Types of reactions — combination, decomposition, displacement, double displacement, redox: Question bank

True or false — justify

Every combination reaction is also a redox reaction.

False.

C a O + H_{2} O \to C a (O H)_{2}

combines two things into one, but every element keeps its oxidation number (

C a

stays

+ 2

O

stays

- 2

H

stays

+ 1

), so no electrons move — it is combination but not redox.

Every displacement reaction is a redox reaction.

True. A free element (oxidation number

0

) becomes an ion inside a compound while another ion becomes a free element — those numbers must change, so electrons are always transferred. See Oxidation Number Rules.

Double displacement reactions are usually redox.

False, and it's almost the opposite. In a partner-swap like

B a C l_{2} + N a_{2} S O_{4}

every ion (

B a^{2 +}, C l^{-}, N a^{+}, S O_{4}^{2 -}

) walks away with the same charge it arrived with. No electrons transfer, so double displacements are almost never redox.

Decomposition always needs heat.

False. Decomposition needs energy, but the energy can come from electricity (electrolysis of water) or light (photolysis of

A g B r

), not only heat. "Needs energy input" ≠ "needs a flame."

A reaction that releases a gas must be redox.

False.

N a_{2} C O_{3} + 2 H C l \to 2 N a C l + H_{2} O + C O_{2}

fizzes off

C O_{2}

but carbon stays

+ 4

throughout — the gas leaving drives a double displacement, no electron transfer.

In every redox reaction, oxidation and reduction both happen.

True. Electrons lost by one species have to be gained by another; they cannot vanish. That is exactly why we write two half-reactions that share the same electron count.

The oxidising agent is the substance that gets oxidised.

False — it's the reverse. The oxidising agent causes oxidation in the other species, so it itself gets reduced. The agent always undergoes the opposite of what its name suggests it does to others.

Spot the error

C u + F e S O_{4} \to C u S O_{4} + F e

is a valid displacement."

Wrong direction. Copper sits below iron in the Reactivity Series of Metals, so it cannot kick iron out of its salt. Only the more reactive metal displaces the less reactive one, so this reaction simply does not occur.

N a C l + K N O_{3} \to N a N O_{3} + K C l

is a double displacement that goes to completion."

No reaction, really. All four species are soluble salts, so nothing leaves the solution (no precipitate, gas, or water). Without a driving force the ions just mingle — no net reaction.

2 H_{2} O \to 2 H_{2} + O_{2}

is a combination because two waters make gases."

It's decomposition. One kind of compound (

H_{2} O

) is broken apart into simpler substances by electricity — that is the

A B \to A + B

pattern, the exact opposite of combination.

"In

2 M g + O_{2} \to 2 M g O

, oxygen is oxidised because it's in the name 'oxidation.'"

Oxygen is reduced here.

O

goes from

0

- 2

(gains electrons), so it is reduced; magnesium goes

0 \to + 2

and is oxidised. "Oxidation" being named after oxygen is historical, not a rule about oxygen's own fate.

H C l + N a O H \to N a C l + H_{2} O

is neutralisation, so it can't be a double displacement."

It is both. Neutralisation is a special case of double displacement:

H^{+}

and

O H^{-}

swap partners to form water, which is the driving force. See Acids, Bases and Salts.

"Free chlorine gas

C l_{2}

has oxidation number

- 1

It's $0$ . Any element bonded only to identical atoms (a free element) has oxidation number

0

. Chlorine is

- 1

only inside a compound like

N a C l

Why questions

Why is redox called an "overlay" rather than a sixth reaction shape?

The other four patterns describe how atoms are arranged (

A + B \to A B

, etc.), while redox describes whether electrons transfer. They are different axes, so one reaction can be, say, "combination and redox" at once.

Why does an iron nail in blue

C u S O_{4}

turn the solution pale green?

Iron displaces copper: the blue

C u^{2 +}

ions leave solution as reddish metal on the nail, while pale-green

F e^{2 +}

ions enter solution. The colour change is direct evidence of the ion swap.

Why must the electrons lost equal the electrons gained in a redox reaction?

Because charge, like mass, is conserved — every electron that leaves one species must land on another. This equality is what forces the coefficients to balance in redox equations.

Why do most combination reactions release heat?

Atoms combine to reach lower-energy, more stable electron arrangements, and the "extra" energy is given off as heat — so combinations are typically exothermic. See Exothermic and Endothermic Reactions.

Why does forming a precipitate, gas, or water "drive" a double displacement forward?

Each of those products leaves the free-ion pool — a solid settles out, a gas escapes, water is un-ionised — so those ions can't recombine into the reactants. Removing a product makes the swap permanent.

Why is the reactivity series the "referee" for displacement?

It ranks metals by how readily they give up electrons. Displacement only happens when the incoming element gives up electrons more readily (sits higher) than the one it evicts, so the series predicts whether the reaction runs at all.

Why do we pretend every bond is fully ionic when assigning oxidation numbers?

It's a bookkeeping trick: by handing each shared pair to the more electronegative atom, we get a clean whole-number "charge" to track. It isn't the real charge — it's a consistent way to spot who lost or gained electrons.

Edge cases

Can a single reaction be decomposition and redox at the same time?

Yes.

2 H_{2} O electricity 2 H_{2} + O_{2}

splits one compound (decomposition) and changes oxidation numbers (

H : + 1 \to 0

O : - 2 \to 0

), so it is redox too. See Electrolysis.

Zinc reacts with dilute

H C l

to give

H_{2}

— is this displacement even though acid isn't a "metal salt"?

Yes. Zinc sits above hydrogen in the reactivity series, so it displaces hydrogen from the acid:

Z n + 2 H C l \to Z n C l_{2} + H_{2}

. Hydrogen behaves as the "less reactive element" being pushed out.

Gold (

A u

) is dropped into

C u S O_{4}

solution — what happens?

Nothing. Gold is far below copper in the reactivity series, so it can't displace

C u^{2 +}

. A metal only displaces one that is less reactive than itself; a less-reactive metal is inert here.

H_{2} + C l_{2} \to 2 H C l

combination, redox, or both?

Both. Two elements join into one product (combination) while

H

goes

0 \to + 1

and

C l

goes

0 \to - 1

(redox). This is a clean example of the two classifications applying together.

Photolysis of

A g B r

produces grey specks — which species is reduced, and why does it matter?

Silver is reduced:

A g^{+} \to A g^{0}

(gains an electron) as light supplies energy, forming the dark metallic image used in photography.

B r^{-}

is oxidised to

B r_{2}

, keeping the electron count balanced.

What is the oxidation number of oxygen in a peroxide such as

H_{2} O_{2}

, and why is this an exception?

It's $- 1$ , not the usual $- 2$ . In peroxides two oxygens share a bond with each other, and that shared pair splits evenly (same electronegativity), leaving each oxygen with only

- 1

. Always check for O–O bonds before assuming

- 2

Recall One-line self-test before you close the page

Give a reaction that is combination-but-NOT-redox, and one that is decomposition-AND-redox. Combination not redox ::: $C a O + H_{2} O \to C a (O H)_{2}$ (no oxidation numbers change). Decomposition and redox ::: $2 H_{2} O electricity 2 H_{2} + O_{2}$ ( $H$ and $O$ both change oxidation number).