Balancing a Redox Reaction

I. Oxidation Number method

II. Ion Electron Method / Half reaction method

Oxidation Number method

Steps: 

1. Write the skeletal equation of all the reactants and products of the reaction.

2. Indicate the oxidation no. of each element

3. Calculate the increase or decrease in oxidation no. per atom (oxidising and reducing agent)

4. Equalise the oxidation no. with the help of multiplication

5.Balance other than H and O

6. Balance H and O by adding H2O

7. acidic medium = H atoms are balanced by adding H+

    basic medium = H atoms are balance by adding OH-

MCQs

Q1 Oxidation No. of sodium in Sodium amalgam

a. +1

b. 0

c. -1

d. +2

Ans.b. 0

Explanation: Na/Hg (both metals), none will gain or lose.

Q2. The oxidation no. of P in Ba(H2PO)2

a. +3

b. +2

c. +1

d. -1

Ans.c. +1

Q.3. The oxidation state of chromium in Cr(CO)6

a. 0

b. +2

c. -2

d. +6

Ans.a.0

Q4. Which is not a redox reaction

a. CaCO3 →  CaO + CO2

b. O2 + 2H2 →  2H2O

c. Na+H2O →  NaOH + 1/2 H2

d. MnCl3 →  MnCl2 + 1/2 Cl2

Ans. a.  CaCO3 →  CaO + CO2

Q5. In which of the following pairs, there is greatest difference in the oxidation no. of the underlined elements?

a. NO2 and N2O4

b. P2O5 and P4O10

c. N2O and NO 

d. SO2 and SO3

Ans.d. SO2 and SO3

Explanation: 

a. in NO2 N has +2, and in  N2O4 N has +2; so there is no difference in +2 and +2.

b. in P2O5 P has +5, and in P4O10 P has +5; so there is no difference in +5 and +5

c. in N2O N has +1, and in NO N has +2; so the difference in +1 and +2 is of 1.

d. in SO2 S has +4, and in SO3 S has +6; so the difference in +4 and +6 is of 2.

Q6. In the reaction: 2FeCl3 + H2S = 2FeCl2 + 2HCl + S

a. FeCl3 acts as an oxidising agent

b. Both H2S and FeCl3 are oxidized

c. FeCl3 is oxidised while H2S is reduced

d. H2S acts as an oxidising agent.

Ans. a. FeCl3 acts as an oxidising agent

Explanation: - Oxidising agent ( जो दुसरो के इलेक्ट्रान खा जाए ) means which itself gets reduced (electron खाना ) and oxidizes the opponent. 

Balancing a Redox Reaction

1.  Phosphorous in basic medium gives phosphine.  Balance the redox.

Ans. P + OH− = PH3 + H₃PO₂−