Balancing Of Redox Reactions

ChemistryRedox ReactionsFor JEE aspirants

Balancing redox reactions means making the electrons lost by the reductant equal to the electrons gained by the oxidant, while atoms and charge balance too. Two methods do it: the oxidation number method, quick for molecular equations, and the half-reaction (ion-electron) method, best for ionic equations in acidic or basic solution. This page works through both, then handles basic medium, disproportionation and reactions in which two elements change. Balancing redox reactions is a direct NEET question and the first step of every JEE Main titration numerical.

On this page1What must balance2Oxidation number method3Half-reaction method4Basic medium5Disproportionation6Two elements change7Standard half reactions

Key Formulas - Quick Reference

★ Must learnElectrons lost by the reductant $=$ electrons gained by the oxidant (total rise in ON $=$ total fall).
Electrons in a half reaction $=$ (change in ON) $\times$ (number of atoms that change); check with the charges.
★ Must learnAcidic medium: balance O with $H_{2} O$ , then H with $H^{+}$ , then charge with $e^{-}$ .
★ Must learnBasic medium: balance as if acidic, then add as many $O H^{-}$ as $H^{+}$ to both sides; $H^{+}$ + $O H^{-}$ → $H_{2} O$ .
★ Must learnPermanganate: acid $Mn O_{4}^{-}$ + $8 H^{+}$ + $5 e^{-}$ → $M n^{2 +}$ + $4 H_{2} O$ ; neutral $Mn O_{4}^{-}$ + $2 H_{2} O$ + $3 e^{-}$ → $Mn O_{2}$ + $4O H^{-}$ ; strong alkali $Mn O_{4}^{-}$ + $e^{-}$ → $Mn O_{4}^{2 -}$ .
Dichromate: $C r_{2} O_{7}^{2 -}$ + $14 H^{+}$ + $6 e^{-}$ → $2C r^{3 +}$ + $7 H_{2} O$ .
★ Must learnDisproportionation: product ratio is the inverse of the electron changes, e.g. $3C l_{2}$ + $6O H^{-}$ → $5C l^{-}$ + $Cl O_{3}^{-}$ + $3 H_{2} O$ .
Final check: every element balances, total charge is equal on both sides, and no electrons remain.

1. What Must Balance in a Redox Equation

Many redox equations cannot be balanced by inspection, because water, $H^{+}$ or $O H^{-}$ take part and the electron count is hidden. A balanced redox equation obeys three conservation rules at once.

A redox equation is balanced only when atoms, charge and electrons (lost = gained) all match.

Figure 1: Three things must balance in every redox equation: atoms, charge (here

+ 24

on each side) and electrons. One dichromate takes the 6 electrons that six

F e^{2 +}

give up.

Oxidation number method

Work on the whole equation. Find the rise and fall in oxidation number and equalise them. Fast for molecular equations (Cu + $HN O_{3}$ , $KMn O_{4}$ + $FeS O_{4}$ ).

Half-reaction method

Split into two half reactions, balance each, add. Best for ionic equations in solution and for disproportionation. The electrons appear explicitly.

Both give the same answer, so use whichever you are faster with. In exams the half-reaction method is safer for ionic equations in basic medium.

Key idea

Balance in this order of thought: electrons first, then charge, then hydrogen and oxygen.

2. Oxidation Number Method

Write the correct formula of every reactant and product (skeleton equation).
Write oxidation numbers and pick out the atoms whose oxidation number changes.
Find the rise or fall per atom and per formula unit. Multiply so that total rise $=$ total fall. (Two species reduced and nothing oxidised means a formula or an ON is wrong.)
If the reaction is in water, balance charge with $H^{+}$ (acidic) or $O H^{-}$ (basic) on the deficient side.
Balance hydrogen with $H_{2} O$ . If oxygen now balances too, the equation is done.

Figure 2: Equalise the rise (

2 \times 3 = 6

) and the fall (

3 \times 2 = 6

). Only 2 of the 8

HN O_{3}

are reduced; the other 6 just supply nitrate, a favourite exam trap.

Exam Trick Cross-multiply the changes. The rise of one species becomes the coefficient of the other: Cu rises by 2 and N falls by 3, so 3 Cu go with 2 N that change. Then add any extra molecules that only supply ions (the 6 nitrate-giving

HN O_{3}

3. Half-Reaction (Ion-Electron) Method

Take the oxidation of $F e^{2 +}$ by dichromate in acid, in which $C r_{2} O_{7}^{2 -}$ becomes $C r^{3 +}$ :

F e^{2 +} + C r_{2} O_{7}^{2 -} F e^{3 +} + C r^{3 +}

Write the unbalanced ionic equation.
Separate it into an oxidation half ( $F e^{2 +} \to F e^{3 +}$ ) and a reduction half ( $C r_{2} O_{7}^{2 -} \to C r^{3 +}$ ).
Balance atoms other than O and H in each half: $C r_{2} O_{7}^{2 -} \to 2C r^{3 +}$ .
In acid, add $H_{2} O$ to the side short of O, then $H^{+}$ to the side short of H: $C r_{2} O_{7}^{2 -}$ + $14 H^{+}$ → $2C r^{3 +}$ + $7 H_{2} O$ .
Add electrons to the more positive side to balance charge: the left is $+ 12$ , the right $+ 6$ , so add $6 e^{-}$ on the left. The iron half needs one: $F e^{2 +}$ → $F e^{3 +}$ + $e^{-}$ .
Multiply the halves so the electrons are equal (iron half $\times 6$ ), add them and cancel the electrons.
Verify that atoms and charges balance.

Figure 3: Each half is balanced on its own in a fixed order: other atoms, O with

H_{2} O

, H with

H^{+}

, then electrons. Multiplying the iron half by 6 makes the electrons cancel.

6F e^{2 +} + C r_{2} O_{7}^{2 -} + 14 H^{+} 6F e^{3 +} + 2C r^{3 +} + 7 H_{2} O

Figure 4: Electrons in a half reaction = (change in ON) × (number of atoms changing). The charge count gives the same 6 for dichromate, which makes it a built-in check.

Exam Trick Electrons always go to the more positive side. Count the charge on each side of the half reaction; the difference is the number of electrons. If it disagrees with (change in ON) × (atoms), a formula or an oxidation number is wrong.

Key idea

Half-reaction order: other atoms, O with water, H with $H^{+}$ , charge with electrons, then equalise electrons and add.

Quick Recall: tap to check

Electrons in the half reaction

Mn O_{4}^{-}

→

M n^{2 +}

(acid)?

5: Mn falls from

+ 7

+ 2

; charge check

(- 1 + 8) - 2 = 5

Which side gets the

H_{2} O

when balancing O in acid?

The side short of oxygen.

Electrons released when one oxalate ion becomes

C O_{2}

C_{2} O_{4}^{2 -}

→

2C O_{2}

2 e^{-}

(two C, each

+ 3 \to + 4

4. Balancing in Basic Medium

In basic solution $H^{+}$ cannot appear in the final equation. The safest route is to balance the half reaction exactly as in acid, then neutralise every $H^{+}$ with $O H^{-}$ added to both sides, combine $H^{+}$ and $O H^{-}$ into water, and cancel water that appears on both sides.

Figure 5: In basic medium finish the acidic balance first, then add as many

O H^{-}

as there are

H^{+}

to both sides and cancel water. No

H^{+}

may remain.

Acidic medium toolkit

Use $H_{2} O$ and $H^{+}$ only. $H^{+}$ usually ends up on the side with the oxidant (more O to remove).

Basic medium toolkit

Use $H_{2} O$ and $O H^{-}$ only. $O H^{-}$ usually ends up on the side opposite to where $H^{+}$ stood in the acidic form.

Exam Trick Basic-medium shortcut: for every extra O atom on one side, add one

H_{2} O

to that side and two

O H^{-}

to the other.

Mn O_{4}^{-}

→

Mn O_{2}

has 2 extra O on the left, so add

2 H_{2} O

on the left and

4O H^{-}

on the right, then

3 e^{-}

for charge.

Figure 6: One route for every ionic redox equation. The only branch is the medium: basic solutions get the extra

O H^{-}

step before the electrons are added.

5. Disproportionation and Comproportionation

In disproportionation one species is both oxidised and reduced, so write it in both half reactions. For chlorine in hot concentrated alkali:

C l_{2} + 12O H^{-} oxidation 2Cl O_{3}^{-} + 6 H_{2} O + 10 e^{-}

C l_{2} + 2 e^{-} reduction, \times 5 2C l^{-}

Adding gives $6C l_{2} + 12O H^{-} \to 10C l^{-} + 2Cl O_{3}^{-} + 6 H_{2} O$ ; dividing by 2,

3C l_{2} + 6O H^{-} hot, conc. 5C l^{-} + Cl O_{3}^{-} + 3 H_{2} O

Figure 7: In disproportionation the product ratio is the inverse of the electron changes: chlorate (5 e

^{-}

) and chloride (1 e

^{-}

) form in the ratio 1 : 5.

Comproportionation is balanced the same way, read backwards. Iodate ( $+ 5$ , gains 5 e $^{-}$ ) and iodide ( $- 1$ , loses 1 e $^{-}$ ) meet at $I_{2}$ in the ratio 1 : 5:

I O_{3}^{-} + 5 I^{-} + 6 H^{+} 3 I_{2} + 3 H_{2} O

The same species can disproportionate differently under different conditions:

C l_{2}

gives

Cl O^{-}

in cold dilute alkali but

Cl O_{3}^{-}

in hot concentrated alkali, because

Cl O^{-}

itself disproportionates on heating:

3Cl O^{-}

→

2C l^{-}

Cl O_{3}^{-}

Key idea

In disproportionation the products form in the inverse ratio of their electron changes: 5 e

^{-}

for

Cl O_{3}^{-}

, 1 e

^{-}

for

C l^{-}

, so 1 : 5.

Quick Recall: tap to check

Balance

B r_{2}

→

B r^{-}

Br O_{3}^{-}

in base.

3B r_{2}

6O H^{-}

→

5B r^{-}

Br O_{3}^{-}

3 H_{2} O

Why must the reactant of a disproportionation be in a middle oxidation state?

It has to go both up and down from where it is.

Convert

Mn O_{4}^{-}

4 H^{+}

3 e^{-}

→

Mn O_{2}

2 H_{2} O

to basic medium.

Mn O_{4}^{-}

2 H_{2} O

3 e^{-}

→

Mn O_{2}

4O H^{-}

6. Molecular Equations and Several Elements Changing

Exams often give the full molecular equation. Balance the ionic core first, then add the spectator ions ( $K^{+}$ , $S O_{4}^{2 -}$ , $N O_{3}^{-}$ ) and count them separately. Watch for one reagent playing two roles, such as $HN O_{3}$ acting as oxidant and as a source of nitrate (Figure 2).

When two elements in one formula change, add their electron changes per formula unit before cross-multiplying. Roasting of iron pyrite is a simple case: Fe rises by 1 and each of the two S atoms rises by 5, so $Fe S_{2}$ loses 11 electrons, while $O_{2}$ gains 4:

4Fe S_{2} + 11 O_{2} Δ 2F e_{2} O_{3} + 8S O_{2}

Figure 8: Add the electron changes of every element in one formula unit.

A s_{2} S_{3}

loses 28, so 3

A s_{2} S_{3}

pair with 28

N O_{3}^{-}

;

Fe S_{2}

loses 11, so 4

Fe S_{2}

need 11

O_{2}

JEE Advanced Arsenic(III) sulphide with nitric acid. In

A s_{2} S_{3}

N O_{3}^{-}

→

H_{3} As O_{4}

S O_{4}^{2 -}

NO

both As (

+ 3 \to + 5

, 2 atoms: 4 e

^{-}

) and S (

- 2 \to + 6

, 3 atoms: 24 e

^{-}

) are oxidised, 28 e

^{-}

per

A s_{2} S_{3}

. Nitrate gains 3. The LCM 84 gives 3

A s_{2} S_{3}

: 28

N O_{3}^{-}

. Oxygen then needs

4 H_{2} O

on the left and hydrogen needs

10 H^{+}

on the left:

3A s_{2} S_{3} + 28N O_{3}^{-} + 10 H^{+} + 4 H_{2} O 6 H_{3} As O_{4} + 9S O_{4}^{2 -} + 28NO

7. Standard Half Reactions to Memorise

Most exam equations are built from a small set of half reactions. Knowing them saves the balancing work entirely: just multiply and add.

Species (role, medium)	Half reaction	e $^{-}$ per formula unit
$Mn O_{4}^{-}$ (oxidant, acid)	$Mn O_{4}^{-}$ + $8 H^{+}$ + $5 e^{-}$ → $M n^{2 +}$ + $4 H_{2} O$	5
$Mn O_{4}^{-}$ (oxidant, neutral / weak base)	$Mn O_{4}^{-}$ + $2 H_{2} O$ + $3 e^{-}$ → $Mn O_{2}$ + $4O H^{-}$	3
$Mn O_{4}^{-}$ (oxidant, strong base)	$Mn O_{4}^{-}$ + $e^{-}$ → $Mn O_{4}^{2 -}$	1
$C r_{2} O_{7}^{2 -}$ (oxidant, acid)	$C r_{2} O_{7}^{2 -}$ + $14 H^{+}$ + $6 e^{-}$ → $2C r^{3 +}$ + $7 H_{2} O$	6
$H_{2} O_{2}$ (oxidant, acid)	$H_{2} O_{2}$ + $2 H^{+}$ + $2 e^{-}$ → $2 H_{2} O$	2
$H_{2} O_{2}$ (reductant)	$H_{2} O_{2}$ → $O_{2}$ + $2 H^{+}$ + $2 e^{-}$	2
dilute $HN O_{3}$ (oxidant)	$N O_{3}^{-}$ + $4 H^{+}$ + $3 e^{-}$ → $NO$ + $2 H_{2} O$	3
concentrated $HN O_{3}$ (oxidant)	$N O_{3}^{-}$ + $2 H^{+}$ + $e^{-}$ → $N O_{2}$ + $H_{2} O$	1
hot conc. $H_{2} S O_{4}$ (oxidant)	$S O_{4}^{2 -}$ + $4 H^{+}$ + $2 e^{-}$ → $S O_{2}$ + $2 H_{2} O$	2
$C_{2} O_{4}^{2 -}$ (reductant)	$C_{2} O_{4}^{2 -}$ → $2C O_{2}$ + $2 e^{-}$	2
$S_{2} O_{3}^{2 -}$ with $I_{2}$ (reductant)	$2 S_{2} O_{3}^{2 -}$ → $S_{4} O_{6}^{2 -}$ + $2 e^{-}$	1
$S_{2} O_{3}^{2 -}$ with $C l_{2}$ / $B r_{2}$ (reductant)	$S_{2} O_{3}^{2 -}$ + $5 H_{2} O$ → $2S O_{4}^{2 -}$ + $10 H^{+}$ + $8 e^{-}$	8
$F e^{2 +}$ , $I^{-}$ , $S O_{3}^{2 -}$ , $H_{2} S$ (reductants)	$F e^{2 +}$ → $F e^{3 +}$ + $e^{-}$ ; $2 I^{-}$ → $I_{2}$ + $2 e^{-}$ ; $S O_{3}^{2 -}$ + $H_{2} O$ → $S O_{4}^{2 -}$ + $2 H^{+}$ + $2 e^{-}$ ; $H_{2} S$ → $S$ + $2 H^{+}$ + $2 e^{-}$	1; 2; 2; 2

Figure 9: The medium decides the product of

Mn O_{4}^{-}

, and so the electrons per ion: 5 (acid), 3 (neutral), 1 (strong alkali). Write the wrong product and every coefficient is wrong.

Key idea

The medium decides the product: the same

Mn O_{4}^{-}

takes 5, 3 or 1 electrons, and every coefficient follows from that number.

Quick Recall: tap to check

Electrons gained per

K_{2} C r_{2} O_{7}

in acid?

6 (two Cr, each

+ 6 \to + 3

Product of

KMn O_{4}

in neutral solution, and its colour?

Mn O_{2}

, a brown precipitate; 3 e

^{-}

per

Mn O_{4}^{-}

Why does thiosulphate give 1 e

^{-}

with

I_{2}

but 8 e

^{-}

with

B r_{2}

Iodine is a mild oxidant and stops at

S_{4} O_{6}^{2 -}

(S average

+ 2.5

); bromine is stronger and takes S to

+ 6

S O_{4}^{2 -}

7.1 The whole concept at a glance

Figure 10: The whole concept on one page. Revise from the centre outwards.

8. Solved Examples

Solved Example 1

Write the net ionic equation for the reaction of potassium dichromate(VI),

K_{2} C r_{2} O_{7}

, with sodium sulphite,

N a_{2} S O_{3}

, in acid solution to give chromium(III) ions and sulphate ions (oxidation number method).

Solution:

Step 1, skeleton: $C r_{2} O_{7}^{2 -}$ + $S O_{3}^{2 -}$ → $C r^{3 +}$ + $S O_{4}^{2 -}$

Step 2, oxidation numbers: Cr $+ 6 \to + 3$ (dichromate is the oxidant); S $+ 4 \to + 6$ (sulphite is the reductant).

Step 3, equalise: each $C r_{2} O_{7}^{2 -}$ has two Cr, so it gains $2 \times 3 = 6$ ; each $S O_{3}^{2 -}$ loses 2. One dichromate pairs with three sulphite: $C r_{2} O_{7}^{2 -}$ + $3S O_{3}^{2 -}$ → $2C r^{3 +}$ + $3S O_{4}^{2 -}$

Step 4, charge: left $- 2 - 6 = - 8$ , right $+ 6 - 6 = 0$ . Add $8 H^{+}$ on the left (acid medium).

Step 5, hydrogen: 8 H on the left need $4 H_{2} O$ on the right. Oxygen: $7 + 9 = 16$ on each side.

C r_{2} O_{7}^{2 -} + 3S O_{3}^{2 -} + 8 H^{+} 2C r^{3 +} + 3S O_{4}^{2 -} + 4 H_{2} O

Solved Example 2

Permanganate ion oxidises bromide ion in basic medium to give manganese dioxide and bromate ion. Write the balanced ionic equation (oxidation number method).

Solution:

Skeleton: $Mn O_{4}^{-}$ + $B r^{-}$ → $Mn O_{2}$ + $Br O_{3}^{-}$ . Mn $+ 7 \to + 4$ (falls 3); Br $- 1 \to + 5$ (rises 6). So two $Mn O_{4}^{-}$ go with one $B r^{-}$ : $2Mn O_{4}^{-}$ + $B r^{-}$ → $2Mn O_{2}$ + $Br O_{3}^{-}$

Charge: left $- 3$ , right $- 1$ . In base, add $2O H^{-}$ on the right. Hydrogen: the right now has 2 H, so add one $H_{2} O$ on the left. Oxygen: $8 + 1 = 4 + 3 + 2 = 9$ .

2Mn O_{4}^{-} + B r^{-} + H_{2} O 2Mn O_{2} + Br O_{3}^{-} + 2O H^{-}

Solved Example 3

Permanganate(VII) ion in basic solution oxidises iodide ion to iodine and is itself reduced to

Mn O_{2}

. Balance by the half-reaction method.

Solution:

Halves: oxidation $I^{-}$ → $I_{2}$ ; reduction $Mn O_{4}^{-}$ → $Mn O_{2}$ .

Oxidation half: $2 I^{-}$ → $I_{2}$ + $2 e^{-}$ .

Reduction half: add $2 H_{2} O$ on the right for O, $4 H^{+}$ on the left for H, then $4O H^{-}$ on both sides for the basic medium and cancel water: $Mn O_{4}^{-}$ + $2 H_{2} O$ + $3 e^{-}$ → $Mn O_{2}$ + $4O H^{-}$ .

Equalise: oxidation × 3, reduction × 2 (6 electrons each), then add:

6 I^{-} + 2Mn O_{4}^{-} + 4 H_{2} O 3 I_{2} + 2Mn O_{2} + 8O H^{-}

Check: charge $- 8 = - 8$ ; O 12 = 12; H 8 = 8.

Solved Example 4

In the balanced equation

x Mn O_{4}^{-} + y C_{2} O_{4}^{2 -} + z H^{+} \to M n^{2 +} + C O_{2} + H_{2} O

, the values of

x

y

z

are
(A) 2, 5, 16
(B) 16, 5, 2
(C) 5, 16, 2
(D) 2, 16, 5

Solution:

Answer: (A). Mn gains 5 e $^{-}$ ; each oxalate loses 2 e $^{-}$ (two C, $+ 3 \to + 4$ ). So 2 $Mn O_{4}^{-}$ : 5 $C_{2} O_{4}^{2 -}$ . Charge: left $- 2 - 10 = - 12$ must reach $+ 4$ on the right ( $2M n^{2 +}$ ), so $z = 16$ and 8 $H_{2} O$ form.

2Mn O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} 2M n^{2 +} + 10C O_{2} + 8 H_{2} O

Solved Example 5

Balance in basic medium:

Cr (OH)_{3}

Cl O_{3}^{-}

→

Cr O_{4}^{2 -}

C l^{-}

Solution:

Cr rises $+ 3 \to + 6$ (3 e $^{-}$ ); Cl falls $+ 5 \to - 1$ (6 e $^{-}$ ). So 2 $Cr (OH)_{3}$ : 1 $Cl O_{3}^{-}$ . Balanced as if acidic:

2Cr (OH)_{3} + Cl O_{3}^{-} 2Cr O_{4}^{2 -} + C l^{-} + H_{2} O + 4 H^{+}

Add $4O H^{-}$ to both sides and combine $4 H^{+}$ + $4O H^{-}$ into $4 H_{2} O$ :

2Cr (OH)_{3} + Cl O_{3}^{-} + 4O H^{-} 2Cr O_{4}^{2 -} + C l^{-} + 5 H_{2} O

Check: charge $- 5 = - 5$ ; O 13 = 13; H 10 = 10.

Solved Example 6

Manganate ion is stable only in strong alkali. In acid it disproportionates into permanganate and manganese dioxide. Write the balanced equation.

Solution:

Mn $+ 6$ rises to $+ 7$ in $Mn O_{4}^{-}$ (1 e $^{-}$ ) and falls to $+ 4$ in $Mn O_{2}$ (2 e $^{-}$ ). Inverse ratio: 2 $Mn O_{4}^{-}$ : 1 $Mn O_{2}$ , from 3 $Mn O_{4}^{2 -}$ .

3Mn O_{4}^{2 -} + 4 H^{+} 2Mn O_{4}^{-} + Mn O_{2} + 2 H_{2} O

Check: charge $- 6 + 4 = - 2$ on each side; O 12 = 12.

Solved Example 7

How many moles of

KMn O_{4}

are needed to oxidise one mole of ferrous oxalate,

Fe C_{2} O_{4}

, in acidic medium?
(A) 0.6
(B) 0.4
(C) 1.67
(D) 0.2

Solution:

Answer: (A). Both parts of $Fe C_{2} O_{4}$ are oxidised: Fe $+ 2 \to + 3$ (1 e $^{-}$ ) and $C_{2} O_{4}^{2 -}$ → $2C O_{2}$ (2 e $^{-}$ ), 3 e $^{-}$ per formula unit. $Mn O_{4}^{-}$ takes 5. Moles of $KMn O_{4}$ $= \frac{3}{5} = 0.6$ .

5Fe C_{2} O_{4} + 3Mn O_{4}^{-} + 24 H^{+} 5F e^{3 +} + 10C O_{2} + 3M n^{2 +} + 12 H_{2} O

Solved Example 8

Balance the molecular equation

KMn O_{4}

FeS O_{4}

H_{2} S O_{4}

→

K_{2} S O_{4}

MnS O_{4}

F e_{2} (S O_{4})_{3}

H_{2} O

Solution:

Ionic core: $Mn O_{4}^{-}$ + $5F e^{2 +}$ + $8 H^{+}$ → $M n^{2 +}$ + $5F e^{3 +}$ + $4 H_{2} O$ . Iron(III) sulphate holds Fe in pairs, so double it: 2 $KMn O_{4}$ , 10 $FeS O_{4}$ , 16 $H^{+}$ (from 8 $H_{2} S O_{4}$ ).

Spectators: 2 K end in one $K_{2} S O_{4}$ . Sulphate: $10 + 8 = 18$ on the left, $1 + 2 + 15 = 18$ on the right.

2KMn O_{4} + 10FeS O_{4} + 8 H_{2} S O_{4} K_{2} S O_{4} + 2MnS O_{4} + 5F e_{2} (S O_{4})_{3} + 8 H_{2} O

Practice Questions

Why is it more appropriate to write (a) $6C O_{2}$ + $12 H_{2} O$ → $C_{6} H_{12} O_{6}$ + $6 H_{2} O$ + $6 O_{2}$ and (b) $O_{3}$ + $H_{2} O_{2}$ → $H_{2} O$ + $O_{2}$ + $O_{2}$ ? Suggest a technique to study their paths.Answer: (a) In photosynthesis all the $O_{2}$ comes from water (12 O from $12 H_{2} O$ ); the 6 $H_{2} O$ formed carry oxygen from $C O_{2}$ . (b) One $O_{2}$ comes from $O_{3}$ , the other from $H_{2} O_{2}$ . Use isotope labelling with $^{18}$ O and follow it by mass spectrometry.
(a) Why is alcoholic $KMn O_{4}$ used to make benzoic acid from toluene? Give the balanced equation. (b) Why does conc. $H_{2} S O_{4}$ give HCl gas with a chloride but red $B r_{2}$ vapour with a bromide?Answer: (a) Alcohol dissolves both toluene and $KMn O_{4}$ , and the neutral medium makes its own $O H^{-}$ : $C_{6} H_{5} C H_{3}$ + $2Mn O_{4}^{-}$ → $C_{6} H_{5} CO O^{-}$ + $2Mn O_{2}$ + $O H^{-}$ + $H_{2} O$ ; acidify to get benzoic acid. (b) HCl is too weak a reductant to reduce $H_{2} S O_{4}$ , but HBr is stronger: $2HBr$ + $H_{2} S O_{4}$ → $B r_{2}$ + $S O_{2}$ + $2 H_{2} O$ .
Balance by the ion-electron method: (a) $Mn O_{4}^{-}$ + $I^{-}$ → $Mn O_{2}$ + $I_{2}$ (basic) (b) $Mn O_{4}^{-}$ + $S O_{2}$ → $M n^{2 +}$ + $HS O_{4}^{-}$ (acidic) (c) $H_{2} O_{2}$ + $F e^{2 +}$ → $F e^{3 +}$ + $H_{2} O$ (acidic) (d) $C r_{2} O_{7}^{2 -}$ + $S O_{2}$ → $C r^{3 +}$ + $S O_{4}^{2 -}$ (acidic).Answer: (a) $2Mn O_{4}^{-}$ + $6 I^{-}$ + $4 H_{2} O$ → $2Mn O_{2}$ + $3 I_{2}$ + $8O H^{-}$ (b) $2Mn O_{4}^{-}$ + $5S O_{2}$ + $2 H_{2} O$ + $H^{+}$ → $2M n^{2 +}$ + $5HS O_{4}^{-}$ (c) $H_{2} O_{2}$ + $2F e^{2 +}$ + $2 H^{+}$ → $2F e^{3 +}$ + $2 H_{2} O$ (d) $C r_{2} O_{7}^{2 -}$ + $3S O_{2}$ + $2 H^{+}$ → $2C r^{3 +}$ + $3S O_{4}^{2 -}$ + $H_{2} O$
Balance in basic medium and name the oxidant and reductant: (a) $P_{4}$ + $O H^{-}$ → $P H_{3}$ + $H_{2} P O_{2}^{-}$ (b) $N_{2} H_{4}$ + $Cl O_{3}^{-}$ → NO + $C l^{-}$ (c) $C l_{2} O_{7}$ + $H_{2} O_{2}$ → $Cl O_{2}^{-}$ + $O_{2}$ + $H^{+}$ .Answer: (a) $P_{4}$ + $3O H^{-}$ + $3 H_{2} O$ → $P H_{3}$ + $3 H_{2} P O_{2}^{-}$ ; $P_{4}$ is both. (b) $3 N_{2} H_{4}$ + $4Cl O_{3}^{-}$ → $6NO$ + $4C l^{-}$ + $6 H_{2} O$ ; oxidant $Cl O_{3}^{-}$ , reductant $N_{2} H_{4}$ . (c) $C l_{2} O_{7}$ + $4 H_{2} O_{2}$ + $2O H^{-}$ → $2Cl O_{2}^{-}$ + $4 O_{2}$ + $5 H_{2} O$ ; oxidant $C l_{2} O_{7}$ , reductant $H_{2} O_{2}$ .
What can you learn from $(CN)_{2}$ + $2O H^{-}$ → $C N^{-}$ + $CN O^{-}$ + $H_{2} O$ ?Answer: Cyanogen disproportionates (C: $+ 3 \to + 2$ in $C N^{-}$ and $+ 4$ in $CN O^{-}$ ), just as $C l_{2}$ does in alkali, so it behaves as a pseudohalogen.
$M n^{3 +}$ disproportionates in solution to $M n^{2 +}$ , $Mn O_{2}$ and $H^{+}$ . Write the balanced ionic equation.Answer: $2M n^{3 +}$ + $2 H_{2} O$ → $M n^{2 +}$ + $Mn O_{2}$ + $4 H^{+}$
Excess chlorine in drinking water is removed with sulphur dioxide. Write the balanced equation.Answer: $C l_{2}$ + $S O_{2}$ + $2 H_{2} O$ → $2C l^{-}$ + $S O_{4}^{2 -}$ + $4 H^{+}$ (that is, $H_{2} S O_{4}$ + 2HCl).
From the periodic table pick (a) non-metals that can disproportionate (b) three metals that can disproportionate.Answer: (a) P, S, Cl, Br, I, N (all show three or more oxidation states). (b) Cu ( $C u^{+}$ ), Mn ( $M n^{3 +}$ , $Mn O_{4}^{2 -}$ ) and Ga or In ( $M^{+}$ ).

Common Mistakes to Avoid

Watch out

Balancing atoms but not charge. Always compare the total charge on the two sides last.
Leaving $H^{+}$ in an equation for a basic medium, or $O H^{-}$ in one for an acidic medium.
Counting electrons per atom instead of per formula unit: $C r_{2} O_{7}^{2 -}$ takes 6, not 3; $C_{2} O_{4}^{2 -}$ gives 2, not 1.
Writing $M n^{2 +}$ as the product of $KMn O_{4}$ in neutral or alkaline solution (it is $Mn O_{2}$ or $Mn O_{4}^{2 -}$ ).
Forgetting the extra reagent that only supplies ions: 3Cu need 8 $HN O_{3}$ , not 2.
Leaving electrons in the final equation, or not cancelling water that appears on both sides.
In disproportionation, writing the reactant only once instead of in both halves, then not dividing by the common factor.
Adding $O_{2}$ or O atoms to balance oxygen. Oxygen is balanced only with $H_{2} O$ (and $O H^{-}$ in base).

Frequently Asked Questions

What are the steps to balance a redox reaction by the half-reaction method?

Write the ionic equation and split it into oxidation and reduction halves. Balance atoms other than O and H, add water for oxygen and hydrogen ions for hydrogen, add electrons to the more positive side, multiply the halves so the electrons are equal, add them and check atoms and charge.

How do you balance a redox reaction in basic medium?

Balance each half exactly as in acid. Then add as many hydroxide ions as there are hydrogen ions to both sides, combine each H+ and OH- into water and cancel any water appearing on both sides. For example MnO4- + 4H+ + 3e- gives MnO2 + 2H2O becomes MnO4- + 2H2O + 3e- giving MnO2 + 4OH-.

Which is better, the oxidation number method or the half-reaction method?

Both give the same result. The oxidation number method is quicker for molecular equations such as copper with nitric acid. The half-reaction method is safer for ionic equations, for basic medium and for disproportionation, because the electrons and charges are written out at every step.

How many electrons does permanganate gain in acidic, neutral and basic solution?

In acidic solution MnO4- gains 5 electrons and becomes nearly colourless Mn2+. In neutral or weakly alkaline solution it gains 3 and forms brown MnO2. In strongly alkaline solution it gains only 1 and forms green manganate, MnO4 2-. The medium therefore fixes every coefficient.

How do you balance a disproportionation reaction?

Write the same reactant in both the oxidation and the reduction half reaction, balance each, equalise the electrons and add. The products form in the inverse ratio of their electron changes; for chlorine in hot alkali this gives 3Cl2 + 6OH- forming 5Cl- + ClO3- + 3H2O.

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

Electrons are neither created nor destroyed in a chemical reaction. Every electron released by the reductant must be taken up by the oxidant, so the total rise in oxidation number must equal the total fall. If they differ, the equation cannot conserve charge.

How are balancing redox questions asked in NEET?

NEET usually asks for a coefficient in a balanced equation, the ratio of oxidant to reductant, or the number of electrons transferred, using NCERT reactions such as dichromate with iron(II), permanganate with oxalate or iodide, and disproportionation of chlorine or phosphorus in alkali.

Why is balancing important for JEE Main redox titration problems?

Every titration calculation in JEE Main needs the mole ratio of oxidant to reductant, which comes from the balanced equation or from electrons transferred per formula unit. For example 2 MnO4- react with 5 oxalate ions; a wrong electron count makes the volume or molarity wrong.

Previous year questions on Balancing Of Redox Reactions

10 questions from past papers, each with a step-by-step solution.

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