Why a Metal Is the Reductant When It Dissolves in Acid

A clean strip of zinc is dropped into dilute hydrochloric acid. Within seconds, bubbles appear, the zinc starts disappearing, and the solution begins to contain zinc ions.

The metal is dissolving, but something more important is happening: electrons are moving from the metal to the acid’s hydrogen ions.

The metal is the reductant because it donates electrons to hydrogen ions, causing them to be reduced to hydrogen gas. The metal itself loses electrons and is oxidised as it enters the solution as positive metal ions.

This can seem backwards at first. If the metal is oxidised, why do we call it the reducing agent?

The answer lies in what the word reducing describes. A reducing agent causes another substance to be reduced. It does not describe what happens to the reducing agent itself.

Let’s establish that using the reaction between zinc and dilute hydrochloric acid.

01What is actually happening when a metal dissolves?

Consider a piece of zinc placed into dilute hydrochloric acid, \(\ce{HCl}\).

The overall reaction is:

\[ \ce{Zn(s) + 2HCl(aq) -> ZnCl2(aq) + H2(g)} \]

Three things happen:

• The zinc metal gradually dissolves.
• Hydrogen gas forms as bubbles.
• Zinc ions enter the solution.

The obvious question is where the hydrogen comes from. Hydrochloric acid separates into hydrogen ions, \(\mathrm{H}^{+}\), and chloride ions, \(\mathrm{Cl}^{-}\), when dissolved in water.

The hydrogen ions are involved in forming hydrogen gas. The chloride ions remain in solution.

But how does a neutral zinc atom become a positively charged zinc ion?

It must lose electrons.

A neutral zinc atom has no overall electrical charge. When it becomes a \(\mathrm{Zn}^{2+}\) ion, it has lost two negatively charged electrons.

That electron loss is oxidation.

Before looking at the formal equations, consider how we could investigate this reaction experimentally.

02Investigating the reaction: where do the electrons go?

Imagine setting up an investigation using a clean zinc strip and dilute hydrochloric acid.

Start with a conical flask containing a measured volume of acid. Add a known mass of zinc, seal the flask with a bung, and connect a delivery tube to a gas syringe.

As the reaction proceeds, hydrogen gas travels through the delivery tube and pushes the gas syringe’s plunger outwards.

The syringe measures the volume of gas produced.

What should the experiment show?

Before starting, make a prediction.

If every zinc atom that dissolves becomes a \(\mathrm{Zn}^{2+}\) ion, how many electrons must each zinc atom lose?

Two.

If hydrogen ions are the only substances accepting those electrons, how many hydrogen molecules should form?

One molecule of hydrogen gas requires two hydrogen atoms, with each hydrogen atom gaining one electron.

Therefore:

• One zinc atom loses two electrons.
• Two hydrogen ions gain those two electrons.
• One hydrogen molecule forms.

The predicted mole ratio is:

\[ 1\text{ mol Zn}:1\text{ mol }\ce{H2} \]

The experimental observations should be consistent with this prediction:

• The zinc strip loses mass as zinc atoms enter the solution.
• Hydrogen gas accumulates in the gas syringe.
• An independent gas-identification test confirms that the gas is hydrogen.

A simple open test tube would show bubbles, but gas would escape before its volume could be measured. The gas syringe improves the design by collecting the gas.

There is an important limitation: a gas syringe measures volume, not gas identity. The hydrogen must be identified separately using an approved school laboratory method.

The experiment also does not directly show electrons moving. Electron transfer is inferred using the chemical changes, oxidation states, and conservation of charge.

For a quantitative investigation, you can weigh the zinc before and after the reaction. If the acid is limiting, some zinc remains and can be rinsed, dried, and reweighed to determine the mass consumed.

A useful prediction from the experiment

Suppose \(0.0100\) mol of zinc reacts completely with excess dilute hydrochloric acid.

The expected amount of hydrogen is:

\[ n(\ce{H2})=0.0100\text{ mol} \]

The amount of electrons transferred is:

\[ n(e^-)=2(0.0100)=0.0200\text{ mol} \]

These quantities are not separate coincidences. They follow directly from the electron transfer occurring during the reaction.

03Following the electrons: the two half-equations

We can now describe the reaction more precisely.

Instead of treating the overall equation as one large change, separate it into two processes.

1. Zinc loses electrons

Each zinc atom starts as neutral solid zinc.

It becomes a zinc ion by losing two electrons:

\[ \ce{Zn(s) -> Zn^2+(aq) + 2e^-} \]

Notice where the electrons appear.

They are on the right-hand side because zinc is releasing them.

This is the oxidation half-equation.

Zinc changes oxidation state from \(0\) to \(+2\).

2. Hydrogen ions gain electrons

The hydrogen ions in the acid accept the electrons released by zinc.

Two hydrogen ions are needed to form one hydrogen molecule:

\[ \ce{2H+(aq) + 2e^- -> H2(g)} \]

The electrons appear on the left-hand side because hydrogen ions are gaining them.

This is the reduction half-equation.

Each hydrogen atom changes oxidation state from \(+1\) to \(0\).

3. Combine the half-equations

Now add the two equations together.

\[ \ce{Zn -> Zn^2+ + 2e^-} \]

\[ \ce{2H+ + 2e^- -> H2} \]

The two electrons appear on opposite sides, so they cancel.

The resulting net ionic equation is:

\[ \boxed{\ce{Zn(s) + 2H+(aq) -> Zn^2+(aq) + H2(g)}} \]

Why does the equation conserve charge?

This is an important check.

On the left-hand side:

• Zinc has a charge of zero.
• Two hydrogen ions contribute a total charge of \(+2\).

Total charge = \(+2\).

On the right-hand side:

• One zinc ion has a charge of \(+2\).
• Hydrogen gas has no overall charge.

Total charge = \(+2\).

The charge is conserved.

The atoms are also conserved:

• One zinc atom appears on each side.
• Two hydrogen atoms appear on each side.

Nothing has been created or destroyed. The atoms have simply changed how they are arranged, and electrons have been transferred between reacting species.

Where did the chloride ions go?

They have not disappeared.

In the original molecular equation:

\[ \ce{Zn + 2HCl -> ZnCl2 + H2} \]

The chloride ions are present on both sides.

They remain as \(\mathrm{Cl}^{-}\) ions in the aqueous solution. They do not gain or lose electrons.

We call them spectator ions because they do not participate in the electron transfer.

This is why the net ionic equation contains hydrogen ions but not chloride ions.

The same reasoning applies to dilute sulfuric acid. The sulfate ions remain unchanged while hydrogen ions accept electrons.

04Why the metal is called the reductant

Now we can answer the original question precisely.

A reductant, also called a reducing agent, is a substance that causes another substance to be reduced by donating electrons.

In the zinc-acid reaction:

• Zinc donates electrons.
• Hydrogen ions accept electrons.
• Hydrogen ions are reduced.
• Zinc is oxidised.

Therefore, zinc is the reductant.

The apparent contradiction comes from confusing the change experienced by a substance with the effect it causes on another substance.

Think of the terminology this way:

The reducing agent is always the electron donor in a redox reaction.

The oxidising agent is always the electron acceptor.

This is the key distinction:

The reductant is oxidised, while the oxidant is reduced.

Why don’t the electrons simply float around in the acid?

You might picture zinc releasing electrons into the solution, where hydrogen ions collect them.

That is not quite what happens.

Electron transfer occurs at the metal-solution interface. Zinc atoms at the metal surface are oxidised, and the electrons are transferred to suitable electron acceptors at that interface.

The metal itself conducts electrons through its metallic structure.

There is no need for an external wire or battery in this direct chemical reaction.

The electrons also do not remain as a collection of free electrons in the solution. They are transferred to the reacting species.

05When does this explanation apply?

There is an important condition behind the simple zinc-acid model.

We are considering a metal reacting with a dilute, non-oxidising acid, such as hydrochloric acid or dilute sulfuric acid, under ordinary conditions.

The metal must be capable of reducing hydrogen ions under those conditions.

A metal above hydrogen in the reactivity series can generally displace hydrogen from a dilute non-oxidising acid.

However, not every metal dissolves in every acid.

The contrasting case: copper in dilute hydrochloric acid

Imagine placing a clean piece of copper into dilute hydrochloric acid.

Would you expect the same vigorous bubbling as zinc?

No.

Copper does not normally react with dilute hydrochloric acid at room temperature.

It is less reactive than hydrogen and cannot spontaneously donate electrons to hydrogen ions to produce hydrogen gas under these conditions.

This comparison exposes an important misconception.

Being a metal does not automatically make a substance a reductant in every reaction. A substance’s role depends on which other substances it reacts with.

Copper does not act as a reductant towards hydrogen ions in dilute hydrochloric acid.

However, copper can act as a reductant towards silver ions.

The reaction is:

\[ \ce{Cu(s) + 2Ag+(aq) -> Cu^2+(aq) + 2Ag(s)} \]

Copper loses two electrons and becomes \(\mathrm{Cu}^{2+}\).

Silver ions gain those electrons and become solid silver.

Copper is therefore the reductant in this reaction.

An important extension: not every acid uses hydrogen ions as the oxidant

The simple model above assumes a non-oxidising acid.

Nitric acid is different. It is an oxidising acid, meaning nitrate ions can accept electrons from a reacting metal.

A metal may therefore dissolve in nitric acid without producing hydrogen gas.

The metal can still be the reductant, but you must identify which species actually accepts the electrons.

Do not automatically assume that every metal-acid reaction follows the hydrogen-producing equation.

For more detail on which metals react with dilute acids and how their reaction rates compare, see Metal Reactivity with Dilute Acids: HSC Chemistry Guide.

06Worked examples

07Questions and solutions

The following questions build from identifying the reductant to balancing electron transfer, interpreting experimental evidence, and dealing with competing oxidising agents.

Question 1

A clean zinc strip is added to excess dilute hydrochloric acid at room temperature. Hydrogen gas is produced, and the zinc gradually dissolves.

Write:

a) The balanced net ionic equation.

b) The oxidation and reduction half-equations.

c) The identity of the reductant and oxidant.

Explain your choices using electron transfer.

Solution 1

Zinc is the reductant, and hydrogen ions are the oxidant. Zinc loses electrons, while hydrogen ions gain them.

The balanced net ionic equation is:

\[ \ce{Zn(s) + 2H+(aq) -> Zn^2+(aq) + H2(g)} \]

The oxidation half-equation is:

\[ \ce{Zn(s) -> Zn^2+(aq) + 2e^-} \]

Zinc loses two electrons and its oxidation state increases from \(0\) to \(+2\).

The reduction half-equation is:

\[ \ce{2H+(aq) + 2e^- -> H2(g)} \]

Hydrogen ions gain two electrons in total and form hydrogen gas.

Therefore:

• Zinc is oxidised and acts as the reductant.
• Hydrogen ions are reduced and act as the oxidant.

The two electrons cancel when the half-equations are combined, conserving charge in the overall reaction.

Chloride ions are spectators and do not participate in the electron transfer.

Question 2

A student reacts a clean iron strip with excess dilute hydrochloric acid.

The reaction produces \(48.0\text{ mL}\) of dry hydrogen gas at \(25^\circ\text{C}\) and \(100\text{ kPa}\).

Assume iron forms \(\mathrm{Fe}^{2+}\).

Calculate:

a) The amount of hydrogen gas produced.

b) The amount and mass of iron consumed.

c) The amount of electrons transferred.

Use \(V_m=24.8\text{ L mol}^{-1}\) and \(M(\ce{Fe})=55.85\text{ g mol}^{-1}\).

Solution 2

The reaction consumes approximately \(0.108\text{ g}\) of iron and transfers \(3.87\times10^{-3}\text{ mol}\) of electrons. This follows from the \(1:1\) mole ratio between iron and hydrogen gas when iron forms \(\mathrm{Fe}^{2+}\).

Step 1: Write the balanced net ionic equation.

\[ \ce{Fe(s) + 2H+(aq) -> Fe^2+(aq) + H2(g)} \]

One mole of iron produces one mole of hydrogen gas.

Step 2: Calculate the amount of hydrogen gas.

Convert the volume into litres:

\[ 48.0\text{ mL}=0.0480\text{ L} \]

Use:

\[ n=\frac{V}{V_m} \]

\[ n(\ce{H2})=\frac{0.0480}{24.8} \]

\[ =1.94\times10^{-3}\text{ mol} \]

Step 3: Calculate the amount of iron consumed.

The mole ratio is \(1:1\).

\[ n(\ce{Fe})=1.94\times10^{-3}\text{ mol} \]

Step 4: Calculate the mass of iron consumed.

\[ m=nM \]

\[ m=1.94\times10^{-3}\times55.85 \]

\[ =0.108\text{ g} \]

Step 5: Calculate the amount of electrons transferred.

The oxidation half-equation is:

\[ \ce{Fe(s) -> Fe^2+(aq) + 2e^-} \]

Each iron atom loses two electrons.

\[ n(e^-)=2(1.94\times10^{-3}) \]

\[ =3.87\times10^{-3}\text{ mol} \]

The result means that approximately \(0.108\text{ g}\) of iron supplied enough electrons to reduce the hydrogen ions responsible for producing \(48.0\text{ mL}\) of hydrogen gas.

Question 3

A student reacts \(0.00250\text{ mol}\) of zinc with exactly \(0.00500\text{ mol}\) of hydrochloric acid in a deaerated solution.

The zinc reacts completely.

The student argues:

“Zinc forms positive ions in the solution, so the solution must have a net positive charge. The chloride ions must have accepted electrons to balance this.”

Evaluate this explanation.

Your response must:

a) Write the balanced net ionic equation.

b) Calculate the amount of electrons transferred.

c) Identify the spectator ion.

d) Explain how the solution remains electrically neutral.

Solution 3

The solution remains electrically neutral, and chloride ions do not accept electrons. The positive charge from the dissolved zinc ions is balanced by the negative charge from the chloride ions.

Step 1: Write the balanced net ionic equation.

\[ \ce{Zn(s) + 2H+(aq) -> Zn^2+(aq) + H2(g)} \]

The \(0.00250\text{ mol}\) of zinc produces:

\[ n(\mathrm{Zn}^{2+})=0.00250\text{ mol} \]

It also produces:

\[ n(\ce{H2})=0.00250\text{ mol} \]

Step 2: Calculate the amount of electrons transferred.

The oxidation half-equation is:

\[ \ce{Zn -> Zn^2+ + 2e^-} \]

Therefore:

\[ n(e^-)=2(0.00250) \]

\[ =0.00500\text{ mol} \]

These electrons are accepted by the hydrogen ions.

Step 3: Identify the spectator ion.

Hydrochloric acid provides:

\[ \ce{HCl -> H+ + Cl-} \]

The initial amount of hydrochloric acid is \(0.00500\text{ mol}\).

Therefore, the solution initially contains \(0.00500\text{ mol}\) of chloride ions.

Chloride remains unchanged throughout the reaction.

Its oxidation state is \(-1\) before and after the reaction.

It does not accept or donate electrons.

Step 4: Check the final solution’s charge.

The dissolved zinc ions contribute:

\[ 0.00250\times(+2)=+0.00500 \]

The chloride ions contribute:

\[ 0.00500\times(-1)=-0.00500 \]

Adding these charges:

\[ +0.00500-0.00500=0 \]

The final solution is electrically neutral.

The mistake in the student’s explanation is confusing charge balancing with electron transfer.

Chloride ions balance the positive charge of the dissolved zinc ions, but that does not make them participants in the redox reaction.

Question 4

Two identical pieces of freshly cleaned copper are placed into separate beakers at room temperature.

The observations are:

• Beaker A contains dilute hydrochloric acid. No visible reaction occurs.
• Beaker B contains silver nitrate solution. Grey silver crystals form on the copper, and the solution develops a blue colour.

A student concludes:

“Copper cannot be a reductant because it does not react with hydrochloric acid.”

Evaluate this statement.

Write the oxidation and reduction half-equations for the reaction in Beaker B, identify the reductant and oxidant, and explain why the two beakers produce different results.

Solution 4

The student’s statement is incorrect. Copper is a reductant in silver nitrate solution, even though it does not act as a reductant towards hydrogen ions in dilute hydrochloric acid.

The role of a substance depends on the reaction in which it participates.

Step 1: Explain the observation in Beaker A.

Copper does not normally react with dilute hydrochloric acid at room temperature.

It is below hydrogen in the reactivity series, so it cannot spontaneously donate electrons to hydrogen ions to produce hydrogen gas.

There is no observable redox reaction under these conditions.

Step 2: Write the oxidation half-equation for Beaker B.

Copper loses two electrons:

\[ \ce{Cu(s) -> Cu^2+(aq) + 2e^-} \]

Copper changes oxidation state from \(0\) to \(+2\).

It is oxidised.

Step 3: Write the reduction half-equation.

Silver ions gain electrons:

\[ \ce{2Ag+(aq) + 2e^- -> 2Ag(s)} \]

Silver changes oxidation state from \(+1\) to \(0\).

It is reduced.

Step 4: Combine the half-equations.

\[ \ce{Cu(s) + 2Ag+(aq) -> Cu^2+(aq) + 2Ag(s)} \]

Copper supplies the electrons that reduce silver ions.

Therefore:

• Copper is the reductant.
• Silver ions are the oxidant.
• Nitrate ions are spectators.

The blue solution is consistent with the formation of aqueous copper(II) ions.

The important distinction is that copper is not permanently classified as a reductant or non-reductant. Its role depends on the species it reacts with.

Question 5

A student investigates the reaction between zinc and dilute hydrochloric acid using a closed experimental apparatus.

Unlike the earlier examples, the acid solution also contains dissolved oxygen.

The following measurements are recorded:

Use:

• \(M(\ce{Zn})=65.4\text{ g mol}^{-1}\)
• \(V_m=24.8\text{ L mol}^{-1}\)

Assume:

• All zinc forms \(\mathrm{Zn}^{2+}\).
• Hydrogen ions and dissolved oxygen are the only electron acceptors.
• Hydrogen ions are reduced to hydrogen gas.
• Oxygen is reduced completely to water according to the half-equation:

\[ \ce{O2(aq) + 4H+(aq) + 4e^- -> 2H2O(l)} \]

• There are no other reactions or gas losses.

Answer the following:

a) Calculate the total amount of electrons released by the zinc.

b) Calculate the amount of electrons accepted by hydrogen ions.

c) Determine the amount of oxygen consumed and the amount remaining.

d) Calculate the amount of hydrogen ions consumed and the amount remaining.

e) Verify that the final solution remains electrically neutral, assuming chloride ions are spectators.

f) Explain why the amount of hydrogen gas is smaller than the amount predicted by the simple \(1:1\) zinc-to-hydrogen mole ratio. Is zinc still the reductant?

Solution 5

Zinc is still the reductant, but not all of its electrons are accepted by hydrogen ions. Some are accepted by dissolved oxygen, which is reduced to water.

This means that the simple \(1:1\) mole ratio between zinc and hydrogen gas no longer applies.

The key is to account for every electron released by the zinc.

Step 1: Calculate the amount of zinc consumed.

Use:

\[ n=\frac{m}{M} \]

\[ n(\ce{Zn})=\frac{0.327}{65.4} \]

\[ =0.00500\text{ mol} \]

Each zinc atom loses two electrons:

\[ \ce{Zn -> Zn^2+ + 2e^-} \]

Therefore:

\[ n(e^-)_{\text{released}}=2(0.00500) \]

\[ =\boxed{0.0100\text{ mol}} \]

This is the total amount of electrons available for reduction reactions.

Step 2: Calculate how many electrons hydrogen ions accept.

First, calculate the amount of hydrogen gas collected.

Convert the volume into litres:

\[ 99.2\text{ mL}=0.0992\text{ L} \]

Then:

\[ n(\ce{H2})=\frac{0.0992}{24.8} \]

\[ =0.00400\text{ mol} \]

The hydrogen reduction half-equation is:

\[ \ce{2H+ + 2e^- -> H2} \]

Each mole of hydrogen gas requires two moles of electrons.

Therefore:

\[ n(e^-)_{\mathrm{H}^{+}}=2(0.00400) \]

\[ =\boxed{0.00800\text{ mol}} \]

Step 3: Calculate the electrons accepted by oxygen.

The zinc supplied \(0.0100\text{ mol}\) of electrons.

Hydrogen ions accepted \(0.00800\text{ mol}\).

The remaining electrons must have been accepted by oxygen:

\[ n(e^-)_{\ce{O2}}=0.0100-0.00800 \]

\[ =\boxed{0.00200\text{ mol}} \]

Step 4: Calculate the amount of oxygen consumed.

The oxygen reduction half-equation is:

\[ \ce{O2 + 4H+ + 4e^- -> 2H2O} \]

Four moles of electrons are needed for every mole of oxygen.

Therefore:

\[ n(\ce{O2})=\frac{0.00200}{4} \]

\[ =\boxed{0.000500\text{ mol}} \]

The initial oxygen amount was \(0.00100\text{ mol}\).

Therefore, the remaining oxygen is:

\[ 0.00100-0.000500 \]

\[ =\boxed{0.000500\text{ mol}} \]

The oxygen supply was sufficient to accept the remaining electrons.

Step 5: Calculate the amount of hydrogen ions consumed.

There are two separate reactions consuming hydrogen ions.

First, hydrogen gas formation:

\[ \ce{2H+ + 2e^- -> H2} \]

This consumes:

\[ 2(0.00400)=0.00800\text{ mol H+} \]

Second, oxygen reduction:

\[ \ce{O2 + 4H+ + 4e^- -> 2H2O} \]

This consumes:

\[ 4(0.000500)=0.00200\text{ mol H+} \]

Total hydrogen ions consumed:

\[ 0.00800+0.00200 \]

\[ =\boxed{0.0100\text{ mol}} \]

The initial amount was \(0.0200\text{ mol}\).

Therefore, the amount remaining is:

\[ 0.0200-0.0100 \]

\[ =\boxed{0.0100\text{ mol H+}} \]

Step 6: Verify charge conservation in the final solution.

All \(0.00500\text{ mol}\) of zinc has formed \(\mathrm{Zn}^{2+}\).

The total positive charge contributed by zinc ions is:

\[ 0.00500\times2=+0.0100 \]

There are also \(0.0100\text{ mol}\) of hydrogen ions remaining, contributing:

\[ +0.0100 \]

The total positive charge is:

\[ +0.0100+0.0100=+0.0200 \]

The original \(0.0200\text{ mol}\) of chloride ions remains unchanged.

Their total negative charge is:

\[ -0.0200 \]

Therefore:

\[ \text{Total charge}=+0.0200-0.0200=0 \]

The final solution is electrically neutral.

Step 7: Explain why the simple mole ratio no longer applies.

In a reaction involving only zinc and hydrogen ions, each mole of zinc produces one mole of hydrogen gas.

However, oxygen provides an additional pathway for electron acceptance.

In this experiment:

• Zinc releases \(0.0100\text{ mol}\) of electrons.
• Hydrogen ions accept \(0.00800\text{ mol}\) of electrons.
• Oxygen accepts the remaining \(0.00200\text{ mol}\) of electrons.

Only 80% of the electrons released by zinc contribute to hydrogen gas production.

Zinc is still the reductant because it supplies all the electrons. The difference is that two different species are accepting them.

This is the important conceptual result: measuring hydrogen gas alone does not always tell you how much metal has been oxidised. You must account for every possible electron acceptor.

Once you can track electrons through a metal-acid reaction, you can apply the same reasoning to more complicated redox systems. The next useful step is learning how to balance half-equations when the reacting species have different electron requirements.

For that technique, see How to Balance a Redox Half-Equation in Acidic Solution.