When electricity is applied to a molten binary salt, the cation will be reduced and the anion will be oxidized. The electrolysis of CaBr2(l), for example, produces Ca(s) at the cathode (from the reduction of Ca2+) and Br2(l) at the anode (from the oxidation of Br-).

If more than one cation is present, only the one with highest reduction potential will be reduced. Similarly, if more than one anion is present, only the one with the highest oxidation potential will be oxidized.

Electrolysis is performed on a mixture of CuBr(l), AgBr(l), MgBr2(l), and NiBr2(l). Which of the following is produced at the cathode?

a) Br2(l)
b) Cu(s)
c) Ag(s)
d) Mg(s)
e) Ni(s)

The Answer is c: Ag(s)

See your post above.

To determine which substance is produced at the cathode during the electrolysis of the mixture, we need to compare the reduction potentials of the cations present.

The reduction potentials can be found in a table of standard reduction potentials. By referring to this table, we can look up the reduction potentials for the cations: Cu2+, Ag+, Mg2+, and Ni2+.

Copper (Cu2+): Cu2+(aq) + 2e- -> Cu(s) = +0.34 V
Silver (Ag+): Ag+(aq) + e- -> Ag(s) = +0.80 V
Magnesium (Mg2+): Mg2+(aq) + 2e- -> Mg(s) = -2.37 V
Nickel (Ni2+): Ni2+(aq) + 2e- -> Ni(s) = -0.25 V

The highest reduction potential among the cations is Ag+(aq) with a reduction potential of +0.80 V.

Therefore, during electrolysis, only Ag+(aq) will be reduced to form Ag(s) at the cathode.

So, the correct answer is c) Ag(s).