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Question:

Hall-Heroult's process is given by:

ZnO+CCoke,1673K−−−−−−−→Zn+CO

Cu2+(aq)+H2(g)→Cu(s)+2H+(aq)

2Al2O3+3C→4Al+3CO2

Cr2O3+2Al→Al2O3+2Cr

Solution:

The correct option is D
2Al2O3+3C→4Al+3CO2
In Hall-Heroult's process is given by
2Al2O3+3C→4Al+3CO2
2Al2O3(l)→4Al3+(l)+6O2-(l)
At cathode: 4Al3+(l)+12e-→4Al(l)
At Anode: 6O2-(l)→2O2(g)+12e-
3C+3O2→3CO2(↑)