Which half-reaction will take place at the cathode during the electrolysis of 1.0 M H2O2 solution containing 1.0 M H2SO4?
2H2O --> O2 + 4H+ + 4e-
2H+ + 2e- --> H2
H2O2 --> O2 + 2H+ + 2e-
H2O2 + 2H+ + 2e- --> 2H2O
I thought the answer was H2 + 2OH^1 --> 2H2O + 2e^- but that isn't an option.
Which half-reaction will take place at the cathode during the electrolysis of molten MgCl2?
Mg2+ + 2e- --> Mg (I think this one is right)
Mg --> Mg2+ + 2e-
2Cl- --> Cl2 + 2e-
Cl2 + 2e- --> 2Cl-
2H2O --> O2 + 4H+ + 4e-
2H+ + 2e- --> H2
H2O2 --> O2 + 2H+ + 2e-
H2O2 + 2H+ + 2e- --> 2H2O
I thought the answer was H2 + 2OH^1 --> 2H2O + 2e^- but that isn't an option.
Which half-reaction will take place at the cathode during the electrolysis of molten MgCl2?
Mg2+ + 2e- --> Mg (I think this one is right)
Mg --> Mg2+ + 2e-
2Cl- --> Cl2 + 2e-
Cl2 + 2e- --> 2Cl-
-
The thing to remember in any cell, electrolytic or galvanic/voltaic, is that oxidation always occurs at the anode.
So reduction occurs at the cathode. In a reduction half-reaction, the electrons appear on the left side. So in the first one:
H2O2 + 2H+ +2e- ===> 2H2O
In the second one:
Mg2+ + 2e- ===> Mg
The other one, Cl2 + 2e- ===> 2Cl- is not an option, because you don't start with any Cl2.
So reduction occurs at the cathode. In a reduction half-reaction, the electrons appear on the left side. So in the first one:
H2O2 + 2H+ +2e- ===> 2H2O
In the second one:
Mg2+ + 2e- ===> Mg
The other one, Cl2 + 2e- ===> 2Cl- is not an option, because you don't start with any Cl2.