Question

Balance the following equations by oxidation number method.
(i) XeO6 4- + Mn2+ + H+→ MnO4 + XeO3

Answer 1

Question is wrong. in a chemical equation any compound does not contain a charge since it reacts

Answer 2

We have to balance the given equation by oxidation number method.

XeO₆⁴⁻ + Mn²⁺ + H⁺ => MnO₄⁻ + XeO₃ + H₂O

oxidation no of Xe in XeO₆⁴⁻ :

let o.n of Xe is x

∴ -4 = x + 6(-2)

⇒ -4 = x - 12

⇒ x = +8

it is clear that, oxidation no of Xe in XeO₃ = + 6  [ ∵ o.n of each O ion = -2 ]

here you see that Xe reduces its oxidation number from +8 to +6.

i.e., change in the oxidation number of Xe = 2

i.e.,

also we see, oxidation number of Mn in MnO₄⁻ is +7. but the oxidation number of Mn²⁺ is +2.

so Mn increases from +2 to +7. so the change in oxidation number is +5

but a balanced equation must have the same value of reduction of oxidation number and increment of oxidation number.

so here two equations :

XeO₆⁴⁻ => XeO₃ + 2e⁻      ...(1) × 5

Mn²⁺ => MnO₄⁻ - 5e⁻       ....(2) × 2  

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5XeO₆⁴⁻ + 2Mn²⁺ => 5XeO₃ + 2MnO₄⁻

you see, oxygen is only 23 in the right side while 30 , oxygen atoms are present in the left side. To balance it, use 7H₂O to the right side.

i.e., 5XeO₆⁴⁻ + 2Mn²⁺ => 5XeO₃ + 2MnO₄⁻ + 7H₂O

now add 14H⁺ to the left side to balance the hydrogen.

i.e., 5XeO₆⁴⁻ + 2Mn²⁺  + 14H⁺ => 5XeO₃ +2MnO₄⁻ + 7H₂O

Therefore the balanced chemical equation is , 5XeO₆⁴⁻ + 2Mn²⁺  + 14H⁺ => 5XeO₃ +2MnO₄⁻ + 7H₂O