Predict half-reactions at each electrochemical electrode

Alberta Grade 12 Chemistry

Electrochemical Changes

Alberta Chemistry 30 — Outcome 30-B2.3k

Place a strip of zinc metal into a blue solution of copper(II) sulfate. Within minutes, the blue colour fades and a reddish-brown solid coats the zinc. Electrons moved — spontaneously — from zinc atoms to copper ions. That transfer did not happen in one single step. It happened at two distinct locations, each with its own chemical change. In electrochemistry, each of those locations is called an electrode, and the chemical change that occurs there is called a half-reaction. This lesson teaches you how to predict exactly which half-reaction occurs at each electrode, whether the cell runs on its own (galvanic) or is driven by an external power supply (electrolytic). Every prediction follows the same logical steps: identify which species gains electrons and which loses them, locate their reduction potentials in the Data Booklet, assign anode and cathode, and write balanced half-reactions with correct states and charges.

What you will learn

Identify which electrode is the anode and which is the cathode in both galvanic and electrolytic cells.
Use the standard reduction potential table from the Alberta Chemistry 30 Data Booklet to predict the half-reaction at each electrode.
Write balanced half-reactions that conserve both atoms and charge, including correct state symbols.
Combine electrode half-reactions to confirm the overall cell reaction makes chemical sense.

Prerequisite Review: Oxidation, Reduction, and Reduction Potentials

Before predicting half-reactions, you need three ideas from earlier in Chemistry 30. First, oxidation means losing electrons and reduction means gaining electrons. The memory aid OIL RIG (Oxidation Is Loss, Reduction Is Gain) captures this. A species that loses electrons is oxidized; the electrode where oxidation occurs is the anode. A species that gains electrons is reduced; the electrode where reduction occurs is the cathode.

Second, every redox half-reaction can be written as a reduction: a species gains electrons. The Alberta Chemistry 30 Data Booklet lists these as standard reduction half-reactions, each paired with a standard reduction potential, E°, measured in volts (V). A large positive E° means the species strongly attracts electrons — it is easily reduced. A large negative E° means the species resists gaining electrons — it is more easily oxidized.

Third, to reverse a half-reaction (turn it from a reduction into an oxidation), you simply flip it and change the sign of E°. This is the oxidation potential. You do not change the magnitude, only the sign. These three ideas are all you need before moving forward.

E°_{\text{ox}} = -E°_{\text{red}}

Oxidation = loss of electrons; occurs at the anode.
Reduction = gain of electrons; occurs at the cathode.
The Data Booklet lists half-reactions as reductions with standard reduction potentials E° in volts.
Reversing a half-reaction flips its sign: $E°_{\text{ox}} = -E°_{\text{red}}$ .

What Is a Half-Reaction and Why Does Each Electrode Have One?

In any electrochemical cell, two separate reactions happen at two separate electrodes. Neither reaction is complete on its own — one produces electrons and the other consumes them. We call each of these partial reactions a half-reaction because each is only half of the full redox process.

At the anode, a species gives up electrons. Those electrons travel through an external wire. At the cathode, a different species picks up those arriving electrons. The two electrodes are physically separated, so the oxidation and reduction half-reactions cannot mix — they each happen at their own electrode surface. This separation is exactly what makes electrochemical cells useful: the electron flow through the wire can do electrical work.

A half-reaction must be balanced for both atoms and charge. For example, if a copper(II) ion gains two electrons and becomes a copper atom, the half-reaction is written as:

\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)

. The two positive charges on the left are exactly cancelled by the two negative charges from the electrons, giving zero net charge on the right — charge is conserved.

\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)

A half-reaction represents either oxidation or reduction alone.
The anode half-reaction is always an oxidation (electrons are a product).
The cathode half-reaction is always a reduction (electrons are a reactant).
Every half-reaction must balance atoms first, then balance charge using electrons.

Predicting Electrode Half-Reactions in a Galvanic Cell

A galvanic cell converts chemical energy into electrical energy spontaneously — no external power is needed. The classic example is the zinc-copper cell. Zinc is dipped into a zinc sulfate solution (one half-cell) and copper is dipped into a copper sulfate solution (the other half-cell). A salt bridge connects the two solutions.

To predict which half-reaction occurs at each electrode, look up both species in the Data Booklet reduction table. Zinc:

\text{Zn}^{2+}(aq) + 2e^- \rightarrow \text{Zn}(s)

,

E° = -0.76\text{ V}

. Copper:

\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)

,

E° = +0.34\text{ V}

. The species with the more positive (or less negative) E° is more easily reduced, so it goes to the cathode. Copper has the higher E°, so copper(II) ions are reduced at the cathode. Zinc has the lower E°, so zinc metal is oxidized at the anode.

Write the cathode half-reaction as a reduction exactly as it appears in the Data Booklet:

\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)

. Write the anode half-reaction by reversing the zinc reduction entry:

\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^-

. Check that each half-reaction is balanced for atoms and for charge before proceeding. In this case, both have equal numbers of zinc or copper atoms on each side, and the charge balances on both sides of each equation.

E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}}

In a galvanic cell, the species with the higher E° is reduced at the cathode.
The species with the lower E° is oxidized at the anode.
The cathode half-reaction is copied from the Data Booklet as written (reduction).
The anode half-reaction is the Data Booklet entry written in reverse (oxidation).
Both half-reactions must conserve atoms and net charge.

Predicting Electrode Half-Reactions in an Electrolytic Cell

An electrolytic cell uses an external power supply to force a non-spontaneous redox reaction. Common examples include electroplating metals and the electrolysis of water. The external supply pushes electrons in a chosen direction, so you cannot rely on spontaneity to tell you which reaction occurs where. Instead, the cell is designed so that a specific reaction is forced at each electrode.

The key rule remains the same: oxidation always occurs at the anode and reduction always occurs at the cathode — this is true for every electrochemical cell, galvanic or electrolytic. What changes is how we decide which species reacts. In an electrolytic cell, the species present in solution (or in the melt) are the candidates, and the applied voltage determines whether the reaction is feasible. For predicting half-reactions at this course level, identify the species at each electrode and write the appropriate oxidation (anode) or reduction (cathode) half-reaction, using the Data Booklet for guidance.

Consider the electrolysis of molten sodium chloride, NaCl(l). At the cathode (reduction): sodium ions gain electrons to form sodium metal,

\text{Na}^+(l) + e^- \rightarrow \text{Na}(l)

. At the anode (oxidation): chloride ions lose electrons to form chlorine gas,

2\text{Cl}^-(l) \rightarrow \text{Cl}_2(g) + 2e^-

. Notice that the anode half-reaction requires two chloride ions to produce one molecule of diatomic chlorine — atoms and charge must both balance.

2\text{Cl}^-(l) \rightarrow \text{Cl}_2(g) + 2e^-

In an electrolytic cell, an external power supply forces the redox reaction.
Oxidation at the anode and reduction at the cathode — this rule never changes.
Identify which species is present at each electrode, then write its half-reaction.
Multiply half-reactions by whole numbers to balance electrons before combining, but do not change E° values when multiplying.

Balancing and Combining Half-Reactions to Confirm the Overall Reaction

Once you have the two half-reactions, you can combine them to check the overall cell reaction. The number of electrons released at the anode must exactly equal the number consumed at the cathode. If they do not already match, multiply one or both half-reactions by the smallest whole number that makes the electrons equal. Multiplying a half-reaction by a number is allowed because it just scales the amount of substance — it does not change the identity of the reaction or the value of E°.

After scaling, add the two half-reactions together. Cancel the electrons (they appear on opposite sides) and cancel any other species that appear identically on both sides. The result is the balanced overall ionic equation. If atoms and charges balance in the overall equation, your half-reactions were written correctly.

For example, combining the zinc-copper half-reactions: the anode gives

\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^-

and the cathode gives

\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)

. Each already involves 2 electrons, so no scaling is needed. Adding them:

\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)

. The

2e^-

cancel. Check: one Zn on each side ✓, one Cu on each side ✓, net charge is

+2

on both sides ✓. The half-reactions were correct.

\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)

Scale half-reactions so that electrons lost at the anode equal electrons gained at the cathode.
Add the two scaled half-reactions and cancel electrons to get the overall equation.
Check the overall equation for atom balance and charge balance.
Scaling a half-reaction does not change its E° value.

Comparing Galvanic and Electrolytic Cells: Electrode Half-Reactions

Feature	Galvanic Cell	Electrolytic Cell
Energy source	Spontaneous chemical reaction	External power supply (forced)
Anode process	Oxidation of the species with lower E°	Oxidation of the species present at the anode
Cathode process	Reduction of the species with higher E°	Reduction of the species present at the cathode
Electron flow (external)	From anode to cathode through wire	From power supply into cathode; from anode back to supply
Half-reaction prediction tool	Compare E° values from Data Booklet	Identify species at each electrode; use Data Booklet entries

Worked example

Predicting Half-Reactions for a Silver-Nickel Galvanic Cell

A galvanic cell is constructed using a nickel electrode in a 1.00 mol/L nickel(II) sulfate solution and a silver electrode in a 1.00 mol/L silver nitrate solution. Using the Alberta Chemistry 30 Data Booklet, predict the half-reaction at each electrode, identify the anode and cathode, and write the balanced overall ionic equation.

Step 1: Look up standard reduction potentials
Find both half-reactions in the Data Booklet reduction table and record their standard reduction potentials.
$\text{Ni}^{2+}(aq) + 2e^- \rightarrow \text{Ni}(s), \quad E° = -0.26\text{ V}$
Step 1 (continued): Silver reduction potential
Record the silver entry from the same table.
$\text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s), \quad E° = +0.80\text{ V}$
Step 2: Identify cathode and anode
The species with the more positive reduction potential is more easily reduced — it goes to the cathode. Silver has $E° = +0.80$ V, which is greater than $-0.26$ V for nickel, so silver ions are reduced at the cathode. Nickel metal is oxidized at the anode.
$+0.80\text{ V} > -0.26\text{ V} \Rightarrow \text{Ag}^+ \text{ reduced at cathode; Ni oxidized at anode}$
Step 3: Write the cathode half-reaction (reduction)
Copy the silver half-reaction from the Data Booklet exactly as written. This is a reduction: silver ions gain one electron each to form silver metal.
$\text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s)$
Step 4: Write the anode half-reaction (oxidation)
Reverse the nickel Data Booklet entry to convert it from a reduction into an oxidation. Nickel metal loses two electrons to form nickel(II) ions.
$\text{Ni}(s) \rightarrow \text{Ni}^{2+}(aq) + 2e^-$
Step 5: Balance electrons between the two half-reactions
The cathode half-reaction involves 1 electron; the anode involves 2 electrons. Multiply the cathode half-reaction by 2 so that both half-reactions transfer exactly 2 electrons. The E° value is not multiplied — only the stoichiometric coefficients change.
$2\text{Ag}^+(aq) + 2e^- \rightarrow 2\text{Ag}(s)$
Step 6: Add the half-reactions to get the overall equation
Add the scaled cathode half-reaction to the anode half-reaction. The $2e^-$ appear on opposite sides, so they cancel exactly.
$\text{Ni}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Ni}^{2+}(aq) + 2\text{Ag}(s)$
Step 7: Verify atom and charge balance
Check atoms: 1 Ni on each side, 2 Ag on each side — atoms balance. Check charge: left side has $2 \times (+1) = +2$ ; right side has $+2$ from $\text{Ni}^{2+}$ — charge balances. The half-reactions are confirmed correct.
$\text{Left charge: } 2(+1) = +2 \qquad \text{Right charge: } +2 \checkmark$

Answer: Cathode (reduction):

2\text{Ag}^+(aq) + 2e^- \rightarrow 2\text{Ag}(s)

. Anode (oxidation):

\text{Ni}(s) \rightarrow \text{Ni}^{2+}(aq) + 2e^-

. Overall:

\text{Ni}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Ni}^{2+}(aq) + 2\text{Ag}(s)

.

Check: Atom count: 1 Ni = 1 Ni and 2 Ag = 2 Ag. Net charge left:

+2

. Net charge right:

+2

. Electrons produced at anode (2) equal electrons consumed at cathode (2). All checks pass.

Common mistakes and how to avoid them

Writing the anode half-reaction as a reduction instead of reversing it to show oxidation.

Correction: The anode is always where oxidation occurs. Take the Data Booklet reduction entry for that species and reverse the arrow. Electrons must appear as a product on the right side for an oxidation half-reaction.

Multiplying the E° value when scaling a half-reaction to balance electrons.

Correction: E° is an intensive property — it does not change when you scale the equation. Only the stoichiometric coefficients (amounts of each species) are multiplied. The E° value stays the same.

Choosing the cathode based on the species with the lower E° instead of the higher E°.

Correction: A higher (more positive) E° means the species is more easily reduced. The more easily reduced species is always reduced at the cathode. The species with the lower E° is harder to reduce — it is more easily oxidized and goes to the anode.

Forgetting to check charge balance after writing each half-reaction.

Correction: Always count the total charge on each side of the half-reaction. The net charge on the left must equal the net charge on the right. If they differ, the number of electrons in the half-reaction is wrong.

Confusing the anode and cathode labels between galvanic and electrolytic cells, thinking they switch.

Correction: The definitions never change: oxidation always occurs at the anode and reduction always occurs at the cathode, in every type of electrochemical cell. What may differ between cell types is which electrode is physically connected to the positive terminal, but the oxidation/reduction assignment stays fixed.

Lesson summary

A half-reaction shows either oxidation (electrons as a product) or reduction (electrons as a reactant) occurring at a single electrode.
Oxidation always occurs at the anode; reduction always occurs at the cathode — this is true for both galvanic and electrolytic cells.
In a galvanic cell, compare standard reduction potentials from the Data Booklet: the higher E° species is reduced at the cathode; the lower E° species is oxidized at the anode.
To write the anode half-reaction, reverse the Data Booklet reduction entry for that species; the sign of E° also reverses but the magnitude stays the same.
Scale half-reactions using the smallest whole-number multipliers so that electrons lost at the anode equal electrons gained at the cathode; E° values are never multiplied.
Confirm correctness by adding the two half-reactions: electrons must cancel and the overall equation must balance for both atoms and net charge.

Check your understanding

Question 1

A galvanic cell contains a magnesium electrode (

E° = -2.37\text{ V}

) and a copper electrode (

E° = +0.34\text{ V}

). Which half-reaction correctly describes the process at the anode?

$\text{Mg}^{2+}(aq) + 2e^- \rightarrow \text{Mg}(s)$
$\text{Mg}(s) \rightarrow \text{Mg}^{2+}(aq) + 2e^-$
$\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)$
$\text{Cu}(s) \rightarrow \text{Cu}^{2+}(aq) + 2e^-$

Show answer and explanation

\text{Mg}(s) \rightarrow \text{Mg}^{2+}(aq) + 2e^-

Magnesium has the lower reduction potential (

-2.37

V compared to

+0.34

V), so it is more easily oxidized. Oxidation occurs at the anode. The anode half-reaction is the reverse of the Data Booklet reduction entry:

\text{Mg}(s) \rightarrow \text{Mg}^{2+}(aq) + 2e^-

. Option A is the reduction form (wrong direction). Options C and D involve copper, which is reduced at the cathode, not the anode.

Question 2

In the electrolysis of molten magnesium chloride,

\text{MgCl}_2(l)

, what is the correct half-reaction at the cathode?

$\text{Mg}(l) \rightarrow \text{Mg}^{2+}(l) + 2e^-$
$2\text{Cl}^-(l) \rightarrow \text{Cl}_2(g) + 2e^-$
$\text{Mg}^{2+}(l) + 2e^- \rightarrow \text{Mg}(l)$
$\text{Cl}_2(g) + 2e^- \rightarrow 2\text{Cl}^-(l)$

Show answer and explanation

\text{Mg}^{2+}(l) + 2e^- \rightarrow \text{Mg}(l)

The cathode is always where reduction occurs — a species gains electrons. In molten

\text{MgCl}_2

, the magnesium(II) ions are reduced at the cathode:

\text{Mg}^{2+}(l) + 2e^- \rightarrow \text{Mg}(l)

. Option A shows oxidation of magnesium (that is the anode reaction in a different cell). Option B shows chloride oxidation, which happens at the anode here. Option D shows chlorine reduction, which is not the reaction occurring under these conditions.

Question 3

A student writes the following two half-reactions for a galvanic cell and wants to combine them into an overall equation. Anode:

\text{Al}(s) \rightarrow \text{Al}^{3+}(aq) + 3e^-

. Cathode:

\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)

. What is the correct overall balanced ionic equation?

$\text{Al}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Al}^{3+}(aq) + \text{Cu}(s)$
$2\text{Al}(s) + 3\text{Cu}^{2+}(aq) \rightarrow 2\text{Al}^{3+}(aq) + 3\text{Cu}(s)$
$3\text{Al}(s) + 2\text{Cu}^{2+}(aq) \rightarrow 3\text{Al}^{3+}(aq) + 2\text{Cu}(s)$
$\text{Al}(s) + 2\text{Cu}^{2+}(aq) \rightarrow \text{Al}^{3+}(aq) + 2\text{Cu}(s)$

Show answer and explanation

2\text{Al}(s) + 3\text{Cu}^{2+}(aq) \rightarrow 2\text{Al}^{3+}(aq) + 3\text{Cu}(s)

The anode releases 3 electrons per Al atom; the cathode consumes 2 electrons per Cu ion. To balance electrons, multiply the anode half-reaction by 2 (giving

6e^-

released) and the cathode half-reaction by 3 (giving

6e^-

consumed). This gives

2\text{Al}(s) \rightarrow 2\text{Al}^{3+}(aq) + 6e^-

and

3\text{Cu}^{2+}(aq) + 6e^- \rightarrow 3\text{Cu}(s)

. Adding and cancelling the

6e^-

:

2\text{Al}(s) + 3\text{Cu}^{2+}(aq) \rightarrow 2\text{Al}^{3+}(aq) + 3\text{Cu}(s)

. Check charge: left

= 3(+2) = +6

; right

= 2(+3) = +6

. Atoms balance. Option C has the Al and Cu coefficients swapped and does not balance charge or atoms.

Question 4

Which statement correctly describes what happens when you reverse a half-reaction from the Data Booklet?

Both the direction of the arrow and the magnitude of E° are reversed.
The direction of the arrow is reversed and the sign of E° changes, but the magnitude stays the same.
Only the direction of the arrow changes; E° remains exactly as listed.
The half-reaction cannot be reversed; a new E° must be looked up separately.

Show answer and explanation

The direction of the arrow is reversed and the sign of E° changes, but the magnitude stays the same.

Reversing a half-reaction converts a reduction into an oxidation. The arrow flips, and the sign of the potential changes (the oxidation potential equals the negative of the reduction potential), but the numerical magnitude is unchanged. For example, if a reduction has

E° = -0.76

V, the oxidation potential is

+0.76

V. Option A is wrong because the magnitude does not change. Option C is wrong because the sign does change. Option D is wrong because you never need a separate lookup — you simply reverse the sign.

Key terms

Half-reaction: A chemical equation that shows either oxidation or reduction alone at one electrode, with electrons written explicitly as a reactant (reduction) or product (oxidation).
Anode: The electrode where oxidation occurs in any electrochemical cell; the species at the anode loses electrons.
Cathode: The electrode where reduction occurs in any electrochemical cell; the species at the cathode gains electrons.
Standard reduction potential (E°): A value in volts (V) listed in the Data Booklet that measures how strongly a species attracts electrons under standard conditions. A more positive E° means the species is more easily reduced.
Galvanic cell: An electrochemical cell in which a spontaneous redox reaction produces electrical energy. No external power supply is needed.
Electrolytic cell: An electrochemical cell in which an external power supply forces a non-spontaneous redox reaction to occur.
Oxidation potential: The tendency of a species to lose electrons, equal to the negative of its standard reduction potential: $E°_{\text{ox}} = -E°_{\text{red}}$ .
Overall ionic equation: The balanced equation obtained by adding the anode and cathode half-reactions together after cancelling the electrons; it must balance for both atoms and net charge.

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Published by DoAssignment. This reviewed lesson follows Alberta Grade 12 Chemistry (CHEM30), expectation 30-B2.3k. It is a study resource, not an official curriculum publication.

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30-B2.3k · Predict half-reactions at each electrochemical electrode

Alberta Grade 12 Chemistry

Alberta Chemistry 30 — Outcome 30-B2.3k

What you will learn

Prerequisite Review: Oxidation, Reduction, and Reduction Potentials

What Is a Half-Reaction and Why Does Each Electrode Have One?

Predicting Electrode Half-Reactions in a Galvanic Cell

Predicting Electrode Half-Reactions in an Electrolytic Cell

Balancing and Combining Half-Reactions to Confirm the Overall Reaction

Comparing Galvanic and Electrolytic Cells: Electrode Half-Reactions

Predicting Half-Reactions for a Silver-Nickel Galvanic Cell

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Question 4

Key terms

Continue through CHEM30

About this lesson and its review