Given the two following half reactions: Ni²⁺(aq) + 2e⁻ → Ni(s) E° = -0.25 V; Hg²⁺(aq) + 2e⁻ → Hg(l) E° = +0.86 V. Calculate E° for the following reaction: Hg²⁺(aq) + Ni(s) → Ni²⁺(aq) + Hg(l)

Ni²⁺(aq) + 2e⁻ → Ni(s) E° = -0.25 V; Hg²⁺(aq) + 2e⁻ → Hg(l) E° = +0.86 V. Calculate E° for the following reaction: Hg²⁺(aq) + Ni(s) → Ni²⁺(aq) + Hg(l)

Hg^2+(aq) + 2e ==> Hg(l) .....Eo = 0.86
.......Ni(s) ==> Ni^2+(aq) + 2e........Eo = +0.25
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......... Hg²⁺(aq) + Ni(s) → Ni²⁺(aq) + Hg(l) Eorxn = 0.86 + (+0.25) = ?
Note that I changed the sign for the Ni half cell BECAUSE the half cell in the reaction is the reverse of the tabular reaction.