6 mol of Cl2 undergo a loss and gain of 10 mol of electrons to form two oxidation states of Cl which are Cl5+ and Cl−
The reduction half reaction is as follows:
5Cl2+10e−→10Cl−
The oxidation half reaction is as follows:
Cl2→2Cl+5+2e−
The overall reaction is given below:
6Cl2→10Cl−+2Cl+5
Hence, the oxidation state of Cl having positive charge is +5.