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2Au+(aq) + Cu(s) → 2Au(s) + Cu2+(aq)
That formula is correct. It represents the net ionic equation of a complex oxidation-reduction reaction. Note that mass (2Au, 1Cu) and charge (2+) are preserved on both sides. The copper is oxidized from 0 to 2+, whereas the gold is reduced from 1+ to 0. This makes copper the reducing agent and gold the oxidizing agent.
In layman's terms, when you put solid copper into a solution with gold in it, the gold will precipitate out. The reason the reaction happens at all is because gold is so resistant to being oxidized. Essentially any metal, including copper, will force gold out of solution.
That equation, as stated above, is a net ionic equation and is the simplified form of a larger equation, such as:
Molecular: 2AuNO3(aq) + Cu(s) → Cu(NO3)2(aq) + 2Au(s)
Ionic: 2Au+(aq) + 2NO3-(aq) + Cu(s) → Cu2+(aq) + 2NO3-(aq) + 2Au(s)
Note that the nitrate ions are the same on both sides and do not participate in the overall reaction, hence their removal to form the net ionic equation.
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What is the voltage of the cell shown?
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-2.71v
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What is the voltage of the electrochemical cell written as Cu(s) Cu2 (aq) Mg2 (aq) Mg(s)?
-2.71v
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