Calculate the cell potential for the following reaction that takes place in an e

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+2.62 V reduction: Cu2+(aq) + 2 e– ---> Cu(s) Eo = 0.340 reduction: Mg2+(aq) + 2 e– ---> Mg(s) Eo = –2.356 the redox would be reduction: Cu2+(aq) + 2 e– ---> Cu(s) Eo = 0.340 oxidation: Mg(s) --> Mg2+(aq) + 2 e– Eo = +2.356 the reaction: Cu2+(aq) & Mg(s) --> Mg2+(aq & Cu(s) would have an Eo = 2.696 volts nernst equation: E = Eo - (0.0592 / n ) (log Q) E = 2.696 - (0.0592 / 2 mol e- ) (log prod / reactant) E = 2.696 - (0.0296) (log [Mg+2] / [Cu+2) nothing else is soluble E = 2.696 - (0.0296) (log [2.74] / [0.0033) E = 2.696 - (0.0296) (log 830.3) E = 2.696 - (0.0296) (2.919) E = 2.696 - (0.086) E = 2.61

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