Answer:
see explanation
Explanation:
Break down the following equations into two half equations(One for oxidation, one or reduction )
1. Cu + 2Ag+ ➡️Cu2+ +2Ag
2. Cl2 +2I- ➡️I2 + 2Cl-
in reaction "1", the copper is being oxidized from 0 to +2 by losing 2 electrons
Cu---->Cu2+ + 2e-in reaction "2
in reaction "1" the silver is being reduced by gaining those 2 e-.
2Ag1+ + 2e----------->2Ag
In reaction "2", the iodine is being oxidized by losing 2 e-
2I- -------------> I2 + 2e
In reaction "2" the chlorine is being reduced by gaining those 2 e
Cl2 + 2e------------> 2Cl-
