Current is applied to a molten mixture of CuF, ZnCl2, and AlBr3.
What is produced at the cathode?
a)F2
b)Al
c)Cu
d)Zn
e)Cl2
f)Br2
What is produced at the anode?
Same options as above.
6 answers
Look up the potentials. The easiest to reduce will be reduced first; the easiest to oxidize will be oxidized first.
haallllp me
Looking for help with this too
8 years later and here I am looking for the answer to this same question lmaOo
smh no one came through
The cations in this mixture are Cu+(l), Zn2+(l), and Ca2+(l), which will all compete for reduction at the cathode. Although standard tables list potentials for the reduction of aqueous metal ions, you can infer the relative reduction potentials of these molten ions from their aqueous counterparts. The reduction potential for Cu+ is greater than that of either Zn2+ or Ca2+, so the reduction of Cu+ is favored at the cathode. The reduction of Cu+ produces Cu.
The anions in this mixture are F−(l), Cl−(l), and S2−(l), which will all compete for oxidaiton at the anode. Although standard tables list potentials for aqueous nonmetal ions, you can infer the relative oxidation potentials of these molten ions from their aqueous counterparts. The reduction potential for S2− is the lowest, and so its oxidation potential is the highest. Thus, the oxidation of S2− is favored at the anode. The oxidation of S2− produces S.
The anions in this mixture are F−(l), Cl−(l), and S2−(l), which will all compete for oxidaiton at the anode. Although standard tables list potentials for aqueous nonmetal ions, you can infer the relative oxidation potentials of these molten ions from their aqueous counterparts. The reduction potential for S2− is the lowest, and so its oxidation potential is the highest. Thus, the oxidation of S2− is favored at the anode. The oxidation of S2− produces S.