Determine the cell potential for an electrochemical cell based on the following two half-reactions: Oxidation: Cu(s)- >>>
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Determine the cell potential for an electrochemical cell based on the following two half-reactions:
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Oxidation: Cu(s)- →>>> Cu²+ (aq, 0.010 M) + 2 e Reduction: MnO4 (aq, 2.0 M) + 4 H(aq, 1.0 M) + 3 e MnO₂ (s) + 2 H₂O(1)
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Oxidation Anode Reduction Cathode 2MnO4 aq 4Haq 3e Ecell ...View the full answer
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