For the following electron-transfer reaction:
2Cr(s) + 3Cu2+(aq) 2Cr3+(aq) + 3Cu(s)
The oxidation half-reaction is:
The reduction half-reaction is:
Cr in Cr has oxidation state of 0
Cr in Cr+3 has oxidation state of +3
So, Cr in Cr is oxidised to Cr+3
Cu in Cu+2 has oxidation state of +2
Cu in Cu has oxidation state of 0
So, Cu in Cu+2 is reduced to Cu
1)
Oxidation half cell:
Cr(s) -> Cr3+(aq) + 3e-
2)
Reduction half cell:
Cu2+(aq) + 2e- -> Cu(s)
For the following electron-transfer reaction: 2Cr(s) + 3Cu2+(aq) 2Cr3+(aq) + 3Cu(s) The oxidation half-reaction is: The...
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