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Divide the following redox reaction into oxidation and reduction half-reaction. SnA2++2HgA2+⟶SnA4++HgA2A2+ - Chemistry

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Question

Divide the following redox reaction into oxidation and reduction half-reaction.

\[\ce{Sn^2+ + 2Hg^2+ -> Sn^4+ + Hg2^2+}\]

Short Note

Solution

Oxidation: \[\ce{Sn^2+ -> Sn^4+ + 2e-}\]

Reduction: \[\ce{2Hg^2+ + 2e- -> Hg^2+2}\]

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