Redox Reactions · Lesson 2 of 8

Electron Transfer and Competition for Electrons

“Explain redox changes through electron transfer and use metal–ion reactions to compare reducing tendencies.”

Learning Objectives

• Describe oxidation as electron loss and reduction as electron gain. • Identify an oxidising agent as an electron acceptor and a reducing agent as an electron donor. • Write and combine simple oxidation and reduction half reactions. • Interpret the observations in zinc–copper, copper–silver, and cobalt–nickel systems. • Infer relative electron-releasing tendencies from displacement observations.

Electrons unite the earlier definitions

The classical descriptions follow the elements that are added or removed. Electron transfer gives us a deeper common explanation. When sodium forms an ionic compound with chlorine, it becomes a positive ion while chlorine becomes a negative ion. The charges show that sodium has lost electrons and chlorine has gained them.

The same reasoning applies to sodium oxide and sodium sulphide: these can be represented as combinations of $N a^{+}$ with $O^{2 -}$ or $S^{2 -}$ . The formation of an ion records a change in electron number. A positive sodium ion has one fewer electron than a neutral sodium atom; it has not gained a proton.

The corresponding balanced changes are $4Na + O_{2} \to 2N a_{2} O$ and $2Na + S \to N a_{2} S$ . In each, sodium is oxidised to +1 ions, while the non-metal is reduced to negative ions. The number of sodium atoms is set by the electron requirement of the non-metal partner.

Definition

Oxidation: electron description

Oxidation is the loss of one or more electrons by a species.

Definition

Reduction: electron description

Reduction is the gain of one or more electrons by a species.

Sodium supplies electronsLaTeX

2 Na \to 2 N a^{+} + 2 e^{-}

Each sodium atom loses one electron. The total electron loss is two electrons for two atoms.

Chlorine accepts the electronsLaTeX

C l_{2} + 2 e^{-} \to 2 C l^{-}

A chlorine molecule contains two chlorine atoms. Each accepts one electron.

Each equation is a half reaction: it isolates the oxidation or reduction part of the overall process. Electrons appear explicitly so that we can track their number. In the complete reaction the electrons lost and gained are equal, so no electron remains unaccounted for.

Example — Reconstructing the overall change

Problem
Combine the two half reactions for sodium and chlorine.

1.Write the oxidation half reaction: $2Na \to 2N a^{+} + 2 e^{-}$ .
2.Write the reduction half reaction: $C l_{2} + 2 e^{-} \to 2C l^{-}$ .
3.Add them and cancel the two electrons because they appear on opposite sides. The ions form solid sodium chloride.
4.The overall equation is $2Na + C l_{2} \to 2NaCl$ . The half reactions are an electron-accounting description; free gaseous ions do not need to accumulate as a separate experimental stage.

The agent causes the opposite change in its partner

The name of an agent describes what it does to another species. An electron donor allows another species to gain electrons and be reduced, so the donor is a reducing agent. An electron acceptor allows another species to lose electrons and be oxidised, so the acceptor is an oxidising agent. Consequently, the reducing agent is itself oxidised, and the oxidising agent is itself reduced.

Definition

Reducing agent

A species that donates electrons to another species and is itself oxidised. It is also called a reductant.

Definition

Oxidising agent

A species that accepts electrons from another species and is itself reduced. It is also called an oxidant.

A matched transfer of electrons— Follow the electrons from the reducing agent to the oxidising agent. The agent names describe what each does to its partner.

Example — Hydrogen becomes hydride

Problem
Use half reactions to justify $2Na + H_{2} \to 2NaH$ .

1.Sodium loses electrons: $2Na \to 2N a^{+} + 2 e^{-}$ .
2.Hydrogen gains them: $H_{2} + 2 e^{-} \to 2 H^{-}$ . The negative hydrogen ion is called hydride.
3.Sodium is the reducing agent and hydrogen is the oxidising agent. This agrees with the electronegativity-based classical interpretation.
4.The example also shows why hydrogen need not always have a positive state in a compound.

Watch zinc compete with copper ions

Imagine placing a clean zinc strip in blue copper nitrate solution and observing it over time. A reddish coating of copper develops on the strip while the blue colour becomes weaker. The visible changes connect the metal being produced with the ions being consumed. The net ionic equation omits nitrate because it does not undergo the redox change.

Zinc reduces copper ionsLaTeX

Zn (s) + C u^{2 +} (a q) \to Z n^{2 +} (a q) + Cu (s)

The symbol (aq) means dissolved in water. Zinc donates two electrons, and one copper(II) ion accepts them.

Reading the zinc–copper observation— The fading blue colour records removal of copper ions; the reddish deposit records formation of copper metal.

The half reactions are $Zn \to Z n^{2 +} + 2 e^{-}$ and $C u^{2 +} + 2 e^{-} \to Cu$ . Thus zinc is oxidised and is the reductant; copper ions are reduced and are the oxidant. The textbook supports the ion identification with laboratory observations: zinc ions can give white zinc sulphide in alkaline conditions, while copper ions give black copper sulphide. These are descriptions of supervised laboratory tests, not instructions to generate hydrogen sulphide yourself.

What if we reverse the materials and put copper metal in zinc sulphate solution? No visible displacement is observed under the stated conditions, and a sensitive test does not detect copper ions. The forward zinc–copper reaction therefore greatly favours zinc ions and copper metal. “No visible reverse reaction” is evidence of a strongly one-sided equilibrium under those conditions, rather than a reason to discard the idea of equilibrium altogether.

Compare copper–silver with cobalt–nickel

Copper metal placed in silver nitrate solution develops a silver deposit while the solution turns blue. Here copper can supply electrons to silver ions even though copper ions accepted electrons from zinc in the previous experiment. Reducing and oxidising behaviour must always be interpreted relative to the reaction partner.

Copper reduces silver ionsLaTeX

Cu (s) + 2 A g^{+} (a q) \to C u^{2 +} (a q) + 2 Ag (s)

Copper loses two electrons; two silver ions each gain one. Formation of copper(II) ions accounts for the blue colour.

Copper in silver nitrate— Compare this with zinc in copper nitrate: here the blue colour grows as copper ions form, while silver deposits on the copper strip.

Zinc supplies electrons to copper ions, and copper supplies electrons to silver ions. These observations give the order of electron-releasing tendency $Zn > Cu > Ag$ for these comparisons. An activity series expands this kind of comparison to more metals. The chemical tendency to release electrons will later help us understand how a cell generates electrical energy.

The cobalt–nickel system shows a useful contrast: $Co + N i^{2 +} ⇌ C o^{2 +} + Ni$ . At equilibrium, both metal ions occur at moderate concentrations. Neither side is overwhelmingly favoured. A redox reaction may therefore be much less one-sided than zinc with copper ions or copper with silver ions.

Example — What can the observations establish?

Problem
A metal M deposits copper from copper(II) solution but does not deposit zinc from zinc(II) solution. What can be inferred?

1.Depositing copper means that M can donate electrons to copper ions under the test conditions. M therefore has a greater electron-releasing tendency than copper in that comparison.
2.Failure to deposit zinc is consistent with M being a weaker electron donor than zinc under those conditions.
3.Taken together, the observations place M between zinc and copper in a simple activity comparison. They do not, by themselves, provide a numerical electrode potential.

Common mistake

An oxidising agent does not lose electrons to “supply oxidation”. It accepts electrons and is reduced. Also, a metal and its ion are different redox species: copper metal can donate electrons to silver ions, whereas copper ions accept electrons from zinc.

Quick check

Which half reaction represents oxidation?

Quick check

A species accepts electrons. What is its role and what happens to it?

Quick check

Why does the blue colour fade when zinc reacts with copper nitrate solution?

Quick check

How many silver ions accept the electrons released when one copper atom becomes $C u^{2 +}$ ?

Quick check

What does the cobalt–nickel observation illustrate?

Practice Problems

Write the two half reactions for $Zn + C u^{2 +} \to Z n^{2 +} + Cu$ and identify the agents.
Explain why silver deposits and a blue colour develops when copper enters silver nitrate solution.
Use $Zn > Cu > Ag$ to predict whether silver metal should readily displace copper from copper(II) solution under comparable conditions.
Explain the difference between “copper is reduced” and “copper(II) ions are the oxidising agent”.
For $2Na + H_{2} \to 2NaH$ , check the atom count, charge balance, and electron count in both half reactions.
Compare the evidence for the equilibrium positions of the zinc–copper and cobalt–nickel reactions.

Key Takeaways

Oxidation releases electrons; reduction consumes them. A reducing agent donates electrons and is oxidised; an oxidising agent accepts them and is reduced. The electron losses and gains cancel when half reactions are combined. Metal deposits and solution colours connect microscopic electron changes with observations. Displacement experiments reveal relative electron-releasing tendencies and show that redox equilibria need not all be equally one-sided.

Previous · Lesson 1

Oxidation and Reduction Always Occur Together

Next · Lesson 3

Oxidation Numbers: Rules, Meaning, and Limits

Electrons unite the earlier definitions

The agent causes the opposite change in its partner

Watch zinc compete with copper ions

Compare copper–silver with cobalt–nickel

Quiz

Practice Problems

Key Takeaways