Identify the substance oxidised reduced, oxidising agent and reducing agent for each of the following reaction

Exercises · 7.13

Q. Identify the substance oxidised reduced, oxidising agent and reducing agent for each of the following reactions:

(a) 2AgBr(s) + C6H6O2(aq) → 2Ag(s) + 2HBr(aq) + C6H4O2(aq)

(b) HCHO(l) + 2[Ag(NH3)2]+(aq) + 3OH–(aq) → 2Ag(s) + HCOO–(aq) + 4NH3(aq) + 2H2O(l)

(d) N2H4(l) + 2H2O2(l) → N2(g) + 4H2O(l)

(e) Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)

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Concept understanding — Oxidation Reduction

Let’s start with something you already know from everyday life.

The intuition: what does “oxidation” really mean?

Think of a piece of iron left out in the rain. Over time, it turns into reddish-brown rust. Or think of a slice of apple turning brown when you leave it on the table. Or a fire burning wood to ash. In all these cases, something is combining with oxygen — that’s the original meaning of “oxidation.” The iron combines with oxygen from the air to form iron oxide (rust). The apple’s chemicals react with oxygen in the air. The wood burns because carbon in the wood combines with oxygen.

So the first, simplest idea: oxidation = adding oxygen. And the reverse — taking oxygen away — was called reduction. For example, if you heat iron oxide with carbon, the carbon steals the oxygen away, leaving pure iron. That’s reduction: removing oxygen.

But chemists soon realised this was too narrow. Many reactions that look like oxidation-reduction don’t involve oxygen at all. For instance, when sodium metal reacts with chlorine gas to make table salt, no oxygen is involved — yet the sodium clearly “rusts” in a sense, and the chlorine “steals” something from it.

So the definition had to be broadened.

The precise modern definition: electron transfer

Here’s the clean, exam-ready statement:

Oxidation is the loss of electrons by a substance.

Reduction is the gain of electrons by a substance.

They always happen together — you cannot have one without the other. That’s why we call them redox reactions (short for reduction-oxidation).

Let’s see this with the sodium-chlorine example:

Sodium atom ( $\text{Na}$ ) loses one electron to become $\text{Na}^+$ . That’s oxidation.
Chlorine atom ( $\text{Cl}$ ) gains that electron to become $\text{Cl}^-$ . That’s reduction.

You can write the two halves separately:

\text{Na} \rightarrow \text{Na}^+ + e^- \quad (\text{oxidation})

\text{Cl} + e^- \rightarrow \text{Cl}^- \quad (\text{reduction})

Add them together:

\text{Na} + \text{Cl} \rightarrow \text{Na}^+ + \text{Cl}^-

That’s table salt.

Tip

A handy mnemonic: OIL RIG — Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons).

How to spot a redox reaction without seeing electrons

You can’t watch electrons move directly. So chemists use oxidation numbers (also called oxidation states) — a bookkeeping system that tracks electrons.

Rules (simplified for first-time learners):

An atom in its elemental form has oxidation number 0.
A monatomic ion has oxidation number equal to its charge (e.g., $\text{Na}^+$ is +1, $\text{Cl}^-$ is -1).
Oxygen is usually -2 (except in peroxides).
Hydrogen is usually +1 (except in metal hydrides).
The sum of oxidation numbers in a neutral compound is 0; in a polyatomic ion, it equals the ion’s charge.

Then:

Oxidation = increase in oxidation number.
Reduction = decrease in oxidation number.

Example: Rusting of iron.

4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3

Fe starts at 0 (elemental). In $\text{Fe}_2\text{O}_3$ , each Fe is +3. So Fe’s oxidation number goes up from 0 to +3 → oxidation.
O starts at 0 (in $\text{O}_2$ ). In $\text{Fe}_2\text{O}_3$ , each O is -2. So O’s oxidation number goes down from 0 to -2 → reduction.

Watch out

A common mistake: thinking that “reduction” means something becomes smaller or less. It doesn’t — it’s about gaining electrons (or losing oxygen, in the old sense). The name comes from metallurgy: when you “reduce” iron ore to iron, you’re taking away oxygen, so the mass reduces.

One more way to think about it …

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Concept: Oxidation–Reduction (Redox) Analysis

Track oxidation states of key atoms in reactants and products. Oxidation = increase in oxidation state (loss of electrons); reduction = decrease (gain of electrons). The substance oxidised is the reducing agent; the substance reduced is the oxidising agent.

(a) $2\text{AgBr (s)} + \text{C}_6\text{H}_6\text{O}_2\text{(aq)} \to 2\text{Ag(s)} + 2\text{HBr (aq)} + \text{C}_6\text{H}_4\text{O}_2\text{(aq)}$

Ag goes from $+1$ (in AgBr) to $0$ (Ag metal) → reduced.

The organic compound (hydroquinone) loses H atoms, effectively oxidising to quinone → oxidised.

Role	Species
Oxidised (reducing agent)	$\text{C}_6\text{H}_6\text{O}_2$
Reduced (oxidising agent)	$\text{AgBr}$ (specifically Ag $^+$ )

(b) $\text{HCHO(l)} + 2[\text{Ag(NH}_3)_2]^+\text{(aq)} + 3\text{OH}^-\text{(aq)} \to 2\text{Ag(s)} + \text{HCOO}^-\text{(aq)} + 4\text{NH}_3\text{(aq)} + 2\text{H}_2\text{O(l)}$

Ag goes from $+1$ to $0$ → reduced.

Formaldehyde (HCHO) oxidises to formate (HCOO $^-$ ); carbon goes from $0$ to $+2$ → oxidised.

Role	Species
Oxidised (reducing agent)	$\text{HCHO}$
Reduced (oxidising agent)	$[\text{Ag(NH}_3)_2]^+$ (Ag $^+$ )

(c) $\text{HCHO (l)} + 2\text{Cu}^{2+}\text{(aq)} + 5\text{OH}^-\text{(aq)} \to \text{Cu}_2\text{O(s)} + \text{HCOO}^-\text{(aq)} + 3\text{H}_2\text{O(l)}$

Cu goes from $+2$ to $+1$ (in Cu $_2$ O) → reduced.

HCHO oxidises to HCOO $^-$ → oxidised.

Role	Species
Oxidised (reducing agent)	$\text{HCHO}$
Reduced (oxidising agent)	$\text{Cu}^{2+}$

(d) $\text{N}_2\text{H}_4\text{(l)} + 2\text{H}_2\text{O}_2\text{(l)} \to \text{N}_2\text{(g)} + 4\text{H}_2\text{O(l)}$ …

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Track oxidation-state changes to identify what loses electrons (oxidised, reducing agent) and what gains electrons (reduced, oxidising agent). In each reaction: (a) Ag⁺ reduced, C₆H₆O₂ oxidised; (b) Ag⁺ reduced, HCHO oxidised; (c) Cu²⁺ reduced, HCHO oxidised; (d) N₂H₄ oxidised, H₂O₂ reduced; (e) Pb oxidised, PbO₂ reduced.

The Core Idea: Following the Electrons

Redox reactions are fundamentally about electron transfer. The substance that loses electrons is oxidised (its oxidation state increases) and acts as the reducing agent because it donates electrons to something else. The substance that gains electrons is reduced (its oxidation state decreases) and acts as the oxidising agent because it accepts electrons.

The strategy is simple: assign oxidation states to key atoms on both sides, spot the changes, then label accordingly.

(a) $2\text{AgBr (s)} + \text{C}_6\text{H}_6\text{O}_2\text{(aq)} \to 2\text{Ag(s)} + 2\text{HBr (aq)} + \text{C}_6\text{H}_4\text{O}_2\text{(aq)}$

Identify oxidation states of silver.

In $\text{AgBr}$ , silver is $+1$ (Br is $-1$ ). In metallic $\text{Ag}$ , it is $0$ .

Change: $+1 \to 0$ — silver is reduced by gaining one electron.
Identify oxidation states in the organic molecule.

$\text{C}_6\text{H}_6\text{O}_2$ is hydroquinone; $\text{C}_6\text{H}_4\text{O}_2$ is benzoquinone. The two hydrogen atoms lost correspond to oxidation of the molecule. Each hydrogen that departs as $\text{H}^+$ leaves behind an electron, so the organic species loses electrons overall.

The carbon-oxygen system is oxidised.
Assign roles.
- Oxidised: $\text{C}_6\text{H}_6\text{O}_2$ (hydroquinone)
- Reduced: $\text{AgBr}$ (specifically, $\text{Ag}^+$ )
- Oxidising agent: $\text{AgBr}$ (the $\text{Ag}^+$ accepts electrons)
- Reducing agent: $\text{C}_6\text{H}_6\text{O}_2$ (donates electrons)

(b) $\text{HCHO(l)} + 2[\text{Ag (NH}_3)_2]^+\text{(aq)} + 3\text{OH}^–\text{(aq)} \to 2\text{Ag(s)} + \text{HCOO}^–\text{(aq)} + 4\text{NH}_3\text{(aq)} + 2\text{H}_2\text{O(l)}$

Silver's oxidation state.

In the complex $[\text{Ag(NH}_3)_2]^+$ , silver is $+1$ . In metallic $\text{Ag}$ , it is $0$ .

Change: $+1 \to 0$ — silver is reduced.
Carbon in formaldehyde.

In $\text{HCHO}$ (formaldehyde), carbon is $0$ (using $\text{H} = +1$ , $\text{O} = -2$ : $2(+1) + C + (-2) = 0 \implies C = 0$ ).

In $\text{HCOO}^-$ (formate), carbon is $+2$ (using $\text{H} = +1$ , two $\text{O} = -4$ , charge $-1$ : $+1 + C - 4 = -1 \implies C = +2$ ).

Change: $0 \to +2$ — carbon is oxidised.
Assign roles.
- Oxidised: $\text{HCHO}$ (formaldehyde)
- Reduced: $[\text{Ag(NH}_3)_2]^+$ (specifically, $\text{Ag}^+$ )
- Oxidising agent: $[\text{Ag(NH}_3)_2]^+$ (Tollens' reagent)
- Reducing agent: $\text{HCHO}$

Tip

This is the classic Tollens' test for aldehydes: the aldehyde is oxidised to a carboxylate while silver ion is reduced to a metallic mirror.

(c) $\text{HCHO (l)} + 2\text{Cu}^{2+}\text{(aq)} + 5\text{OH}^–\text{(aq)} \to \text{Cu}_2\text{O(s)} + \text{HCOO}^–\text{(aq)} + 3\text{H}_2\text{O(l)}$

Copper's oxidation state.

In $\text{Cu}^{2+}$ , copper is $+2$ . In $\text{Cu}_2\text{O}$ , oxygen is $-2$ and the two coppers share $+2$ total, so each copper is $+1$ .

Change: $+2 \to +1$ — copper is reduced.
Carbon in formaldehyde.

Same as part (b): $\text{HCHO}$ has carbon at $0$ , $\text{HCOO}^-$ has carbon at $+2$ .

Change: $0 \to +2$ — carbon is oxidised.
Assign roles.
- Oxidised: $\text{HCHO}$
- Reduced: $\text{Cu}^{2+}$
- Oxidising agent: $\text{Cu}^{2+}$ (Fehling's or Benedict's reagent)
- Reducing agent: $\text{HCHO}$

Note

This is Fehling's test: aldehydes reduce $\text{Cu}^{2+}$ (blue) to $\text{Cu}_2\text{O}$ (brick-red precipitate).

(d) $\text{N}_2\text{H}_4\text{(l)} + 2\text{H}_2\text{O}_2\text{(l)} \to \text{N}_2\text{(g)} + 4\text{H}_2\text{O(l)}$

Nitrogen in hydrazine.

In $\text{N}_2\text{H}_4$ , each nitrogen is $-2$ (four hydrogens contribute $+4$ , two nitrogens share $-4$ ).

In $\text{N}_2$ , nitrogen is $0$ .

Change: $-2 \to 0$ — nitrogen is oxidised.
Oxygen in hydrogen peroxide.

In $\text{H}_2\text{O}_2$ , oxygen is $-1$ (peroxide state).

In $\text{H}_2\text{O}$ , oxygen is $-2$ .

Change: $-1 \to -2$ — oxygen is reduced.
Assign roles.
- Oxidised: $\text{N}_2\text{H}_4$ (hydrazine)
- Reduced: $\text{H}_2\text{O}_2$ (hydrogen peroxide)
- Oxidising agent: $\text{H}_2\text{O}_2$
- Reducing agent: $\text{N}_2\text{H}_4$ …

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Q.Identify the substance oxidised reduced, oxidising agent and reducing agent for each of the following reactions:

The Core Idea: Following the Electrons

(a) 2AgBr (s)+C6H6O2(aq)→2Ag(s)+2HBr (aq)+C6H4O2(aq)2\text{AgBr (s)} + \text{C}_6\text{H}_6\text{O}_2\text{(aq)} \to 2\text{Ag(s)} + 2\text{HBr (aq)} + \text{C}_6\text{H}_4\text{O}_2\text{(aq)}2AgBr (s)+C6​H6​O2​(aq)→2Ag(s)+2HBr (aq)+C6​H4​O2​(aq)

(c) HCHO (l)+2Cu2+(aq)+5OH–(aq)→Cu2O(s)+HCOO–(aq)+3H2O(l)\text{HCHO (l)} + 2\text{Cu}^{2+}\text{(aq)} + 5\text{OH}^–\text{(aq)} \to \text{Cu}_2\text{O(s)} + \text{HCOO}^–\text{(aq)} + 3\text{H}_2\text{O(l)}HCHO (l)+2Cu2+(aq)+5OH–(aq)→Cu2​O(s)+HCOO–(aq)+3H2​O(l)

(d) N2H4(l)+2H2O2(l)→N2(g)+4H2O(l)\text{N}_2\text{H}_4\text{(l)} + 2\text{H}_2\text{O}_2\text{(l)} \to \text{N}_2\text{(g)} + 4\text{H}_2\text{O(l)}N2​H4​(l)+2H2​O2​(l)→N2​(g)+4H2​O(l)

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(a) $2\text{AgBr (s)} + \text{C}_6\text{H}_6\text{O}_2\text{(aq)} \to 2\text{Ag(s)} + 2\text{HBr (aq)} + \text{C}_6\text{H}_4\text{O}_2\text{(aq)}$

(c) $\text{HCHO (l)} + 2\text{Cu}^{2+}\text{(aq)} + 5\text{OH}^–\text{(aq)} \to \text{Cu}_2\text{O(s)} + \text{HCOO}^–\text{(aq)} + 3\text{H}_2\text{O(l)}$

(d) $\text{N}_2\text{H}_4\text{(l)} + 2\text{H}_2\text{O}_2\text{(l)} \to \text{N}_2\text{(g)} + 4\text{H}_2\text{O(l)}$

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