Displacement reactions

To choose correctly, we must separate different categories of impacts. In IB science, a social impact is about how an action affects people—their health, jobs, living conditions, community life, or culture.

That is not the same as environmental impacts (effects on air, water, land, and ecosystems) or economic impacts (money, costs, and trade).

Now identify choices focused on the natural world rather than humans.

Option A is about changing the atmosphere by increasing $C{O}_2$. Option D is about polluting water with $heavymetals$. Both describe damage to Earth systems, so they are environmental impacts, not social ones.

Option B talks about habitats being destroyed and biodiversity falling. Even though this is serious, it is still an ecological/environmental impact because it mainly concerns living things and ecosystems other than humans.

Option C describes communities being pushed out of their homes. Because it involves human populations and disrupts where they live and how they maintain their way of life, it matches the definition of a social impact.

To get this right, we need to tell apart social impacts from environmental ones. A social impact is an effect that changes how people live and work, how communities function, and how groups maintain their culture and meet their needs.

We can test each option by asking: is it mainly about nature, or mainly about human society?

Option A describes extra $C{O}_2$ emissions from mining equipment. This changes the atmosphere and contributes to the greenhouse effect. Since it primarily concerns climate and Earth systems, it is an environmental (atmospheric) impact, not a social one.

Option B focuses on habitat destruction and reduced biodiversity—changes to plants, animals, and ecosystem stability. That is an environmental (ecological) impact because it is about non-human species and their habitats.

Option D is about heavy metals entering groundwater. This is pollution that harms a natural resource and degrades water quality. Even if people could later be affected by drinking it, the impact described here is classified directly as an environmental (hydrological) impact.

After removing the environmental options, Option C remains. Forcing communities to move away from their land can break up livelihoods, traditions, and social networks. Because it directly affects humans and society, it is the correct social impact.

To explain the energy gap, we need to set the two pathways side by side: Primary Extraction (making aluminium from an ore) versus Recycling (reprocessing aluminium that is already a metal).

In primary extraction, aluminium is not found as pure metal; it is present as aluminium oxide ($A{l}_2{O}_3$). The $A{l}^{3+}$ and $O^{2-}$ ions attract strongly, forming very strong ionic bonding. Separating these ions and turning $A{l}^{3+}$ into aluminium requires electrolysis, which needs a very large electrical energy input to break those chemical bonds.

Recycling aluminium is mainly a physical change. The aluminium in a can or foil is already metallic aluminium, so we do not need to break it apart from oxygen. The main energy cost is heating it until it melts, which means overcoming metallic lattice forces. That takes far less energy than breaking the ionic bonds in an oxide.

The other statements don’t match what we know about aluminium and recycling:

• A is incorrect because aluminium is too reactive to be obtained by carbon reduction.
• B is incorrect because catalysts cannot make aluminium come out of its oxide at room temperature.
• C is incorrect because recycled aluminium is already metallic, not made of ions in a compound.

The reason recycling saves about 95% of the energy is that melting aluminium is only a change of state (physical), while extracting aluminium from $A{l}_2{O}_3$ is a chemical process requiring electrolysis to break extremely strong bonds.

To see why recycling is cheaper in energy, first think about how aluminium is produced in the first place. The ore is bauxite, and aluminium is locked up as aluminium oxide ($A{l}_2{O}_3$) because aluminium is very reactive and forms strong ionic attraction with oxygen.

To get aluminium metal, industry must use electrolysis. The oxide has to be molten (high melting point) and then a large electric current is used to reduce $A{l}^{3+}$ ions to aluminium metal ($Al$). This consumes a lot of energy.

Option A suggests carbon is used as the reducing agent. But in the reactivity series, aluminium is above carbon, so carbon cannot remove oxygen from aluminium oxide.

If carbon reduction worked for aluminium, it would be used for primary extraction (as it is for iron), but it does not. Also, recycling is not about reducing an oxide back to metal; it is about reprocessing aluminium that is already metallic.

Option B claims catalysts allow aluminium to be obtained at room temperature during recycling. That doesn’t fit the chemistry:

1. Recycling is mainly melting and reshaping metal, not separating aluminium from an oxide.
2. There is no industrial catalyst that makes aluminium come out of $A{l}_2{O}_3$ at room temperature; extracting it still requires a high-energy electrolytic method.

Option C says recycled aluminium is already ionic. But objects like aluminium foil or a drinks can are aluminium metal, $Al(s)$, not $A{l}^{3+}$ ions.

If the aluminium really were present as ions (as in a salt), electrons would still have to be supplied to form the metal again, which would take nearly as much energy as primary extraction. Because recycled aluminium is already metal, that electron transfer step isn’t needed.

After removing the incorrect ideas, Option D matches the science.

Recycling is mostly a physical process: heat the metal until it melts and then shape it again. Primary extraction is a chemical process: electricity is used to break the very strong bonding in $A{l}_2{O}_3$ and reduce ions to aluminium.

So, the energy to melt aluminium is much smaller than the energy needed to break chemical bonds and carry out electrolysis.

We’re told the new metal is more reactive than carbon but less reactive than potassium.

In extraction chemistry, whether a metal is above or below carbon in the reactivity series is the key detail that decides if we can use carbon-based reduction or if we need electrolysis. This links directly to the Reactivity series and Extraction ideas in your course.

A core metallurgy rule is: if a metal is more reactive than carbon, then carbon (or carbon monoxide) cannot remove oxygen from that metal’s oxide.

So any method that depends on blast-furnace style reduction using CO will not work here, so that option is eliminated.

Now consider the remaining choices:

• Hydrogen gas (Option A): Hydrogen is less reactive than carbon, so if carbon can’t reduce this metal’s oxide, hydrogen won’t be able to either.
• Physical separation (Option B): This only works when a metal is so unreactive that it can be found as the pure element (like gold). A metal above carbon will be too reactive to exist naturally as a native metal—it will be in compounds.

Because the metal is above carbon, its compounds are too stable for chemical reducing agents like carbon or hydrogen to break apart.

To extract it, we must supply energy using electrolysis of a molten compound. This is the approach used for other fairly reactive metals such as aluminum and magnesium.

Start by locating the metal in the reactivity series. The prompt tells us it is:

• More reactive than carbon
• Less reactive than potassium

Metals in this middle-high region (like magnesium or aluminum) usually form very stable compounds because they react strongly with oxygen and other non-metals. We’ll apply Elimination of Extraction Techniques to decide the best method.

Methods such as panning only succeed for native metals, meaning metals found naturally as pure elements. Only very unreactive metals (like gold or silver) can be native.

Since our metal is quite reactive, it will be present as compounds in ores, not as pure metal, so this option is eliminated.

Carbon monoxide reduction and using hydrogen gas both require the reducing agent (C, CO, or $H_2$) to be able to take oxygen away from the metal compound.

But because the metal is more reactive than carbon, carbon (and CO made from it) cannot remove the oxygen from the oxide. Hydrogen is also not reactive enough to reduce oxides of metals in this range.

After removing physical separation and chemical reduction, the remaining method is electrolysis.

For metals above carbon, electricity is needed to break the strong bonds in the molten ionic compound and release the metal. That is why this is the only practical and chemically possible extraction route.

To choose an extraction method, we begin with the Reactivity Series, which orders metals from most reactive to least reactive. Carbon is often placed in this list as a useful benchmark because it is an inexpensive substance that can remove oxygen from some metal oxides.

The metal is stated to be below carbon. That tells us it is less reactive than carbon. A more reactive element holds onto oxygen more strongly, so carbon can take oxygen from the metal oxide. In other words, carbon can reduce the oxide, freeing the metal in a redox process.

Getting a metal from its ore usually requires a chemical reaction because the metal is bonded to oxygen (often as an oxide).

• Fractional distillation is used to separate liquids in a mixture based on boiling points (for example, fractions from crude oil).
• Filtration can remove solid waste rock (gangue) from an ore but it cannot undo the chemical bonding in a metal oxide.

Electrolysis is an energy-demanding (and costly) approach mainly needed for metals that are more reactive than carbon (such as sodium or aluminum). Since this metal is below carbon, carbon can do the job, so electrolysis is not the typical method.

Because the metal is less reactive than carbon, extraction is usually done by reduction. In a hot furnace, carbon (or carbon monoxide) reacts with the metal oxide: carbon is oxidized and the metal ions gain electrons to become the metal.

$\text{Metal Oxide}+\text{Carbon}\to \text{Metal}+\text{Carbon Dioxide}$

Which extraction process works depends on where the metal sits in the reactivity series. Carbon is treated as a reference point: if a metal is below carbon, carbon is reactive enough to remove it from its compounds.

Fractional distillation separates liquids with different boiling points (like the components of crude oil). Metal oxides are generally ionic solids with very high melting/boiling points, so this approach does not work for extracting a metal from its oxide.

Electrolysis is effective but expensive, so it is mainly used for metals above carbon in the reactivity series (for example, aluminum and sodium). Those metals are too reactive to be displaced by carbon. Since this metal is below carbon, a cheaper chemical method is normally used instead.

Filtration can help remove insoluble waste or separate gangue from ore, but it only separates physical mixtures. It cannot separate a metal from oxygen when they are chemically combined in a metal oxide.

Since the metal is less reactive than carbon, carbon (or carbon monoxide) can act as a reducing agent and remove oxygen from the metal oxide, producing the metal in a redox reaction:

$\text{Metal Oxide}+\text{Carbon}\to \text{Metal}+\text{Carbon Dioxide}$

This is the usual, cost-effective technique for metals such as iron, zinc, and lead.

Producing aluminium from its ore, bauxite, takes a huge amount of energy. The aluminium compounds must be broken apart using electrolysis of molten aluminium oxide (alumina), which needs very high temperatures and large amounts of electrical energy to overcome strong bonding in the compound.

When aluminium is recycled, the metal already exists as aluminium atoms in a solid piece of scrap. Recycling mainly requires melting and cleaning, not splitting a compound by electrolysis. Because aluminium melts at about $660^{\circ }\text{C}$, the process can be done with far less energy than primary extraction.

A commonly learned sustainability statistic is that recycling aluminium uses only 5% of the energy needed to obtain the same mass of aluminium from bauxite. In other words, recycling avoids about 95% of the energy demand.

Check each option against the known value:

• 95% describes energy saved (roughly), not the energy used.
• 25% and 50% still suggest recycling uses a large chunk of the original energy, which doesn’t fit how efficient aluminium recycling is.

Recycling aluminium typically needs about 5% of the energy compared with extracting it from bauxite, which is why recycling aluminium is considered extremely sustainable.

The prompt emphasizes that recycling aluminium is highly sustainable. In chemistry and industry, that usually means the process uses much less energy and creates less environmental harm. So the energy needed for recycling should be a small fraction of the energy needed for extraction.

To match “highly sustainable,” we should look for the smallest energy percentage.

• $95\%$ would mean recycling is almost as energy-hungry as extraction.
• $50\%$ and $25\%$ are better, but still not the dramatic reduction aluminium recycling is known for.

If recycling really required $95\%$, $50\%$, or even $25\%$ of the extraction energy, it wouldn’t be the standout example of energy saving that it is. So those higher percentages can be eliminated.

The smallest option is $5\%$. That corresponds to saving about $95\%$ of the energy compared with the Hall-Héroult extraction from bauxite, so $5\%$ is the correct choice.