1. Balance the following chemical equation: $C r_{2} O_{7}^{- 2} (a q) + S O_{3}^{- 2} (a q) \to C r^{+ 3} (a q) + S O_{4}^{- 2} (a q)$ (Acidic medium)
⇒ (2023)

A. $C r_{2} O_{7}^{- 2} (a q) + 3 S O_{3}^{- 2} (a q) + H^{+} (a q) \to 2 C r^{+ 3} (a q) + 3 S O_{4}^{- 2} (a q) + 4 H_{2} O (a q)$

B. $C r_{2} O_{7}^{- 2} (a q) + S O_{3}^{- 2} (a q) + 8 H^{+} (a q) \to 2 C r^{+ 3} (a q) + 3 S O_{4}^{- 2} (a q) + 4 H_{2} O (a q)$

C. $C r_{2} O_{7}^{- 2} (a q) + 3 S O_{3}^{- 2} (a q) + 8 H^{+} (a q) \to 2 C r^{+ 3} (a q) + 3 S O_{4}^{- 2} (a q) + 4 H_{2} O (a q)$

D. $C r_{2} O_{7}^{- 2} (a q) + 3 S O_{3}^{- 2} (a q) + 8 H^{+} (a q) \to C r^{+ 3} (a q) + 3 S O_{4}^{- 2} (a q) + 4 H_{2} O (a q)$

The correct option is (C)

$C r_{2} O_{7}^{- 2} (a q) + 3 S O_{3}^{- 2} (a q) + 8 H^{+} (a q) \to 2 C r^{+ 3} (a q) + 3 S O_{4}^{- 2} (a q) + 4 H_{2} O (a q)$
Steps for Balancing redox reactions:

Identify the oxidation and reduction half .
Find the oxidising and reducing agent.
Find the n-factor of oxidising and reducing agent.
Balance atom undergoing oxidation and reduction.
Cross multiply the oxidising or reducing agent with simplified n-factor values.
Balance atoms other than oxygen and hydrogen.
Balancing oxygen atoms
Balancing hydrogen atoms
Balance charge

For acidic medium:
As soon as we add x

H_{2} O

units, we add 2x

H^{+}

ions on the opposite side.

n_{f} = (| O . S ._{P r o d u c t} - O . S ._{R e a c t a n t} | \times number of atom

Following the above mentioned steps:

Balance atom undergoing oxidation and reduction.

C r_{2} O_{7}^{- 2} + S O_{3}^{- 2} \to 2 C r^{+ 3} + S O_{4}^{- 2}

Cross multiply the oxidising or reducing agent with simplified n-factor number,

C r_{2} O_{7}^{- 2} + 3 S O_{3}^{- 2} \to 2 C r^{+ 3} + 3 S O_{4}^{- 2}

Balance oxygen atoms.

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

Balance Hydrogen atoms.

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

balance charge
charge in reactant side = 0
charge in product side = 0
so the balanced equation is

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

2. For the redox reaction,
MnO^-₄ + C₂O^2-₄ + H⁺ → Mn²⁺ + CO₂ + H₂O, the correct coefficients of the reactant, i.e. MnO^-₄, C₂O₄^2-, H⁺ for the balanced reaction are respectively :
⇒ (2018)

A. 2, 5, 16

B. 16, 3, 12

C. 15, 16, 12

D. 2, 16, 5

The correct option is (A)

$2, 5, 16$
$M n O_{4}^{-} \to M n^{2 +} M n O_{4}^{-} + H^{+} \to M n^{2 +} + H_{2} O M n O_{4}^{-} + 8 H^{+} \to M n^{2 +} + 4 H_{2} O M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O . . . . . . . . . (1) C_{2} O_{4}^{2 -} \to 2 C O_{2} + 2 e^{-} . . . . . . . . (2)$
For balacing the reaction
$2 \times equation(1) + 5 \times equation (2)$ , we get
$2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O$

3. Consider the following reaction :
xMnO₄^- + yC₂O₄^2- + zH⁺ → xMn²⁺ + 2yCO₂ + z/2 H₂O
The values of x, y and z in the reaction are, respectively :⇒ ()

A. 2, 5, 16

B. 5, 2, 8

C. 5, 2, 16

D. 2, 5, 8

Correct option is A)

Reaction:
$2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O$ The above reaction is the balanced chemical reaction. Thus the value of x,y, and z is 2,5, and 16.

⇒Redox

Topic 5 : Balancing of Redox Reactions