What is the standard emf of a cell with the reaction Pb(s) + Hg₂²⁺(aq) → Pb²⁺(aq) + 2Hg(l) ? (Given: E°Pb²⁺/Pb = -0.13 V

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What is the standard emf of a cell with the reaction Pb(s) + Hg₂²⁺(aq) → Pb²⁺(aq) + 2Hg(l) ? (Given: E°Pb²⁺/Pb = -0.13 V , E°Hg_₂²⁺/Hg = 0.79 V )

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0.92 V Correct 0.66 V 1.05 V -0.92 V

E°cell = E°cathode - E°anode = 0.79 - (-0.13) = 0.92 V .

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