Question

In: Chemistry

Balance the following by the half-reaction method. The reaction occurs in acid. How many water molecules...

Balance the following by the half-reaction method. The reaction occurs in acid. How many water molecules are in the final balanced equation?

Mn⁺²_(aq) + I_2(s) --> I^-1_(aq) + MnO₄^-1_(aq)

	A.	4
	B.	12
	C.	8
	D.	2

Expert Solution

Given unbalanced redox reaction is,

Mn⁺²(aq) + I₂(s) --> I^-1(aq) + MnO₄^-1(aq)

The oxidation state of Mn increases from +2 in Mn⁺²(aq) to +7 in MnO₄^-1(aq) hence its an oxidation half reaction,

Oxidation Half Reaction OHR: Mn⁺²(aq) MnO₄^-1(aq)

WHereas O.S. of I decreased from 0 in I2 to -1 in I^-1.Hence its a Reduction Half Reaction.

Reduction Half Reaction RHR: I₂(s) I^-1(aq)

Let's balance these Half reactions separately,

Step 1) Balancing of atoms other than O & H.

OHR: Mn⁺²(aq) MnO₄^-1(aq) No need.

RHR: I₂(s) 2 I^-1(aq)

Step 2) balancing of O & H.

To balanc O we add that may number of H2O by that other sideis O deficient. Then we add H+ on H deficient side.

OHR: Mn⁺²(aq) + 4 H₂O (l) MnO₄^-1(aq) + 8H⁺ (aq)

RHR: I₂(s) 2 I^-1(aq) No need.

Step 3) Balancing of charge by adding e^- on appropriate side.

OHR: Mn⁺²(aq) + 4 H₂O (l) MnO₄^-1(aq) + 8H⁺ (aq) + 5e^-.

RHR: I₂(s) + 2e^- 2 I^-1(aq)

Step 4) balancing of electron count.

Least common multiple of 5 and 2 is 10 hence we make e count 10 in both. For this we need to do 2*OHR and 5*RHR. Hence,

OHR: 2 Mn⁺²(aq) + 8 H₂O (l) 2 MnO₄^-1(aq) + 16 H⁺ (aq) + 10 e^-.

RHR: 5 I₂(s) + 10 e^- 10 I^-1(aq)

These balanced OHR and RHR added and common species are cancelled out mutually.

2 Mn⁺²(aq) + 8 H₂O (l) + 5 I₂(s) 2 MnO₄^-1(aq) + 10 I^-1(aq) +16 H⁺ (aq).

hence the balanced redox equation in acdic medium.

8 H₂O molecules need to be added to balance the redox equation.

Answer Option: C. 8.

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