How to Predict Metal Displacement Reactions in Solution
Learn how to predict which metal displaces another using the activity series and standard reduction potentials. Work through balanced equations, cell potential calculations, and examples showing why surface conditions can affect experimental observations.
Place a clean zinc strip in blue copper(II) sulfate solution. A reddish-brown coating forms on the zinc, and the blue colour gradually fades. Now swap the metals: put copper into zinc sulfate solution. No zinc appears.
Which metal displaces which?
Zinc displaces copper because zinc is more reactive than copper. The general rule is that a solid metal displaces the ions of another metal when it is higher in the metal activity series, meaning it has a greater tendency to lose electrons.
If you are given standard reduction potentials instead, the same prediction can be made by calculating the standard cell potential. A positive value means the proposed displacement is thermodynamically favourable under standard conditions.
There is one important complication: a reaction can be favourable but still produce little visible change because of the condition of the metal’s surface. You need to understand both the prediction and the experimental observation.
01Start by identifying the solid metal and the metal ions
A displacement reaction involves two different metals in two different forms:
- One metal is present as a solid strip or piece.
- The other metal is present as dissolved positive ions.
For example, consider zinc metal placed into a solution containing copper(II) ions.
The zinc is the solid metal. The copper is initially present as dissolved \(\mathrm{Cu^{2+}}\) ions.
If displacement occurs, zinc atoms lose electrons and enter the solution as zinc ions. Copper ions gain those electrons and become solid copper.
The reaction is:
\[ \ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)} \]
Two processes occur simultaneously.
Oxidation: zinc loses electrons.
\[ \ce{Zn(s) -> Zn^2+(aq) + 2e^-} \]
Reduction: copper ions gain electrons.
\[ \ce{Cu^2+(aq) + 2e^- -> Cu(s)} \]
The electrons transfer from the zinc to the copper ions at the reacting surface. No external battery or connecting wire is needed.
The important relationship is:
\[ \boxed{\ce{Zn > Cu}} \]
Here, the greater-than sign means more reactive than.
If metal A displaces metal B from its ions, A is more reactive than B.
This is the starting point for predicting displacement reactions.
02Use the metal activity series
The metal activity series ranks metals according to their tendency to undergo oxidation.
For the common metals used in HSC displacement questions, a simplified series is:
\[ \ce{Mg > Al > Zn > Fe > Ni > Sn > Cu > Ag} \]
The metals on the left are more reactive. They have a greater tendency to lose electrons and form positive ions.
The metals on the right are less reactive.
The decision rule
When a solid metal is placed into a solution containing another metal’s ions:
| Comparison | Prediction |
|---|---|
| Solid metal is higher in the activity series | Displacement is predicted |
| Solid metal is lower in the activity series | No forward displacement is predicted |
| Both species are the same metal | No net displacement of that metal occurs |
For example:
Case 1: Magnesium in zinc sulfate
Magnesium is above zinc in the activity series.
Therefore, magnesium displaces zinc ions:
\[ \ce{Mg(s) + Zn^2+(aq) -> Mg^2+(aq) + Zn(s)} \]
Case 2: Copper in magnesium sulfate
Copper is below magnesium in the activity series.
Therefore, copper cannot displace magnesium ions:
\[ \ce{Cu(s) + Mg^2+(aq) -> No\ displacement} \]
The direction matters. Reversing the solid metal and the dissolved ions can completely change your prediction.
The activity series is the quickest method when a question supplies a ranking or asks you to use experimental reactivity information.
However, some questions provide a table of standard reduction potentials instead. You need a slightly different method for those.
03Predict displacement using standard reduction potentials
A standard reduction potential, \(E^\circ\), measures the tendency of a chemical species to gain electrons under standard conditions.
A more positive reduction potential means that the species is more likely to undergo reduction.
This is where students often make a mistake.
The standard reduction potential table describes the reduction of metal ions into solid metals. It does not directly rank how readily the solid metals lose electrons.
Consider these approximate standard reduction potentials:
| Reduction half-equation | \(E^\circ\) (V) |
|---|---|
| \(\ce{Mg^2+ + 2e^- -> Mg}\) | -2.37 |
| \(\ce{Al^3+ + 3e^- -> Al}\) | -1.66 |
| \(\ce{Zn^2+ + 2e^- -> Zn}\) | -0.76 |
| \(\ce{Fe^2+ + 2e^- -> Fe}\) | -0.44 |
| \(\ce{Ni^2+ + 2e^- -> Ni}\) | -0.25 |
| \(\ce{Sn^2+ + 2e^- -> Sn}\) | -0.14 |
| \(\ce{Cu^2+ + 2e^- -> Cu}\) | +0.34 |
| \(\ce{Ag+ + e^- -> Ag}\) | +0.80 |
These are rounded standard reference values at 25°C.
Notice the relationship:
- A more negative reduction potential corresponds to a metal that is more readily oxidised.
- A more positive reduction potential corresponds to ions that are more readily reduced.
This means magnesium is more reactive as a metal than copper, even though the copper reduction potential is more positive.
The standard cell potential equation
You can use the standard reduction potentials to calculate whether a displacement reaction is thermodynamically favourable.
The equation is:
\[ E^\circ_{\mathrm{cell}} = E^\circ_{\mathrm{cathode}} - E^\circ_{\mathrm{anode}} \]
Where:
- \(E^\circ_{\mathrm{cell}}\) is the standard cell potential, measured in volts (V).
- \(E^\circ_{\mathrm{cathode}}\) is the standard reduction potential of the half-cell undergoing reduction.
- \(E^\circ_{\mathrm{anode}}\) is the standard reduction potential of the half-cell undergoing oxidation.
Both values must come directly from the standard reduction potential table.
For a proposed displacement in which solid metal A reacts with ions of metal B:
\[ E^\circ_{\mathrm{cell}} = E^\circ_{\mathrm{red}}(\mathrm{B}) - E^\circ_{\mathrm{red}}(\mathrm{A}) \]
The interpretation is straightforward:
| Calculated value | Prediction |
|---|---|
| \(E^\circ_{\mathrm{cell}}>0\) | Forward displacement is thermodynamically favourable |
| \(E^\circ_{\mathrm{cell}}<0\) | Forward displacement is not thermodynamically favourable |
| \(E^\circ_{\mathrm{cell}}=0\) | No net thermodynamic driving force under standard conditions |
A useful shortcut: subtract the reduction potential of the solid metal from the reduction potential of the dissolved metal ions.
There are two common mistakes to avoid:
- Do not reverse the sign of the anode potential before substituting it into this formula. The subtraction already accounts for oxidation.
- Do not multiply a standard reduction potential by the coefficients in a balanced equation. Electrode potentials are not multiplied when half-equations are scaled.
Worked example: Will zinc displace nickel from solution?
A zinc strip is placed into a solution containing nickel(II) ions.
The standard reduction potentials are:
\[ E^\circ_{\ce{Zn^2+/Zn}}=-0.76\text{ V} \]
\[ E^\circ_{\ce{Ni^2+/Ni}}=-0.25\text{ V} \]
Determine whether zinc displaces nickel and calculate the standard cell potential.
Step 1
Zinc is the solid metal, so it can be oxidised:
\[ \ce{Zn -> Zn^2+ + 2e^-} \]
Nickel ions can be reduced:
\[ \ce{Ni^2+ + 2e^- -> Ni} \]
Step 2
Nickel ions undergo reduction, so nickel is the cathode.
Zinc undergoes oxidation, so zinc is the anode.
Therefore:
\[ E^\circ_{\mathrm{cathode}}=-0.25\text{ V} \]
\[ E^\circ_{\mathrm{anode}}=-0.76\text{ V} \]
Step 3
\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=E^\circ_{\mathrm{cathode}} -E^\circ_{\mathrm{anode}}\\ &=-0.25-(-0.76)\\ &=-0.25+0.76\\ &=\boxed{+0.51\text{ V}} \end{aligned} \]
Step 4
The standard cell potential is positive.
Therefore, the forward displacement reaction is thermodynamically favourable under standard conditions.
The balanced net ionic equation is:
\[ \ce{Zn(s) + Ni^2+(aq) -> Zn^2+(aq) + Ni(s)} \]
Zinc is oxidised, while nickel ions are reduced.
You would expect a grey nickel deposit to form on the zinc strip as zinc gradually dissolves.
The important point is that the positive cell potential predicts the direction of the reaction. It does not tell you how many seconds the reaction will take.
04What if the metals form ions with different charges?
You cannot assume that every metal forms \(2+\) ions.
Aluminium forms \(\mathrm{Al^{3+}}\), while silver forms \(\mathrm{Ag^+}\). You must balance the electrons before writing the overall equation.
The standard cell potential calculation remains the same.
Worked example: Will tin displace silver from silver nitrate?
A tin strip is placed into silver nitrate solution.
The supplied standard reduction potentials are:
\[ E^\circ_{\ce{Sn^2+/Sn}}=-0.14\text{ V} \]
\[ E^\circ_{\ce{Ag+/Ag}}=+0.80\text{ V} \]
Determine whether silver will be displaced and write the balanced net ionic equation.
Step 1
Tin is oxidised:
\[ \ce{Sn(s) -> Sn^2+(aq) + 2e^-} \]
Silver ions are reduced:
\[ \ce{Ag+(aq) + e^- -> Ag(s)} \]
Step 2
Silver ions are reduced, so their reduction potential is the cathode potential.
\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=E^\circ_{\ce{Ag+/Ag}} -E^\circ_{\ce{Sn^2+/Sn}}\\ &=0.80-(-0.14)\\ &=\boxed{+0.94\text{ V}} \end{aligned} \]
The positive result means that tin can displace silver under standard conditions.
Step 3
Tin loses two electrons per atom.
Each silver ion gains one electron.
Therefore, multiply the silver reduction half-equation by two:
\[ \ce{2Ag+ + 2e^- -> 2Ag} \]
Add the two half-equations:
\[ \boxed{ \ce{Sn(s) + 2Ag+(aq) -> Sn^2+(aq) + 2Ag(s)} } \]
Step 4
Silver metal forms as a solid deposit, while tin enters the solution as tin(II) ions.
The nitrate ions are spectators. They do not participate in the electron transfer.
Notice that the coefficient of two in front of silver does not change the calculated cell potential. The potential remains \(+0.94\text{ V}\).
05Why a predicted reaction might not be visible
So far, we have assumed that the metal surface allows the reaction to proceed.
Real metal surfaces are not always that cooperative.
A thermodynamically favourable reaction does not necessarily occur quickly. This is the difference between thermodynamics and kinetics.
- Thermodynamics predicts whether the reaction is favourable in a particular direction.
- Kinetics describes how quickly the reaction proceeds.
A positive standard cell potential tells you about the first question, not the second.
Surface area affects reaction rate
Imagine two samples of zinc with the same mass.
One is a single strip. The other is finely divided zinc powder.
The powder exposes a greater surface area to the solution. More metal particles are available at the surface for interactions with dissolved ions.
As a result, the powdered zinc will generally react faster with the same metal-ion solution.
However, both samples are still zinc.
Their standard reduction potentials are identical, and their position in the activity series does not change.
Greater surface area increases reaction rate. It does not make zinc intrinsically more reactive.
Oxide layers can prevent displacement from being observed
Aluminium is a particularly important example.
Aluminium is relatively high in the activity series. You would predict that it can displace ions of less reactive metals.
However, aluminium rapidly develops a thin layer of aluminium oxide when exposed to oxygen:
\[ \ce{Al2O3} \]
This layer adheres strongly to the surface and protects the underlying metal.
It can prevent dissolved metal ions from making effective contact with the aluminium underneath.
The result is that an aluminium strip may show little or no visible reaction, even when the proposed displacement is thermodynamically favourable.
This protective effect is called passivation.
Compare the possible situations
| Situation | Thermodynamic prediction | Likely observation |
|---|---|---|
| Favourable reaction, clean metal surface | Forward reaction favoured | Visible displacement may occur |
| Favourable reaction, protective oxide layer | Forward reaction still favoured overall | Little or no visible change |
| Favourable reaction, very small surface area | Forward reaction favoured | Reaction may be slow |
| Unfavourable reaction | Forward displacement not favoured | No sustained forward displacement |
A metal coating can also change the reaction rate. If the deposited metal covers much of the original strip, it may restrict access to the underlying metal. The effect depends on how the deposit forms and whether the coating is continuous.
The key distinction is that surface conditions can change the observed rate without reversing the underlying activity-series prediction.
What should you conclude from experimental observations?
Do not rely on one observation in isolation.
| Observation | What it can tell you |
|---|---|
| A new solid metal coating forms | Metal ions may have been reduced to solid metal |
| The original strip becomes rough or starts dissolving | Consistent with oxidation of the original metal |
| The solution changes colour | The concentration of a coloured dissolved species may have changed |
| No visible change occurs | No detectable reaction occurred during the observation period |
| A coating forms but the reaction slows | A change in exposed surface or a blocking deposit may be affecting the rate |
A colour change alone does not establish the identity of a deposited metal. Ideally, identify the solid product and look for supporting evidence that the original metal is being oxidised.
Likewise, no visible change after a short observation period is not sufficient evidence that a reaction is thermodynamically impossible.
If the experiment involves aluminium, consider passivation before concluding that the activity series prediction is wrong.
For a more detailed treatment of the difference between metal reactivity and reaction rate, see How to Compare Metal Reactivity Without Confusing Reaction Rate.
06What if the solution contains more than one type of metal ion?
This is a useful extension of the basic prediction method.
Suppose you place an iron strip into a solution containing both copper(II) ions and silver ions.
Which metal will form?
It is tempting to choose just one. However, iron can potentially displace both metals.
The standard reduction potentials are:
| Reduction half-equation | \(E^\circ\) (V) |
|---|---|
| \(\ce{Fe^2+ + 2e^- -> Fe}\) | -0.44 |
| \(\ce{Cu^2+ + 2e^- -> Cu}\) | +0.34 |
| \(\ce{Ag+ + e^- -> Ag}\) | +0.80 |
For copper deposition:
\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=0.34-(-0.44)\\ &=+0.78\text{ V} \end{aligned} \]
For silver deposition:
\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=0.80-(-0.44)\\ &=+1.24\text{ V} \end{aligned} \]
Both reactions have positive standard cell potentials.
The possible reactions are:
\[ \ce{Fe(s) + Cu^2+(aq) -> Fe^2+(aq) + Cu(s)} \]
\[ \ce{Fe(s) + 2Ag+(aq) -> Fe^2+(aq) + 2Ag(s)} \]
Both copper and silver can therefore be deposited by the iron strip.
Silver ions have a greater standard tendency to gain electrons than copper ions. However, this does not prove that silver must deposit first, or that copper cannot deposit while silver ions remain.
Standard reduction potentials predict thermodynamic favourability. They do not rank the reaction rates of competing processes.
The actual deposits and their relative amounts can depend on ion concentrations, surface conditions, and the rates of the competing reactions.
This is particularly useful when interpreting experiments: the presence of one visible deposit does not automatically rule out another reduction process occurring at the same time.
For a broader explanation of how several displacement results can be combined into an activity series, see How to Build a Metal Activity Series from Displacement Results.
07A reliable decision process for HSC questions
When you encounter a metal displacement question, follow this sequence.
| Step | What to do | What you establish |
|---|---|---|
| 1 | Identify the solid metal and dissolved metal ions | Which species can be oxidised and reduced |
| 2 | Compare the metals using the activity series | Whether displacement is predicted |
| 3 | If potentials are provided, calculate \(E^\circ_{\mathrm{cell}}\) | Whether the reaction is favourable under standard conditions |
| 4 | Balance the oxidation and reduction half-equations | The correct electron transfer |
| 5 | Write the balanced net ionic equation | The overall chemical reaction |
| 6 | Consider surface conditions and reaction rate | Whether the reaction should be readily observable |
Your final explanation should connect the prediction to electron transfer.
A strong exam-style response might read:
Zinc is more reactive than nickel because zinc is more readily oxidised. Zinc atoms lose electrons, which are gained by nickel(II) ions. The nickel ions are reduced to solid nickel, while zinc enters the solution as zinc ions. Therefore, zinc displaces nickel from solution.
If a standard cell potential is supplied, include its sign and interpretation.
If the experiment produces an unexpected observation, distinguish a thermodynamic limitation from a kinetic or surface-related limitation.
08Questions and solutions
The following questions move from straightforward displacement predictions to problems involving standard potentials, surface effects, stoichiometry, and competing reduction reactions.
Question 1
An iron nail is placed into a solution of copper(II) sulfate.
After several minutes, a reddish-brown solid forms on the nail, and the blue solution becomes paler.
(a) Identify which metal displaces the other.
(b) Write the balanced net ionic equation.
(c) Identify the species undergoing oxidation and reduction.
(d) Explain why the solution becomes paler.
Solution 1
Iron displaces copper from solution because iron is more reactive than copper.
(a) Identify the displacement.
Iron is above copper in the activity series:
\[ \ce{Fe > Cu} \]
Therefore, iron atoms can transfer electrons to copper(II) ions.
(b) Write the balanced net ionic equation.
\[ \boxed{ \ce{Fe(s) + Cu^2+(aq) -> Fe^2+(aq) + Cu(s)} } \]
There is a 1:1 mole ratio between iron atoms oxidised and copper ions reduced.
(c) Identify oxidation and reduction.
Iron loses two electrons:
\[ \ce{Fe(s) -> Fe^2+(aq) + 2e^-} \]
Iron is oxidised.
Copper(II) ions gain two electrons:
\[ \ce{Cu^2+(aq) + 2e^- -> Cu(s)} \]
Copper ions are reduced.
(d) Explain the colour change.
Copper(II) ions give the original solution its blue colour.
As these ions are consumed during the reaction, their concentration decreases. The blue colour therefore becomes less intense.
The reddish-brown deposit is solid copper formed from the reduction of copper(II) ions.
Question 2
A student places a strip of tin into a solution containing nickel(II) ions.
The following standard reduction potentials are provided:
| Reduction half-equation | \(E^\circ\) (V) |
|---|---|
| \(\ce{Sn^2+ + 2e^- -> Sn}\) | -0.14 |
| \(\ce{Ni^2+ + 2e^- -> Ni}\) | -0.25 |
The student predicts that tin will displace nickel because tin has the more positive standard reduction potential.
(a) Calculate the standard cell potential for the proposed reaction.
(b) Determine whether tin displaces nickel under standard conditions.
(c) Predict what happens if the experiment is reversed and a nickel strip is placed into a solution containing tin(II) ions.
(d) Explain the student’s misconception.
Solution 2
Tin does not displace nickel under standard conditions. The reverse reaction, in which nickel displaces tin, is thermodynamically favourable.
(a) Calculate the standard cell potential.
For the proposed reaction, tin would be oxidised and nickel ions would be reduced.
Therefore:
\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=E^\circ_{\ce{Ni^2+/Ni}} -E^\circ_{\ce{Sn^2+/Sn}}\\ &=-0.25-(-0.14)\\ &=\boxed{-0.11\text{ V}} \end{aligned} \]
(b) Determine whether displacement occurs.
The calculated standard cell potential is negative.
Therefore, the proposed forward reaction is not thermodynamically favourable under standard conditions.
Tin cannot displace nickel ions under these conditions.
(c) Predict the reversed experiment.
Now nickel is the solid metal, and tin ions are dissolved in solution.
The standard cell potential becomes:
\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=-0.14-(-0.25)\\ &=\boxed{+0.11\text{ V}} \end{aligned} \]
The positive value predicts a favourable displacement reaction:
\[ \boxed{ \ce{Ni(s) + Sn^2+(aq) -> Ni^2+(aq) + Sn(s)} } \]
Nickel is oxidised, and tin ions are reduced.
(d) Explain the misconception.
The student has confused the reduction tendency of metal ions with the oxidation tendency of solid metals.
Tin has the more positive reduction potential. This means tin(II) ions are more readily reduced to tin than nickel(II) ions are to nickel.
However, the question involves solid tin losing electrons, not tin ions gaining them.
Nickel has the more negative standard reduction potential and is therefore more readily oxidised in this comparison.
The metal with the more positive reduction potential is not automatically the more reactive solid metal.
Question 3
An aluminium strip is placed into a solution containing iron(II) ions.
The strip shows no visible change after five minutes.
A second aluminium strip of identical dimensions is freshly polished immediately before being placed into an identical solution. This time, a dark grey deposit forms.
The following standard reduction potentials are provided:
| Reduction half-equation | \(E^\circ\) (V) |
|---|---|
| \(\ce{Al^3+ + 3e^- -> Al}\) | -1.66 |
| \(\ce{Fe^2+ + 2e^- -> Fe}\) | -0.44 |
Assume the solutions have the same concentration, volume, and temperature, and that no competing reactions occur.
(a) Calculate the standard cell potential for aluminium reacting with iron(II) ions.
(b) Write the balanced net ionic equation.
(c) Explain why the first strip showed no visible change despite the positive thermodynamic prediction.
(d) Explain what the comparison between the two strips suggests about the role of surface conditions.
Solution 3
Aluminium can displace iron from iron(II) ions, but an oxide layer can prevent the reaction from becoming visibly observable.
(a) Calculate the standard cell potential.
Aluminium is oxidised, while iron(II) ions are reduced.
Therefore:
\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=E^\circ_{\ce{Fe^2+/Fe}} -E^\circ_{\ce{Al^3+/Al}}\\ &=-0.44-(-1.66)\\ &=\boxed{+1.22\text{ V}} \end{aligned} \]
The positive value indicates that the forward displacement reaction is thermodynamically favourable under standard conditions.
(b) Write the balanced net ionic equation.
The oxidation half-equation is:
\[ \ce{Al -> Al^3+ + 3e^-} \]
The reduction half-equation is:
\[ \ce{Fe^2+ + 2e^- -> Fe} \]
The lowest common multiple of three and two is six.
Multiply the aluminium half-equation by two and the iron half-equation by three:
\[ \ce{2Al -> 2Al^3+ + 6e^-} \]
\[ \ce{3Fe^2+ + 6e^- -> 3Fe} \]
Combining them gives:
\[ \boxed{ \ce{2Al(s) + 3Fe^2+(aq) -> 2Al^3+(aq) + 3Fe(s)} } \]
Two aluminium atoms are oxidised for every three iron(II) ions reduced.
(c) Explain the absence of visible change.
Aluminium naturally develops a protective aluminium oxide layer when exposed to oxygen.
This layer prevents the iron(II) ions from making effective contact with the aluminium underneath.
Although the overall displacement reaction is thermodynamically favourable, the oxide coating restricts electron transfer at the surface.
The reaction can therefore be too slow to produce an observable deposit during the five-minute observation period.
(d) Explain the difference between the strips.
Fresh polishing removes or disrupts the protective surface layer.
This exposes the underlying aluminium and allows iron(II) ions to receive electrons more readily.
The formation of the dark grey deposit on the polished strip is consistent with the production of solid iron.
The two experiments do not contradict each other. They demonstrate that surface condition affects reaction rate without reversing the thermodynamic prediction.
A stronger investigation would use several pairs of strips, control their dimensions and exposed surface areas, and measure the amount of deposited iron over the same time period.
Question 4
A zinc strip with a mass of 0.325 g is placed into 50.0 mL of 0.100 mol L\(^{-1}\) silver nitrate solution.
The relevant standard reduction potentials are:
| Reduction half-equation | \(E^\circ\) (V) |
|---|---|
| \(\ce{Zn^2+ + 2e^- -> Zn}\) | -0.76 |
| \(\ce{Ag+ + e^- -> Ag}\) | +0.80 |
The molar masses are:
- Zinc: \(65.38\text{ g mol}^{-1}\)
- Silver: \(107.87\text{ g mol}^{-1}\)
Assume the zinc surface is clean, silver ions are the only oxidising agent, and no competing reactions occur.
(a) Calculate the standard cell potential and predict whether displacement occurs.
(b) Write the balanced net ionic equation.
(c) Determine which reactant is limiting.
(d) Calculate the maximum mass of silver deposited.
Solution 4
Zinc displaces silver from solution, and silver ions are the limiting reactant. The maximum mass of silver deposited is approximately 0.539 g.
(a) Calculate the standard cell potential.
Zinc is oxidised, and silver ions are reduced.
\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=E^\circ_{\ce{Ag+/Ag}} -E^\circ_{\ce{Zn^2+/Zn}}\\ &=0.80-(-0.76)\\ &=\boxed{+1.56\text{ V}} \end{aligned} \]
The positive value predicts that the forward displacement reaction is thermodynamically favourable.
(b) Write the balanced net ionic equation.
The half-equations are:
\[ \ce{Zn -> Zn^2+ + 2e^-} \]
\[ \ce{2Ag+ + 2e^- -> 2Ag} \]
Therefore:
\[ \boxed{ \ce{Zn(s) + 2Ag+(aq) -> Zn^2+(aq) + 2Ag(s)} } \]
One mole of zinc reacts with two moles of silver ions.
(c) Determine the limiting reactant.
First, calculate the amount of zinc available.
Use:
\[ n=\frac{m}{M} \]
Where \(n\) is the amount in moles, \(m\) is mass in grams, and \(M\) is molar mass in g mol\(^{-1}\).
\[ \begin{aligned} n(\ce{Zn}) &=\frac{0.325}{65.38}\\ &=0.00497\text{ mol} \end{aligned} \]
Now calculate the amount of silver ions.
Convert the volume into litres:
\[ 50.0\text{ mL}=0.0500\text{ L} \]
Use:
\[ n=cV \]
\[ \begin{aligned} n(\mathrm{Ag}^{+}) &=0.100\times0.0500\\ &=0.00500\text{ mol} \end{aligned} \]
According to the balanced equation, one mole of zinc requires two moles of silver ions.
To react with all the available zinc, the required amount of silver ions would be:
\[ 2\times0.00497=0.00994\text{ mol} \]
However, only 0.00500 mol of silver ions is available.
Therefore, silver ions are the limiting reactant.
(d) Calculate the mass of silver deposited.
The balanced equation shows a 1:1 mole ratio between silver ions consumed and silver atoms produced.
Therefore:
\[ n(\ce{Ag})=0.00500\text{ mol} \]
Use:
\[ m=nM \]
\[ \begin{aligned} m(\ce{Ag}) &=0.00500\times107.87\\ &=\boxed{0.539\text{ g}} \end{aligned} \]
The maximum theoretical mass of silver deposited is:
\[ \boxed{0.539\text{ g}} \]
This assumes complete reaction of the available silver ions and no loss of deposited silver.
Notice that the standard cell potential established whether the reaction was favourable. The balanced equation and mole calculations established how much silver could form. These are two separate parts of the problem.
Question 5
An iron strip is immersed in a solution containing both copper(II) ions and silver ions.
The experiment is conducted at approximately standard conditions, with a clean iron surface and no complexing agents.
The relevant standard reduction potentials are:
| Reduction half-equation | \(E^\circ\) (V) |
|---|---|
| \(\ce{Fe^2+ + 2e^- -> Fe}\) | -0.44 |
| \(\ce{Cu^2+ + 2e^- -> Cu}\) | +0.34 |
| \(\ce{Ag+ + e^- -> Ag}\) | +0.80 |
After 30 seconds, a reddish-brown coating is visible on the iron. No distinct silver crystals can be identified by visual inspection.
A student makes the following claims:
- Silver ions must be reduced first because they have the most positive standard reduction potential.
- Copper cannot be deposited until all the silver ions have been consumed.
- The reddish-brown coating proves that no silver has formed.
Answer the following:
(a) Write the two possible balanced net ionic displacement equations involving iron.
(b) Calculate the standard cell potential for each reaction.
(c) Determine which of the two solid metals, copper or silver, is more reactive. Explain why this conclusion is not the same as simply choosing the ion with the more positive reduction potential.
(d) Evaluate each of the student’s three claims. In particular, explain whether both displacement reactions can occur in the same solution.
(e) Suggest one direct experiment that would establish the relative activity of copper and silver without using the iron strip.
Solution 5
Both copper and silver can be deposited by the iron strip. Copper is more reactive as a solid metal than silver, even though silver ions have the more positive standard reduction potential. The observed coating alone cannot establish which reduction processes occurred.
(a) Write the two balanced displacement equations.
The first possible reaction involves the reduction of copper(II) ions:
\[ \boxed{ \ce{Fe(s) + Cu^2+(aq) -> Fe^2+(aq) + Cu(s)} } \]
The second possible reaction involves the reduction of silver ions:
\[ \boxed{ \ce{Fe(s) + 2Ag+(aq) -> Fe^2+(aq) + 2Ag(s)} } \]
In both reactions, iron supplies the electrons.
(b) Calculate the standard cell potentials.
For copper deposition:
\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=E^\circ_{\ce{Cu^2+/Cu}} -E^\circ_{\ce{Fe^2+/Fe}}\\ &=0.34-(-0.44)\\ &=\boxed{+0.78\text{ V}} \end{aligned} \]
For silver deposition:
\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=E^\circ_{\ce{Ag+/Ag}} -E^\circ_{\ce{Fe^2+/Fe}}\\ &=0.80-(-0.44)\\ &=\boxed{+1.24\text{ V}} \end{aligned} \]
Both standard cell potentials are positive.
Therefore, both forward displacement reactions are thermodynamically favourable under standard conditions.
(c) Determine which solid metal is more reactive.
Copper is more reactive than silver.
The standard reduction potentials are:
\[ E^\circ_{\ce{Cu^2+/Cu}}=+0.34\text{ V} \]
\[ E^\circ_{\ce{Ag+/Ag}}=+0.80\text{ V} \]
The copper reduction potential is less positive.
This means copper is more readily oxidised than silver when the two metals are compared using their respective metal-ion couples.
Consequently, copper metal can displace silver ions:
\[ \ce{Cu(s) + 2Ag+(aq) -> Cu^2+(aq) + 2Ag(s)} \]
Its standard cell potential is:
\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=0.80-0.34\\ &=\boxed{+0.46\text{ V}} \end{aligned} \]
The reaction is thermodynamically favourable.
This is the important distinction: the more positive reduction potential belongs to the species that is more readily reduced, not to the solid metal that is more readily oxidised.
(d) Evaluate the student’s three claims.
The first claim is not justified.
Silver ions have a greater standard tendency to gain electrons than copper(II) ions. However, a larger standard reduction potential does not establish which reaction will occur first or proceed faster in a real mixture.
The second claim is also incorrect.
Both reactions are thermodynamically favourable. Copper(II) ions do not have to wait until every silver ion has been consumed before they can receive electrons.
The two reduction processes can occur at different locations on the same iron surface, potentially at the same time.
Their relative rates depend on factors such as ion concentrations, transport of ions to the surface, and surface conditions.
The third claim is not justified either.
A reddish-brown coating is consistent with the formation of copper, but visual inspection does not establish that the coating contains only copper.
Silver may also have deposited in a form or location that cannot be distinguished by the observation.
The correct conclusion is that the visible deposit provides evidence consistent with copper formation. It does not prove that silver ions were not reduced.
(e) Suggest a direct experiment.
Place a clean copper strip into a separate solution containing silver ions.
If silver is deposited and copper enters solution as copper(II) ions, the reaction is:
\[ \ce{Cu(s) + 2Ag+(aq) -> Cu^2+(aq) + 2Ag(s)} \]
This directly establishes that copper displaces silver.
Therefore:
\[ \boxed{\ce{Cu > Ag}} \]
This experiment compares the two metals directly rather than relying on their competing reactions with a third metal.
The deeper lesson is that a standard reduction potential table tells you which direction electron transfer is thermodynamically favoured. It does not tell you the order, speed, or exact appearance of competing reactions.
Once you can predict which metal displaces another, the next useful step is to use the same oxidation and reduction half-equations to understand galvanic cells. The same electron-transfer processes that drive displacement reactions are responsible for the voltage produced by an electrochemical cell.
Syllabus context: This article covers metal reactivity, displacement reactions, standard reduction potentials, and reaction-rate considerations from the Chemistry Stage 6 Syllabus (2017), Year 11 Module 3: Reactive Chemistry. NESA’s Chemistry 11-12 Syllabus (2025) is scheduled for implementation from 2028, with its first HSC examination in 2029.