ChemistryAdvanced22 min read

Electrochemistry

Turning redox into voltage — cells, electrodes and the electrochemical series

This topic appears in:

1st Year Chemistry · ch.10A Level Chemistry (9701) · ch.6

Practise the full question bank:

1st Year Chemistry — MCQs & exam QsA Level Chemistry (9701) — MCQs & exam Qs
01

Redox and oxidation numbers

Electrochemistry is redox put to work. A quick tool first: the oxidation number tracks electrons formally. It rises when a species is oxidised (loses electrons) and falls when reduced (gains them), which lets you spot redox even when no obvious electron transfer is visible. Uncombined elements are 0; simple ions equal their charge; oxygen is usually −2 and hydrogen +1.

02

Electrochemical (galvanic) cells

If a redox reaction is spontaneous, you can harness the electron flow as electricity — that is a galvanic (voltaic) cell, i.e. a battery. Separate the two half-reactions into two half-cells joined by a wire (for electrons) and a salt bridge (to complete the circuit and keep charge balanced).

Oxidation happens at the anode (electrons leave here — it is the negative terminal in a galvanic cell) and reduction at the cathode (positive terminal). Electrons flow through the external wire from anode to cathode. Note this is the reverse polarity of electrolysis, because here the reaction drives the current rather than the current driving the reaction.

Galvanic vs electrolytic — mind the polarity

Oxidation is always at the anode and reduction at the cathode in BOTH. But in a galvanic cell the anode is negative (the reaction pushes electrons out); in electrolysis the cathode is negative (the power supply pushes electrons in). Same electrode names, opposite signs.

03

Electrode potentials and the electrochemical series

Each half-cell has a tendency to be reduced, measured as its standard electrode potential (E°) against a hydrogen reference electrode (defined as 0 V). Listing these values gives the electrochemical series — essentially a quantitative reactivity series.

The cell voltage is the difference: E°cell = E°(cathode) − E°(anode). A positive E°cell means the reaction is spontaneous — the more positive, the more the reaction "wants" to go and the more voltage the cell delivers.

E°cell = E°(cathode) − E°(anode)positive E°cell ⇒ spontaneous reaction
04

A worked example

Worked example

A cell uses Zn²⁺/Zn (E° = −0.76 V) and Cu²⁺/Cu (E° = +0.34 V). Which is the anode, and what is the cell voltage?

  1. The more negative electrode is oxidised (anode): zinc, at −0.76 V.The species with the lower (more negative) E° gives up electrons more readily.
  2. Copper (+0.34 V) is reduced at the cathode.The more positive electrode is reduced.
  3. E°cell = E°(cathode) − E°(anode) = 0.34 − (−0.76) = 1.10 V.The difference of the two electrode potentials.
  4. E°cell is positive, so the reaction is spontaneous — this is the classic Daniell cell.Positive cell potential means the redox reaction proceeds on its own.

Zinc is the anode; the cell delivers +1.10 V, and the positive value confirms a spontaneous reaction.

05

Galvanic and electrolytic cells

Both kinds of cell involve redox at two electrodes, but they run in opposite directions. A galvanic (voltaic) cell uses a spontaneous reaction to produce electricity — as in a battery — with the more reactive metal as the negative electrode. An electrolytic cell does the reverse: it uses an external supply to force a non-spontaneous reaction, as in electrolysis or electroplating.

In both, oxidation happens at the anode and reduction at the cathode — but the electrode signs flip: the anode is negative in a galvanic cell and positive in an electrolytic one.

06

Using standard electrode potentials

Each half-cell has a standard electrode potential, E°, measured against the standard hydrogen electrode. A more positive E° means the species is more easily reduced (a better oxidising agent); a more negative E° means it is more easily oxidised. The cell voltage is the difference: E°cell = E°(cathode) − E°(anode), and a positive value means the reaction is feasible.

This lets you predict which metal displaces another, which way electrons flow, and whether a proposed redox reaction will actually happen.

E°cell = E°(cathode) − E°(anode)positive E°cell = spontaneous; the more positive electrode is reduced

Practice questions

6 questions · 21 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

4 · 9 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
Define oxidation and reduction in terms of oxidation number.
Model answer

Oxidation is an increase in oxidation number; reduction is a decrease in oxidation number.

Examiner tip. Track oxidation numbers to spot redox even when no electrons are shown explicitly.

SQ2[2 marks]
State what the standard electrode (reduction) potential of a half-cell measures.
Model answer

The voltage of that half-cell measured against a standard hydrogen electrode under standard conditions (298 K, 1 mol dm⁻³, 1 atm).

Examiner tip. More positive E° means the species is more easily reduced.

SQ3[3 marks]
Given E°(Cu²⁺/Cu) = +0.34 V and E°(Zn²⁺/Zn) = −0.76 V, calculate the EMF of the cell and state which electrode is the negative terminal.
Model answer

EMF = E°(cathode) − E°(anode) = +0.34 − (−0.76) = +1.10 V. The zinc electrode is the negative terminal (it is oxidised).

Examiner tip. The more negative electrode is the anode (negative terminal); EMF = E°cathode − E°anode.

SQ4[2 marks]
Using E°(Ag⁺/Ag) = +0.80 V and E°(Cu²⁺/Cu) = +0.34 V, determine which metal is oxidised in a cell made from these half-cells, and calculate the EMF.
Model answer

Silver has the more positive E°, so it is reduced (cathode) and copper is oxidised (anode). E°cell = 0.80 − 0.34 = +0.46 V.

Examiner tip. The more positive electrode is reduced; EMF = E°(cathode) − E°(anode).

Long questions

2 · 12 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[6 marks]
A simple cell is made from a zinc half-cell and a copper half-cell.
  1. Write the half-equation occurring at the zinc electrode and state whether it is oxidation or reduction. [2]
  2. State the function of the salt bridge. [2]
  3. State the direction of electron flow in the external circuit and explain why. [2]
Mark scheme
  1. Zn → Zn²⁺ + 2e⁻[1]
  2. This is oxidation (loss of electrons)[1]
  3. The salt bridge completes the circuit[1]
  4. and allows ions to flow to balance the charge in each half-cell[1]
  5. Electrons flow from zinc to copper[1]
  6. because zinc is more easily oxidised / has the more negative electrode potential[1]

Zn → Zn²⁺ + 2e⁻ (oxidation); salt bridge completes circuit and balances charge; electrons flow zinc → copper (zinc more easily oxidised).

Examiner tip. Electrons leave the more reactive (more negative) electrode through the wire.

LQ2[6 marks]
A cell is set up from a zinc half-cell and a copper half-cell. E°(Zn²⁺/Zn) = −0.76 V, E°(Cu²⁺/Cu) = +0.34 V.
  1. State which electrode is negative and write the half-equation there. [2]
  2. Calculate the EMF of the cell. [2]
  3. State the direction of electron flow and explain why. [2]
Mark scheme
  1. The zinc electrode is negative[1]
  2. Zn → Zn²⁺ + 2e⁻ (oxidation)[1]
  3. E°cell = E°(cathode) − E°(anode) = 0.34 − (−0.76)[1]
  4. = +1.10 V[1]
  5. Electrons flow from zinc to copper in the external circuit[1]
  6. because zinc has the more negative electrode potential, so it is oxidised and releases electrons[1]

Zinc electrode negative (Zn → Zn²⁺ + 2e⁻); EMF = +1.10 V; electrons flow zinc → copper (zinc more easily oxidised).

Examiner tip. The more negative electrode is the anode and the electron source.