Answer to Question #287666 in Chemistry for Isha

Question

State with a reason, whether each of the following shows oxidation or reduction a)1²(Aq)+2e -----------21(Aq)

B)cu(AQ)-------------------cu²(aq)+e

C)2Br(aq)-----------------br²(aq+2e)

Answer 1

Solution:

a)

I₂(s) + 2e⁻ → 2I⁻(aq) (gain electrons = reduction half reaction)

The oxidation state of I changes from 0 to −1. Hence, I₂ is reduced.

b)

Cu(s) → Cu²⁺(aq) + 2e⁻

Cu(s) − 2e⁻ → Cu²⁺(aq) (lose electrons = oxidation half reaction)

The oxidation state of Cu changes from 0 to +2. Hence, Cu is oxidized.

c)

2Br⁻(aq) → Br₂(l) + 2e⁻

2Br⁻(aq) − 2e⁻ → Br₂(l) (lose electrons = oxidation half reaction)

The oxidation state of Br changes from −1 to 0. Hence, Br⁻ ion is oxidized.