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.

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.

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Looking for help with this too

To determine what is produced at the cathode and anode in this electrochemical reaction, we need to consider the standard reduction potentials of the species involved. The species with a higher reduction potential will be reduced (produced) at the cathode, while the species with a lower reduction potential will be oxidized (produced) at the anode.

The reduction potential values you'll need for this analysis can be found in standard tables such as the Standard Electrode Potentials table.

Let's go through each option and determine their reduction potential:

a) F2: The reduction potential for F2 is +2.87 V.
b) Al: The reduction potential for Al is -1.66 V.
c) Cu: The reduction potential for Cu is +0.34 V.
d) Zn: The reduction potential for Zn is -0.76 V.
e) Cl2: The reduction potential for Cl2 is +1.36 V.
f) Br2: The reduction potential for Br2 is +1.09 V.

Now we can compare the reduction potentials to determine what is produced at the cathode and anode:

Cathode: The species with the highest reduction potential is F2 (+2.87 V). Therefore, F2 will be reduced and produced at the cathode. So the correct option is a) F2.

Anode: The species with the lowest reduction potential is Zn (-0.76 V). Therefore, Zn will be oxidized and produced at the anode. So the correct option is d) Zn.

To summarize:
- The product at the cathode is F2 (Option a).
- The product at the anode is Zn (Option d).

Look up the potentials. The easiest to reduce will be reduced first; the easiest to oxidize will be oxidized first.