Oxidized: I⁻ → I₂, E° (as a reduction) = +0.54 V.
  2. E°cell = 0.77 − 0.54 = +0.23 V.

Yes, E°cell = +0.23 V > 0. Iron(III) ions oxidize iodide to brown iodine.

When conditions are not standard

E° values assume 1.0 mol/L solutions. Real cells drift away from that as they run. In the zinc–copper cell, [Cu²⁺] falls and [Zn²⁺] rises. Thinking with Le Châtelier's principle (from the sibling unit Chemical Equilibrium): using up a reactant and building up a product reduces the push toward products, so the cell voltage gradually decreases.

When the cell voltage reaches zero, the cell reaction is at equilibrium: the battery is "dead". The reverse is also true: raising [Cu²⁺] or lowering [Zn²⁺] increases the voltage a little. (In first-year university chemistry you will calculate this exactly with the Nernst equation, and connect E°cell to free energy with ΔG° = −nFE°cell.)

The voltage you measure while a large current flows is also a bit lower than E°cell, because of the cell's internal resistance.

Just for fun

A dead battery isn't empty. It's at equilibrium. That's the most peaceful way a battery has ever been described, and also the least useful when your bike light goes out on a dark November evening.

Hands-on activity

Battery label audit

You need: batteries already in your home (AA, AAA, 9 V, a coin cell, and the label on a phone or laptop battery if you can see it; a grown-up can read a car battery label), paper and a pencil. Do not open, cut or puncture any battery. Keep coin cells away from young children.

  1. Make a table: battery, voltage on the label, chemistry on the label (alkaline, lithium, lithium-ion, NiMH, lead-acid), rechargeable (yes/no).
  2. A cell's voltage depends on its chemistry. Use these typical cell voltages: alkaline about 1.5 V; NiMH about 1.2 V; lithium coin cell about 3 V; lithium-ion about 3.6 to 3.85 V; lead-acid about 2 V.
  3. For each battery, estimate how many cells must be connected in series inside it (voltages add in series). For example, a 9 V alkaline battery: 9 ÷ 1.5 = 6 cells. A 12 V car battery: 12 ÷ 2 = 6 cells.
  4. Calculate the lead-acid cell voltage from the table: PbO₂/PbSO₄ is +1.69 V and PbSO₄/Pb is −0.36 V. E°cell = 1.69 − (−0.36) = 2.05 V. Six cells: about 12.3 V, which is why a fully charged "12 V" car battery reads a little over 12 V.

What to look for: label voltages are rounded "nominal" values. You will study each of these battery types in the lesson Batteries: Primary and Secondary Cells.

Your turn

1

Calculate E°cell for a Ni–Cu cell. Which electrode is the anode?

2

Calculate E°cell for Mg(s) | Mg²⁺(aq) || Ag⁺(aq) | Ag(s).

3

Is Cu(s) + Zn²⁺(aq) → Cu²⁺(aq) + Zn(s) spontaneous? Use E°cell.

4

Is Cl₂(g) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(l) spontaneous?

5

A student calculates E°cell for Al + 3Ag⁺ by writing 3 × 0.80 − (−1.66) = 4.06 V. Find the error.

6

Will iron metal dissolve in 1.0 mol/L HCl? Use E° values.

7

What is the cell potential of a battery whose reaction has reached equilibrium?

More practice: warm-up

1

Write the formula for standard cell potential.

2

In a galvanic cell, is the cathode the half-cell with the higher or lower E°?

3

What does a negative E°cell mean?

4

E°cathode = +0.80 V, E°anode = +0.34 V. Find E°cell.

5

E°cathode = 0.00 V, E°anode = −0.76 V. Find E°cell.

More practice: on your own

1

A cell is made from Fe²⁺/Fe and Ag⁺/Ag half-cells. Write the cell notation, the overall equation and E°cell.

2

Can acidified potassium permanganate oxidize chloride ions to chlorine under standard conditions? Calculate E°cell.

3

Which pair of half-cells from the table would give the largest possible E°cell? What is it?

4

A Zn–Cu cell reads 1.10 V when first connected. After running for hours it reads 1.04 V. Give two reasons the reading has fallen.

5

Predict, with E°cell, whether bromine water can oxidize iron(II) ions to iron(III).

Challenge

1

An unknown half-cell X²⁺/X is paired with Cu²⁺/Cu. The cell gives E°cell = 0.47 V and copper is the cathode. Paired with Ag⁺/Ag instead, what will E°cell be, and which electrode is the anode?

2

Gold does not dissolve in nitric acid alone, but it dissolves in aqua regia (a mixture of nitric and hydrochloric acids). Using Au³⁺ + 3e⁻ → Au (E° = +1.50 V), explain the first fact. Then suggest how the extra chloride could help (the gold ends up as the ion AuCl₄⁻).

3

In the zinc–copper cell, [Cu²⁺] is raised to 2.0 mol/L while [Zn²⁺] stays 1.0 mol/L. Will the measured voltage be slightly above or slightly below 1.10 V? Explain without a formula.

Check your understanding

Quiz

1.

What is E°cell for Zn(s) | Zn²⁺(aq) || Ag⁺(aq) | Ag(s)?

2.

Which reaction is spontaneous under standard conditions?

3.

In a galvanic cell made from Ni²⁺/Ni (E° = −0.26 V) and Pb²⁺/Pb (E° = −0.13 V), the cathode is:

4.

A redox reaction has E°cell = −0.42 V. Which statement is true?

5.

As a zinc–copper cell runs, its voltage slowly falls because:

6.

E°cathode = +1.36 V and E°anode = +0.54 V. What is E°cell?

For parents and teachers

Big idea (F2.6, F2.5): E°cell = E°cathode − E°anode, using reduction potentials straight from the table. A positive value means the reaction is spontaneous and can power a galvanic cell; a negative value means it is not spontaneous as written.

How to help: Give your teen two half-cells from the table and ask for the anode, the cathode and E°cell. Then ask for the overall equation. Insist on the formula written out each time; it prevents sign errors.

Common difficulties:

  • Changing the sign of the anode's E° and subtracting it (a double negative error). Use the table values as they are, and subtract once.
  • Multiplying E° by the coefficient used to balance electrons.
  • Thinking a negative E°cell means "impossible". It means not spontaneous; electrolysis can still force it.
  • Forgetting that real concentrations change the voltage.

Safety: in the battery audit, never open or puncture batteries, and keep coin cells away from small children (swallowed coin cells can cause serious internal burns).