What are half reactions for an electrolytic cell with aluminum and gold?

Answer 1

Type your answer here... Al3+(aq) + 3e- Al(s) and Au(s) Au+(aq) + e-

Answer 2

Oh, dude, we're talking about aluminum and gold here! So, for aluminum, you'd have Al3+ gaining 3 electrons to form Al, and for gold, Au3+ would pick up 3 electrons to become Au. It's like a fancy electron exchange party where aluminum and gold are the guests of honor.

Answer 3

In an electrolytic cell with aluminum and gold, the half-reaction for the oxidation of aluminum is: Al(s) -> Al^3+(aq) + 3e^-. The half-reaction for the reduction of gold is: Au^3+(aq) + 3e^- -> Au(s). These half-reactions represent the flow of electrons during the electrolysis process, with aluminum being oxidized and gold being reduced.

Answer 4

At the cathode: 2Au^3+ + 6e^- -> 2Au
At the anode: 2Al -> 2Al^3+ + 6e^-

Answer 5

C. Fe3+(aq) + e- -> Fe2+(aq) and Al(s) -> Al3+(aq) + 3e-

Answer 6

Fe3+(aq) + e- → Fe2+(aq) and Al(s) → Al3+(aq) + 3e-