How to Predict Whether a Metal Can Reduce Hydrogen Ions

A clean strip of zinc bubbles when placed in dilute hydrochloric acid. A clean strip of copper, placed in the same acid, produces no visible bubbles. Both are metals, and both are in contact with hydrogen ions. So what determines whether hydrogen gas forms?

The key is the metal's position relative to hydrogen in the activity series. A metal can reduce hydrogen ions if it is sufficiently reactive to lose electrons to them. Metals above hydrogen can do this; metals below hydrogen generally cannot in a non-oxidising acid.

There is one important complication: a reaction that is thermodynamically favourable might still be extremely slow or appear not to happen. The condition of the metal's surface can make a big difference to what you observe.

## The quick prediction: find the metal relative to hydrogen

Consider what happens when zinc is added to dilute hydrochloric acid.

Zinc atoms lose electrons and enter the solution as zinc ions. Hydrogen ions gain those electrons and combine to form hydrogen gas.

The overall reaction is:

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

Zinc is oxidised because it loses electrons. Hydrogen ions are reduced because they gain electrons.

This gives you the first decision rule:

- Metal above hydrogen in the activity series: hydrogen ions can generally be reduced to hydrogen gas.
- Metal below hydrogen in the activity series: hydrogen ions are generally not reduced by that metal in a non-oxidising acid.
- Metal close to hydrogen: use the relevant activity-series evidence or standard electrode potentials rather than relying on an uncertain visual comparison.

[IMAGE: Two labelled beakers side by side, one containing dilute hydrochloric acid and a zinc strip with hydrogen bubbles forming at its surface, the other containing the same acid and a copper strip with no visible bubbles. Include a clear close-up of electron transfer at the zinc surface, showing zinc atoms entering solution as Zn2+ and H+ ions forming H2 gas. Keep the copper beaker visibly unchanged.]

A simplified activity series helps you make the prediction:

| More reactive metals | Position |
|---|---|
| Potassium | Above hydrogen |
| Sodium | Above hydrogen |
| Calcium | Above hydrogen |
| Magnesium | Above hydrogen |
| Aluminium | Above hydrogen |
| Zinc | Above hydrogen |
| Iron | Above hydrogen |
| Tin | Above hydrogen |
| Lead | Above hydrogen |
| Hydrogen | Reference position |
| Copper | Below hydrogen |
| Silver | Below hydrogen |
| Gold | Below hydrogen |

The exact order of metals matters when comparing two metals, but for this question, the most important feature is where hydrogen sits.

For example:

- Magnesium is above hydrogen, so it can reduce hydrogen ions.
- Iron is above hydrogen, so it can reduce hydrogen ions.
- Copper is below hydrogen, so it does not normally reduce hydrogen ions in dilute hydrochloric acid.
- Silver is below hydrogen, so it does not normally produce hydrogen gas in dilute hydrochloric acid.

This prediction assumes an ordinary, non-oxidising acid and suitable conditions. It does not mean every metal above hydrogen will produce bubbles immediately.

## Why the reaction happens: following the electrons

The activity series is useful because it summarises how readily metals undergo oxidation.

Think of the metal as an electron donor. A sufficiently reactive metal can transfer electrons to hydrogen ions. The hydrogen ions then form hydrogen molecules.

There are two half-equations:

**Oxidation:**

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

**Reduction:**

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

Adding the half-equations cancels the two electrons, giving the overall reaction.

Notice that the electrons do not simply disappear when zinc atoms become zinc ions. They are transferred to another chemical species. In this reaction, that species is \(\ce{H+}\).

[IMAGE: An annotated particle-level illustration of a zinc metal surface in contact with an acidic solution. Show zinc atoms at the solid boundary becoming Zn2+ ions in solution, electrons remaining available for transfer at the metal surface, and two H+ ions gaining electrons before forming an H2 molecule. Distinguish the solid metal, aqueous ions and gas without implying that electrons travel freely through the solution.]

The same reasoning works for other metals. For magnesium:

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

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

Therefore:

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

The metal changes into positive ions, while hydrogen ions change into neutral hydrogen molecules.

## A closer look: thermodynamic prediction versus reaction rate

Here is where students often make an incorrect prediction.

Suppose you place two pieces of aluminium into separate beakers of dilute hydrochloric acid. One piece has a freshly cleaned surface. The other has been exposed to air and has its usual thin oxide coating.

You might expect the two pieces to react at similar rates because they are both aluminium. However, their visible behaviour can be very different.

Aluminium is above hydrogen in the activity series. Its oxidation, coupled with hydrogen-ion reduction, is thermodynamically favourable under standard conditions.

But aluminium is normally covered by a thin, adherent layer of aluminium oxide, \(\ce{Al2O3}\). This layer restricts contact between the acid and the underlying metal. It can slow the reaction substantially until the coating is disrupted or removed.

[IMAGE: Side-by-side cross-sections of two aluminium strips in dilute hydrochloric acid. The first has an intact, thin aluminium oxide coating separating the metal from the solution, with very limited visible hydrogen formation. The second has a disrupted oxide coating, exposing aluminium to the acid, with hydrogen bubbles forming at exposed regions. Label the aluminium metal, oxide layer and acid solution.]

This is the distinction:

| Question | What it tells you |
|---|---|
| Is the reaction thermodynamically favourable? | Whether the reaction has a driving force to occur. |
| Is the reaction kinetically fast enough to observe? | How quickly the reaction proceeds. |
| Is the metal surface accessible? | Whether the reactants can reach the metal and exchange electrons effectively. |

The activity series addresses the first question. It does not, by itself, tell you how quickly bubbles will appear.

A useful analogy is pushing a heavy trolley down a slope. The slope represents the thermodynamic driving force. Friction represents a kinetic barrier. A downward slope makes the trolley's motion favourable, but high friction can prevent it from moving easily.

The analogy has a limit: chemical reactions involve changes in bonding, electron transfer, and activation energy, not simply the motion of a solid object. Still, it captures why a favourable reaction need not be fast.

01Check which species is actually being reduced

So far, we’ve assumed the acid supplies hydrogen ions that accept electrons. That is correct for common non-oxidising acids, such as dilute hydrochloric acid and dilute sulfuric acid.

However, not every reaction between a metal and an acid produces hydrogen gas.

Consider copper in concentrated nitric acid. Copper is below hydrogen in the activity series, so you would normally predict that it cannot reduce hydrogen ions.

Yet copper reacts with concentrated nitric acid. The copper dissolves, and a brown gas, nitrogen dioxide, forms.

Does this mean the activity series is wrong?

No. The important detail is that nitric acid contains another oxidising species: the nitrate ion, \(\mathrm{NO_{3}}^{-}\).

In this reaction, nitrate ions accept electrons from copper. Hydrogen ions provide the acidic conditions, but they are not reduced to hydrogen gas.

The overall ionic equation is:

\[ \ce{Cu(s) + 2NO3^-(aq) + 4H+(aq)} \]

\[ \ce{-> Cu^2+(aq) + 2NO2(g) + 2H2O(l)} \]

The oxidation half-equation is:

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

The reduction half-equation is:

\[ \ce{2NO3^-(aq) + 4H+(aq) + 2e^-} \]

\[ \ce{-> 2NO2(g) + 2H2O(l)} \]

Copper loses electrons, while nitrate gains electrons.

This gives you an important rule:

A metal reacting with an acid does not automatically mean that it can reduce hydrogen ions.

You must identify the species accepting the electrons.

This distinction is particularly useful in questions that deliberately give you an unexpected observation.

02Using standard reduction potentials for a more precise prediction

The activity series gives you a quick comparison. Standard reduction potentials let you make the prediction more precisely.

A standard reduction potential, \(E^\circ\), measures how readily a species gains electrons under standard conditions.

The standard hydrogen electrode provides the reference point:

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

Its standard reduction potential is defined as:

\[ E^\circ = 0.00\ \mathrm{V} \]

This means you can compare a metal’s reduction potential with hydrogen’s to determine whether the metal can be oxidised while hydrogen ions are reduced.

Consider these illustrative standard reduction potential values at \(25^\circ\mathrm{C}\):

These values are written as reduction potentials. Notice that aluminium and lead have negative values, while copper has a positive value.

For a metal reacting with hydrogen ions:

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

the standard cell potential is:

\[ E^\circ_{\mathrm{cell}} = E^\circ_{\mathrm{red,cathode}} – E^\circ_{\mathrm{red,anode}} \]

Here:

• The cathode is where reduction occurs. Hydrogen ions are reduced.
• The anode is where oxidation occurs. The metal is oxidised.
• \(E^\circ_{\mathrm{cell}}\) is the overall standard cell potential, measured in volts.

Because the hydrogen reduction potential is \(0.00\ \mathrm{V}\), the calculation becomes:

\[ \boxed{ E^\circ_{\mathrm{cell}} = 0.00-E^\circ_{\mathrm{M^{n+}/M}} } \]

Use the sign of the result:

A positive value means the metal oxidation and hydrogen-ion reduction are thermodynamically favourable when coupled.

A negative value means the proposed electron transfer is not thermodynamically favourable under those conditions.

There is an important limitation here. Standard potentials describe standard conditions, including a hydrogen-ion concentration of \(1.0\ \mathrm{mol\\,L^{-1}}\), hydrogen gas at \(100\ \mathrm{kPa}\), and a temperature of \(25^\circ\mathrm{C}\).

Actual reaction conditions can change the measured cell potential. A standard potential also does not tell you how quickly the reaction will occur.

For the usual HSC prediction, use the supplied standard values and conditions. Don’t confuse a positive cell potential with a guarantee of rapid hydrogen production.

04A practical decision method for HSC questions

When you encounter an unfamiliar metal and acid combination, use this sequence.

If the reaction is favourable but no hydrogen is observed, don’t immediately reverse your prediction.

Consider:

• Is there an oxide coating?
• Has an insoluble product formed on the metal surface?
• Is the exposed surface area very small?
• Is the acid sufficiently concentrated?
• Has enough time passed for a measurable amount of gas to form?
• Is another species, such as nitrate, accepting the electrons instead?

For a closer look at how experimental displacement results establish relative metal reactivity, see How to Build a Metal Activity Series from Displacement Results.

Also remember that a metal reacting with water is a different chemical test from a metal reacting with hydrogen ions in an acid. The relevant electron acceptor changes. See Metal Reactivity with Water: HSC Chemistry Guide.

05Questions and solutions

The following questions move from straightforward acid-metal predictions to situations where the activity series, acid strength, surface chemistry, and competing oxidising agents must be considered together.

Question 1

A clean strip of magnesium is placed in excess dilute hydrochloric acid at room temperature.

(a) Predict whether magnesium can reduce hydrogen ions.

(b) Write the balanced ionic equation for the reaction.

(c) Identify the species oxidised and the species reduced.

Solution 1

(a) Yes. Magnesium is above hydrogen in the activity series, so it can lose electrons to hydrogen ions.

(b) The balanced ionic equation is:

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

(c) Magnesium is oxidised, while hydrogen ions are reduced.

Magnesium loses two electrons:

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

Hydrogen ions gain those two electrons:

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

The electrons lost by magnesium are transferred to hydrogen ions. This produces hydrogen gas, which can be observed as bubbles at the magnesium surface.

The important reasoning is the position of magnesium relative to hydrogen, not simply the fact that hydrochloric acid is present.

Question 2

A student reacts \(0.280\ \mathrm{g}\) of iron with \(25.0\ \mathrm{mL}\) of \(0.200\ \mathrm{mol\\,L^{-1}}\) hydrochloric acid.

The experiment is conducted at \(25^\circ\mathrm{C}\) and \(100\ \mathrm{kPa}\).

Assume iron forms \(\mathrm{Fe}^{2+}\), and use a molar gas volume of \(24.8\ \mathrm{L\\,mol^{-1}}\).

(a) Write the balanced ionic equation.

(b) Determine the limiting reactant.

(c) Calculate the maximum volume of hydrogen gas produced.

Use \(M(\ce{Fe})=55.8\ \mathrm{g\\,mol^{-1}}\).

Solution 2

Iron can reduce hydrogen ions, and the hydrochloric acid is the limiting reactant. The maximum hydrogen volume is 62.0 mL.

(a) Write the balanced ionic equation.

Iron is above hydrogen in the activity series.

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

The mole ratio is:

\[ \ce{Fe:H+:H2}=1:2:1 \]

(b) Determine the limiting reactant.

Calculate the amount of iron:

\[ \begin{aligned} n(\ce{Fe}) &=\frac{m}{M}\\\ &=\frac{0.280}{55.8}\\\ &=5.02\times10^{-3}\ \mathrm{mol} \end{aligned} \]

Calculate the amount of hydrochloric acid:

\[ \begin{aligned} n(\ce{HCl}) &=cV\\\ &=0.200\times0.0250\\\ &=5.00\times10^{-3}\ \mathrm{mol} \end{aligned} \]

The reaction requires two moles of hydrogen ions per mole of iron.

Therefore, the amount of iron that can react is:

\[ \begin{aligned} n(\ce{Fe}) &=\frac{5.00\times10^{-3}}{2}\\\ &=2.50\times10^{-3}\ \mathrm{mol} \end{aligned} \]

Only \(2.50\times10^{-3}\) moles of the available iron can react.

Since more iron is available than the acid can react with, hydrochloric acid is the limiting reactant.

(c) Calculate the hydrogen volume.

From the balanced equation:

\[ n(\ce{H2})=2.50\times10^{-3}\ \mathrm{mol} \]

Use:

\[ V=nV_m \]

Substitute:

\[ \begin{aligned} V(\ce{H2}) &=2.50\times10^{-3}\times24.8\\\ &=0.0620\ \mathrm{L}\\\ &=62.0\ \mathrm{mL} \end{aligned} \]

The maximum theoretical volume is \(62.0\ \mathrm{mL}\).

A common mistake is to assume that the metal must be the limiting reactant because it is a solid. The limiting reactant is determined by the mole ratio, not by whether a reactant is a solid or a solution.

Question 3

Two students investigate the reaction between magnesium and two different acids.

Each experiment uses:

• \(0.100\ \mathrm{g}\) of magnesium.
• \(20.0\ \mathrm{mL}\) of \(0.200\ \mathrm{mol\\,L^{-1}}\) acid.
• A temperature of \(25^\circ\mathrm{C}\).
• Magnesium samples with the same exposed surface area.

The first experiment uses hydrochloric acid. The second uses ethanoic acid, \(\ce{CH3COOH}\), which is a weak acid.

The following illustrative gas-volume measurements are obtained:

(a) Calculate the average hydrogen production rate for each experiment over the first 20 seconds.

(b) Explain why both acids can produce hydrogen even though ethanoic acid is weak.

(c) Explain why the final hydrogen volumes are similar, despite the large difference in their initial reaction rates.

Use \(M(\ce{Mg})=24.3\ \mathrm{g\\,mol^{-1}}\) and a molar gas volume of \(24.8\ \mathrm{L\\,mol^{-1}}\).

Solution 3

Both acids can supply hydrogen ions for reduction. Hydrochloric acid produces hydrogen more quickly at the start, but the two experiments can produce almost the same final quantity of hydrogen because they contain the same total number of moles of monoprotic acid.

(a) Calculate the average rates.

Use:

\[ \text{Average rate} = \frac{\text{change in gas volume}}{\text{time}} \]

For hydrochloric acid:

\[ \begin{aligned} \text{Rate} &=\frac{38.0}{20.0}\\\ &=1.90\ \mathrm{mL\\,s^{-1}} \end{aligned} \]

For ethanoic acid:

\[ \begin{aligned} \text{Rate} &=\frac{5.0}{20.0}\\\ &=0.250\ \mathrm{mL\\,s^{-1}} \end{aligned} \]

Hydrochloric acid produces hydrogen approximately 7.6 times faster over this interval.

(b) Explain why both acids produce hydrogen.

Hydrochloric acid is a strong acid. It ionises essentially completely in water:

\[ \ce{HCl(aq) -> H+(aq) + Cl^-(aq)} \]

Ethanoic acid is weak, meaning it only partially ionises at any given instant:

\[ \ce{CH3COOH(aq) <=> H+(aq) + CH3COO^-(aq)} \]

However, weak does not mean incapable of supplying hydrogen ions.

As magnesium consumes \(\mathrm{H}^{+}\), the ethanoic acid equilibrium shifts towards the products, releasing more hydrogen ions into solution.

These ions can then accept electrons from magnesium.

The overall reaction can be represented as:

\[ \ce{Mg(s) + 2CH3COOH(aq)} \]

\[ \ce{-> Mg(CH3COO)2(aq) + H2(g)} \]

Therefore, both acids can produce hydrogen.

(c) Explain the similar final volumes.

The two acid solutions contain the same total number of moles of monoprotic acid.

Calculate the amount of acid in either experiment:

\[ \begin{aligned} n(\text{acid}) &=cV\\\ &=0.200\times0.0200\\\ &=0.00400\ \mathrm{mol} \end{aligned} \]

Each mole of acid can supply one mole of hydrogen ions.

The reaction requires two moles of hydrogen ions to produce one mole of hydrogen gas.

Therefore:

\[ n(\ce{H2})=\frac{0.00400}{2} \]

\[ n(\ce{H2})=0.00200\ \mathrm{mol} \]

The theoretical maximum hydrogen volume is:

\[ \begin{aligned} V(\ce{H2}) &=0.00200\times24.8\\\ &=0.0496\ \mathrm{L}\\\ &=49.6\ \mathrm{mL} \end{aligned} \]

Magnesium is present in excess, so the acid limits the total hydrogen produced.

The measured final volumes, \(49.0\ \mathrm{mL}\) and \(48.5\ \mathrm{mL}\), are both close to the theoretical value.

The difference in initial reaction rates comes from the different initial hydrogen-ion concentrations. The difference does not mean magnesium has a different reactivity in the two experiments.

This is a useful distinction between acid strength, the total amount of acid available, and reaction rate.

Question 4

A student places copper metal in concentrated nitric acid.

The copper dissolves, and a brown gas is produced. The gas is identified as nitrogen dioxide, \(\ce{NO2}\), rather than hydrogen.

The student argues:

“Copper must be above hydrogen in the activity series because it reacts with an acid.”

Evaluate this claim.

Write the balanced net ionic equation and identify:

(a) The species oxidised.

(b) The species reduced.

(c) Whether hydrogen ions are reduced in this reaction.

Solution 4

The student’s claim is incorrect. Copper reacts with concentrated nitric acid because nitrate ions act as the oxidising agent, not because copper reduces hydrogen ions.

The balanced net ionic equation is:

\[ \ce{Cu(s) + 2NO3^-(aq) + 4H+(aq)} \]

\[ \ce{-> Cu^2+(aq) + 2NO2(g) + 2H2O(l)} \]

(a) Identify the species oxidised.

Copper loses two electrons:

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

Copper is oxidised because its oxidation number increases from 0 to +2.

(b) Identify the species reduced.

Nitrogen in nitrate has an oxidation number of +5.

In nitrogen dioxide, nitrogen has an oxidation number of +4.

Therefore, nitrogen is reduced.

The reduction half-equation is:

\[ \ce{2NO3^- + 4H+ + 2e^-} \]

\[ \ce{-> 2NO2 + 2H2O} \]

Nitrate ions accept the electrons released by copper.

(c) Are hydrogen ions reduced?

No.

Hydrogen ions appear in the overall equation, but their hydrogen atoms have an oxidation number of +1 both before and after the reaction. They form water rather than hydrogen gas.

The electron acceptor is nitrate, not hydrogen ions.

Copper therefore remains below hydrogen in the activity series. Its reaction with nitric acid does not contradict the prediction that it cannot normally reduce hydrogen ions in dilute hydrochloric acid.

The key exam skill is to identify the actual reduction half-equation rather than assuming that every acid reaction produces hydrogen.

Question 5

A student investigates the behaviour of lead in three different acid tests.

Lead is known to form \(\mathrm{Pb}^{2+}\) ions.

The following illustrative observations are recorded at \(25^\circ\mathrm{C}\):

Experiment A: An untreated lead strip is placed in \(0.50\ \mathrm{mol\\,L^{-1}}\) hydrochloric acid. No visible gas forms during the first five minutes.

Experiment B: A freshly abraded lead strip is placed in the same hydrochloric acid. Bubbles form at the exposed metal surface. The gas is identified as hydrogen. After approximately 40 seconds, the reaction slows considerably as a white lead(II) chloride coating forms over the surface.

Experiment C: A freshly abraded lead strip is placed in concentrated nitric acid. The lead dissolves, and a brown gas is produced. The gas is identified as nitrogen dioxide, not hydrogen.

The following illustrative standard reduction potential data are supplied:

Assume the potential values apply under standard conditions, the stated gas identifications are reliable, and the white coating in Experiment B is \(\ce{PbCl2}\).

A student makes two claims:

• Claim 1: Lead must be below hydrogen because Experiment A produces no visible gas.
• Claim 2: Lead must be above hydrogen because Experiment C produces a gas when placed in an acid.

Evaluate both claims.

In your response:

(a) Calculate the standard cell potential for lead reacting with hydrogen ions. State whether this reaction is thermodynamically favourable under standard conditions.

(b) Explain the difference between Experiments A and B. Write the ionic equations for hydrogen production and lead(II) chloride formation.

(c) Explain why Experiment C does not demonstrate the reduction of hydrogen ions. Write the balanced net ionic equation.

(d) Explain how the formation of a solid coating can favour product formation while simultaneously slowing the observed reaction.

(e) Identify which experimental observation provides direct evidence that lead can reduce hydrogen ions.

Solution 5

Lead can reduce hydrogen ions under standard conditions. Neither the absence of visible gas in Experiment A nor the production of nitrogen dioxide in Experiment C disproves this conclusion.

The observations become consistent once thermodynamic feasibility, surface effects, and the identity of the oxidising species are considered separately.

(a) Calculate the standard cell potential.

Hydrogen ions are reduced at the cathode:

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

Lead is oxidised at the anode:

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

Using the supplied reduction potentials:

\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=E^\circ_{\mathrm{cathode}} -E^\circ_{\mathrm{anode}}\\\ &=0.00-(-0.13)\\\ &=+0.13\ \mathrm{V} \end{aligned} \]

The positive standard cell potential means the reaction is thermodynamically favourable under standard conditions.

The overall ionic equation is:

\[ \ce{Pb(s) + 2H+(aq)} \]

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

Therefore, lead is above hydrogen in the activity series.

(b) Explain Experiments A and B.

Experiment A does not provide sufficient evidence to conclude that lead is below hydrogen.

The untreated lead surface may have a protective coating or contamination that prevents effective contact between the metal and the acid.

Experiment B provides stronger evidence. Once the surface is abraded, hydrogen gas forms. This demonstrates that hydrogen ions are accepting electrons from lead.

The oxidation half-equation is:

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

The reduction half-equation is:

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

The lead ions then react with chloride ions to form an insoluble coating:

\[ \ce{Pb^2+(aq) + 2Cl^-(aq)} \]

\[ \ce{-> PbCl2(s)} \]

Combining these reactions gives:

\[ \ce{Pb(s) + 2H+(aq) + 2Cl^-(aq)} \]

\[ \ce{-> PbCl2(s) + H2(g)} \]

The lead(II) chloride coating forms over the exposed metal surface.

This coating reduces the contact between the remaining lead metal and the acid, slowing further electron transfer and hydrogen production.

(c) Explain Experiment C.

Experiment C does not demonstrate that lead reduces hydrogen ions.

Instead, nitrate ions from nitric acid accept electrons from lead.

The balanced net ionic equation is:

\[ \ce{Pb(s) + 2NO3^-(aq) + 4H+(aq)} \]

\[ \ce{-> Pb^2+(aq) + 2NO2(g) + 2H2O(l)} \]

Lead is oxidised:

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

Nitrate is reduced to nitrogen dioxide.

Hydrogen ions remain at oxidation number +1 as they form water. They are not reduced to hydrogen gas.

Therefore, the production of a gas in Experiment C cannot be used as evidence that hydrogen ions have been reduced.

(d) Explain the competing effects of the coating.

This is the most important interaction in the question.

The formation of lead(II) chloride removes \(\mathrm{Pb}^{2+}\) ions from the solution by forming an insoluble solid.

Removing a dissolved reaction product can favour the forward reaction thermodynamically.

However, the same solid can cover the metal surface and restrict access to the reactants. This creates a kinetic barrier.

The two effects are not contradictory:

• Thermodynamically, removing \(\mathrm{Pb}^{2+}\) can favour further lead oxidation.
• Kinetically, the coating can prevent the acid from reaching the remaining lead metal.

The reaction can therefore remain thermodynamically favourable while its observed rate becomes extremely small.

This is why the formation of a precipitate must not automatically be interpreted as evidence that the redox reaction has become thermodynamically unfavourable.

(e) Identify the direct evidence.

Experiment B provides direct evidence that lead can reduce hydrogen ions because the gas produced is independently identified as hydrogen.

The initial lack of visible gas in Experiment A is explained by the condition of the metal surface. Experiment C involves a different oxidising species, nitrate.

Claim 1 is therefore incorrect because a negative observation from an untreated surface does not establish that the reaction is thermodynamically impossible.

Claim 2 is also incorrect because a metal reacting with an acid does not necessarily mean hydrogen ions are being reduced.

The correct conclusion comes from combining the standard potential calculation with the identified reaction products and the observed surface changes.

06Where to go next

The central skill is to separate three decisions:

1. Is the metal capable of transferring electrons to hydrogen ions?
2. Is the acid actually using hydrogen ions as its oxidising species?
3. If the reaction is favourable, can it proceed quickly enough to produce observable evidence?

Once you can answer those questions independently, you can interpret negative experimental results much more reliably and use them to build or test a metal activity series.

This is also the reasoning you’ll need when comparing metal reactions with water, where the electron-accepting species is different and the predicted products can change.

Syllabus context: This article is framed around the NSW Chemistry Stage 6 Syllabus (2017), which is in use in 2026. Its Year 11 Module 3 includes investigating metals in dilute acid, constructing activity series, using standard reduction potentials, and predicting redox spontaneity. Year 12 Module 6 includes predicting and balancing acid-metal reactions. The Chemistry 11-12 Syllabus (2025) is scheduled to begin Year 11 teaching in 2028; this article does not claim that its content is a requirement of that new syllabus. See the NSW Chemistry syllabus and implementation information.