How in copper sulphate electrolysis with copper metal (electrode) the number of Cu+2 ions oxidised are to the one reduced?

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in this case one of the electrode is of copper, and this copper electrode is taken as anode. so the oxidation takes place here.

The half reaction taking place on anode – Oxidation

Cu → Cu2+   +   2e

Similarly on cathode side reduction takes place as follows

Cu2+   +   2e   →  Cu

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