The given net reaction is:
Tl³⁺(aq) + 2Cr²⁺ (aq) → Tl⁺(aq) + 2Cr³⁺(aq)
The half cell reaction are:
Anode(Oxidation): Involves loss of electrons
Cr²⁺(aq) ↔ Cr³⁺(aq) + e⁻ ----------(1)
Cathode(reduction) : Involves gain of electrons
Tl³⁺ (aq) + 2e⁻ ↔ Tl⁺ (aq) ---------(2)
The net equation can be obtained by: 2* eq(1) + eq(2)
Cr²⁺(aq) ↔ Cr³⁺(aq) + e⁻ ----------(1)*2
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2Cr²⁺(aq) ↔ 2Cr³⁺(aq) + 2e⁻
Tl³⁺ (aq) + 2e⁻ ↔ Tl⁺ (aq) --
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Tl³⁺(aq) + 2Cr²⁺ (aq) → Tl⁺(aq) + 2Cr³⁺(aq)