The reactivity series

We need to decide which mathematical relationship links how reactive two electrode metals are to the voltage you measure in a simple cell.

A simple electrochemical cell uses two different metals as electrodes. The Reactivity Series tells us metals vary in how strongly they tend to lose electrons.

When the metals are more widely separated in the reactivity series, electrons are driven more strongly from one metal to the other, so the measured voltage is larger.

The standard cell potential ($E_{cell}^{\theta }$) is found from the standard electrode potentials ($E^{\theta }$) of the cathode and anode using:

$E_{cell}^{\theta }=E_{cathode}^{\theta }-E_{anode}^{\theta }$

Because this is a subtraction, the value you get is the difference between the two potentials—showing the voltage comes from how far apart they are.

If we used a sum, we would be combining the two reactivities by addition, which doesn’t represent how separated the metals are.

A product (multiplying) or a ratio (dividing) are not the operations used to find the voltage of simple electrochemical cells. Using them would not match standard chemistry methods and would give values with no sensible meaning here.

From the reactivity series idea and the subtraction in the cell potential formula, the voltage increases with the difference in reactivity between the two metals.

Focus on the phrase potential difference. In both maths and science, “difference” means the result of subtracting one value from another.

A simple electrochemical cell uses two different metals, and each has its own reactivity level. Since the output is described as a “difference,” it points to the voltage coming from the difference (the gap) between the two reactivity levels.

This matches the Reactivity Series (in the Types of chemical reaction topic): metals that are farther apart on the series produce a larger voltage.

The other operations don’t match what “difference” means:

• Sum: would mean adding reactivities ($A+B$), not subtracting.
• Product: would mean multiplying ($A\times B$).
• Ratio: would mean dividing ($\frac{A}{B}$).

Because the question explicitly refers to “potential difference,” the best match is the difference in reactivity between the two metals.

In an electrochemical cell, two different metals act as electrodes. Each metal has a different tendency to lose electrons—its reactivity. The voltage (potential difference) reflects the electron-driving force created when one metal gives up electrons much more readily than the other.

This connects to the Reactivity Series and Redox reactions, which are covered further in the Types of chemical reaction topic of your syllabus.

Suppose voltage were a sum of reactivities. Then combining any two metals would always boost the voltage compared with just one metal.

But if both electrodes are the same metal (for example, two Copper strips), the voltage is $0\text{ V}$ because there is no driving force between identical electrodes. If addition were correct, you wouldn’t get zero. So sum cannot be right.

Now try product and ratio—they don’t fit how voltage behaves.

1. Product: If one metal’s reactivity were treated as ‘0’ on some scale, multiplying would force the voltage to be 0 no matter how reactive the other metal is, which doesn’t reflect real cells.
2. Ratio: A ratio would suggest that scaling both reactivities up by the same factor wouldn’t change the voltage.

In reality, voltage corresponds to a separation (a ‘gap’) between two positions on the reactivity series.

So the correct operation is difference. The greater the separation between the metals in the reactivity series (and therefore between their electrode potentials), the larger the voltage.

For instance, in a standard Zinc-Copper cell, we subtract the anode potential from the cathode potential:

$\text{Cell Potential}=E_{\text{cathode}}-E_{\text{anode}}$

If Copper is $+0.34\text{ V}$ and Zinc is $-0.76\text{ V}$, the voltage comes from the difference between those two numbers.

Because adding, multiplying, or dividing can’t explain zero voltage with identical metals or the effect of being farther apart in the reactivity series, the voltage must depend on the difference.

Potassium ($K$) and sodium ($Na$) sit in Group 1, the alkali metals. They each have one outer electron, and they tend to lose it extremely easily to reach a more stable electron arrangement.

From the IB content (Topic 3: Periodicity), Group 1 metals become more reactive down the group. Sodium and potassium are therefore very reactive, so they react almost immediately with oxygen ($O_2$) and water vapor ($H_{2}O$) in air. That is why they are stored under oil to keep air away from the metal.

If sodium and potassium reacted slowly, there wouldn’t be a strong safety reason to keep them covered by oil. Instead, sodium quickly dulls/tarnishes when exposed to air, and potassium reacts even more quickly than sodium.

We can also reject the remaining descriptions using chemistry:

1. Endothermically: These reactions release lots of energy, so they are strongly exothermic. Potassium can release enough heat that the hydrogen produced may ignite. $\begin{array}{rl}2\text{K(s)}+2{\text{H}}_2\text{O(l)}& \to 2\text{KOH(aq)}+{\text{H}}_2\text{(g)}+\text{heat}\end{array}$

2. Reversibly: These are forceful redox reactions that proceed to products; the oxides/hydroxides formed do not naturally convert back into the pure metal on their own.

Because the reaction with air begins almost at once and is very intense, the most accurate choice is rapidly.

The key detail is that potassium and sodium are stored under oil. In labs, unusual storage like this is typically used when a substance must be protected from reacting with its surroundings.

The oil layer blocks the metal from contacting the oxygen and moisture in the air. If leaving these metals in air were safe, the oil wouldn’t be necessary. Needing that barrier suggests the air reaction starts quickly and can be hazardous.

If the metals reacted slowly, there would be no pressing need to keep them submerged. For instance, iron reacts slowly with air (rusting), yet we don’t store iron nails under oil in school laboratories. So sodium and potassium must not react slowly.

Consider the other choices:

• Endothermically: That would mean absorbing heat, but the reason for oil storage is to stop the reaction from happening at all, not to manage whether it takes in or gives out heat.
• Reversibly: Reactions of alkali metals with air form stable oxides/hydroxides and generally do not run backward to regenerate the metal, so this doesn’t match what we observe.

Since the metal must be shielded immediately by oil to remain unchanged, it implies it reacts rapidly as soon as it is exposed to air. Therefore, 'rapidly' is the only option that fits the storage method.

Start by focusing on what was observed. Heating lead(II) oxide with zinc produces zinc oxide and lead metal.

We can represent this with a chemical equation:

$\text{Lead(II) oxide}+\text{Zinc}\to \text{Zinc oxide}+\text{Lead}$ $PbO(s)+Zn(s)\to ZnO(s)+Pb(l/s)$

Here, zinc removes oxygen from the lead compound. That oxygen-transfer is a typical displacement reaction where one metal takes over from another in an oxide.

The Reactivity Series Principle says: a more reactive metal can force a less reactive metal out of its oxide (or salt).

A helpful way to picture it is a tug-of-war for oxygen. The metal that is more reactive can “grab” the oxygen and form its own oxide.

Because zinc successfully takes oxygen from PbO, zinc must have the stronger attraction for oxygen in this pair.

If lead were actually more reactive than zinc, then PbO would not give up its oxygen to zinc. Heating the mixture would not lead to zinc oxide forming.

Since the reaction does happen, any choice stating that lead is more reactive must be incorrect.

Now consider the other distractors:

• Stability: A very stable oxide is less likely to be converted into a different oxide. The energetic, vigorous nature of the change supports the idea that forming $ZnO$ is favorable, meaning $ZnO$ is the more stable oxide here (not $PbO$).
• Noble metals: Noble metals are known for hardly reacting or oxidizing. A metal that causes a strong displacement reaction on heating is not behaving like a noble metal.

Because zinc displaces lead from lead(II) oxide (forming $ZnO$ and leaving lead metal), the evidence shows that zinc is more reactive than lead. That means zinc is positioned above lead in the reactivity series.

Begin with the reaction equation given by the description:

$\text{PbO (s)}+\text{Zn (s)}\to \text{ZnO (s)}+\text{Pb (s)}$

This fits a displacement reaction pattern: a more reactive metal replaces a less reactive metal in a compound. That idea links directly to the reactivity series.

A noble metal (such as gold or platinum) is very unreactive and does not readily form oxides. The situation describes a strong, energetic reaction, which is the opposite of what you would expect for a noble metal.

So the statement about zinc being a noble metal cannot be right.

If lead were more reactive than zinc, then lead would keep oxygen more strongly and zinc would fail to take it away. The products would not show zinc oxide being formed from PbO.

Because zinc does take the oxygen, lead cannot be the more reactive metal.

Chemical changes usually proceed toward forming more stable products. Since the process converts $PbO$ into $ZnO$, it suggests that $ZnO$ is the more stable oxide in this comparison.

Therefore, the claim that $PbO$ is more stable than $ZnO$ contradicts what the reaction indicates.

After eliminating the incorrect choices, the remaining conclusion matches the displacement rule: zinc replaces lead from its oxide, so zinc must be higher in the reactivity series than lead.

Galvanizing is a way to stop corrosion of iron or steel by covering it with a layer of zinc ($Zn$). Like paint, the coating blocks oxygen and water from reaching the metal surface. However, zinc also adds an extra layer of protection that depends on redox reactions, not just a physical seal.

To see why zinc helps, we use the reactivity series. Zinc is placed above iron ($Fe$), so zinc is more reactive. That means zinc more readily loses electrons (is oxidized) compared with iron.

Since zinc is more reactive than iron, it can act as a sacrificial metal. If the object gets scratched and iron is exposed, the zinc will oxidize in preference to the iron.

The zinc atoms lose electrons to form ions:

$Zn(s)\to Z{n}^{2+}(aq)+2e^{-}$

Those electrons go to the iron, stopping the iron from being oxidized by oxygen and water (which would produce rust). Paint only works as a barrier, so once the paint layer is broken, it cannot keep protecting the exposed iron.

Check the remaining answers using the reactivity series and basic chemical behavior:

• Option A talks about price, which isn’t a chemical reason for better corrosion protection.
• Option C is wrong because paint is intended to be mostly unreactive; it shouldn’t chemically attack iron to make rusting faster.
• Option D has the reactivity backwards: zinc is not less reactive than iron. Zinc’s higher reactivity is exactly what makes sacrificial protection possible.

Because zinc oxidizes instead of iron, the steel stays protected even if the coating is scratched. This sacrificial protection is the key advantage of galvanizing compared with paint.

Begin with the reactivity series. Zinc ($Zn$) is positioned above iron ($Fe$), so zinc is more reactive than iron.

Option D says zinc is less reactive and also claims it can resist every chemical attack. Since zinc is actually more reactive, and no common metal is completely immune to all chemical reactions, we can remove this option.

Paint works as a passive physical barrier that blocks air and water from the metal.

Option C suggests paint reacts with iron and makes rusting faster. That would mean painting metal objects would cause them to corrode more quickly, which contradicts why paint is used. So this option doesn’t make sense and should be eliminated.

When asked for the 'primary advantage' in science, the expected answer is usually about how it works chemically or functionally, not about cost.

Prices can change, and galvanizing is not chosen mainly because it is cheaper than paint. So Option A doesn’t address the key scientific advantage and can be removed.

After eliminating the others, Option B remains. It describes sacrificial protection (also called cathodic protection).

Because $Zn$ is more reactive than $Fe$, zinc will oxidize first if the surface is damaged, supplying electrons that help prevent the iron from oxidizing and forming rust. With paint, a scratch exposes bare iron, and there is no backup chemical protection.