Topic 16 · page 6 of 19

Cell emf and cell diagrams

Subtract the less positive electrode potential from the more positive one to get the standard emf of a cell, and write the cell in the conventional shorthand with the negative electrode on the left.

15 min read16.716.8

Read first: The electrochemical series

Calculating the standard emf 16.7

When two half-cells are joined, the one with the more negative $E^\ominus$ is the one that is oxidised. It pushes electrons out, so it is the negative electrode. The other half-cell is reduced and is the positive electrode.

MethodFinding $E^\ominus_\text{cell}$ and the cell reaction

Find the two $E^\ominus$ values. The more negative half-cell is the negative electrode and is oxidised.
Calculate $E^\ominus_\text{cell}$ = more positive − more negative.
Write the oxidation (reverse the half-equation of the negative electrode) and the reduction.
Multiply so the electrons match, add, and cancel them.

Worked exampleZinc and copper

$E^\ominus$ (Zn²⁺/Zn) = −0.76 V and $E^\ominus$ (Cu²⁺/Cu) = +0.34 V.

Zinc is more negative: it is oxidised and is the negative electrode. Electrons flow through the wire from Zn to Cu.
$E^\ominus_\text{cell}$ = +0.34 − (−0.76) = +1.10 V
Oxidation: Zn → Zn²⁺ + 2e⁻. Reduction: Cu²⁺ + 2e⁻ → Cu.

\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}

Worked exampleMagnesium and silver

$E^\ominus$ (Mg²⁺/Mg) = −2.37 V and $E^\ominus$ (Ag⁺/Ag) = +0.80 V.

Magnesium is more negative, so it is oxidised: Mg → Mg²⁺ + 2e⁻.
$E^\ominus_\text{cell}$ = +0.80 − (−2.37) = +3.17 V
Silver is reduced: Ag⁺ + e⁻ → Ag. Multiply by 2 so the electrons match.

\small \ce{Mg(s) + 2Ag+(aq) -> Mg^2+(aq) + 2Ag(s)}

Cell diagrams 16.8

A cell diagram is a shorthand for a cell. The conventions are fixed, so learn them as a list.

The negative electrode (where oxidation happens) is on the left. The positive electrode (where reduction happens) is on the right.
A single vertical line | marks a phase boundary, for example between a solid electrode and a solution.
A double vertical line ‖ is the salt bridge.
The electrodes are at the outer ends.
Species in the same phase are separated by a comma, with the oxidised forms next to the salt bridge.
Use Pt(s) as the electrode when the couple has no metal, and include state symbols.

Read a diagram from left to right as the reaction that happens: the left-hand half-cell is oxidised and the right-hand half-cell is reduced.

\ce{Zn(s)} \mid \ce{Zn^2+(aq)} \parallel \ce{Cu^2+(aq)} \mid \ce{Cu(s)}

E^\ominus_\text{cell} = +0.34 - (-0.76) = +1.10\ \text{V}

With a gas, platinum goes at the outer end, then the gas, then the ions next to the salt bridge:

\ce{Pt(s)} \mid \ce{H2(g)} \mid \ce{2H+(aq)} \parallel \ce{Ag+(aq)} \mid \ce{Ag(s)}

E^\ominus_\text{cell} = +0.80 - 0.00 = +0.80\ \text{V}

When both couples have no metal, each side needs its own platinum. The oxidised form (I₂, then Fe³⁺) is next to the salt bridge on each side:

\footnotesize \ce{Pt(s)} \mid \ce{2I-(aq)},\ \ce{I2(aq)} \parallel \ce{Fe^3+(aq)},\ \ce{Fe^2+(aq)} \mid \ce{Pt(s)}

E^\ominus_\text{cell} = +0.77 - (+0.54) = +0.23\ \text{V}

Worked exampleWriting a cell diagram from data

Write the cell diagram and calculate $E^\ominus_\text{cell}$ for a cell made from Fe²⁺/Fe (−0.44 V) and Ag⁺/Ag (+0.80 V).

Iron is more negative, so it goes on the left as the negative electrode.
Both couples have a metal, so no platinum is needed.

\ce{Fe(s)} \mid \ce{Fe^2+(aq)} \parallel \ce{Ag+(aq)} \mid \ce{Ag(s)}

$E^\ominus_\text{cell}$ = +0.80 − (−0.44) = +1.24 V

Check yourself

Calculate $E^\ominus_\text{cell}$ for a cell made from Mg²⁺/Mg (−2.37 V) and Cu²⁺/Cu (+0.34 V). State which electrode is negative.
Write the overall equation for the cell in question 1, with state symbols.
Write the cell diagram for question 1.
A student calculates $E^\ominus_\text{cell}$ for Zn/Ag as +0.80 + (−0.76) = +0.04 V. What mistake did they make, and what is the correct value?
Write the cell diagram for a cell made from Cu²⁺/Cu (+0.34 V) and Fe³⁺/Fe²⁺ (+0.77 V). Give $E^\ominus_\text{cell}$ .
A cell diagram, when evaluated as right minus left, gives −0.46 V. What does this tell you?

Answers cover these until you have tried the questions

$E^\ominus_\text{cell}$ = +0.34 − (−2.37) = +2.71 V. Magnesium is more negative, so magnesium is the negative electrode.
$\ce{Mg(s) + Cu^2+(aq) -> Mg^2+(aq) + Cu(s)}$
Magnesium goes on the left:
$\small \ce{Mg(s)} \mid \ce{Mg^2+(aq)} \parallel \ce{Cu^2+(aq)} \mid \ce{Cu(s)}$
They added instead of subtracting. Correct: $E^\ominus_\text{cell}$ = +0.80 − (−0.76) = +1.56 V.
Copper is more negative (+0.34 against +0.77), so it is on the left. Fe³⁺/Fe²⁺ has no metal, so it needs platinum:
$\begin{aligned} &\ce{Cu(s)} \mid \ce{Cu^2+(aq)} \\ &\parallel \ce{Fe^3+(aq)},\ \ce{Fe^2+(aq)} \mid \ce{Pt(s)} \end{aligned}$
$E^\ominus_\text{cell}$ = +0.77 − (+0.34) = +0.43 V.
The diagram is written the wrong way round. The reaction that really happens is the reverse, so the left-hand electrode is actually the positive one, and the true $E^\ominus_\text{cell}$ is +0.46 V.