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7th Edit 6L5 c

Posted: Sat Feb 23, 2019 8:59 pm
by Bijan Mehdizadeh 1B
For 6L5 c, how do you determine whether you make the half reaction h2 --> hcl or cl2 --> hcl? There is one product, but cl2 is reduced while h2 is oxidized, so I was wondering how come hcl is only used in the anode half rxn eq for h2

Re: 7th Edit 6L5 c

Posted: Sat Feb 23, 2019 9:27 pm
by Chloe Qiao 4C
Consider HCl to be H+ and Cl-. The anode side will be H2 and H+, and the cathode side will be CL2 and CL-.