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)
I answered this before but I can't find my answer; therefore, I will repost it.
I suggest you look up the reduction potentials of each of the cations, then follow the directions in the problem.
Br2
To determine which substance is produced at the cathode during the electrolysis of the given mixture, we need to compare the reduction potentials of the cations present.
First, let's identify the cations: CuBr(l) contains Cu2+ ions, AgBr(l) contains Ag+ ions, MgBr2(l) contains Mg2+ ions, and NiBr2(l) contains Ni2+ ions.
Now, we need to find the reduction potentials of these cations. Reduction potential is a measure of the tendency of a species to be reduced (gain electrons). The species with the highest reduction potential will be reduced during electrolysis.
We can consult a table of reduction potentials to find the values. One such table can be found in electrochemistry textbooks or online resources.
Assuming we have access to a table of reduction potentials, we can compare the values for Cu2+, Ag+, Mg2+, and Ni2+. The species with the highest reduction potential will be reduced and produced at the cathode.
Considering the examples given, let's assume that Cu2+ has the highest reduction potential among the cations present, followed by Ag+, Mg2+, and Ni2+.
Therefore, during the electrolysis of the given mixture, Cu2+ will be reduced and Cu(s) will be produced at the cathode.
So, the correct answer is:
b) Cu(s)