electrolytic cells

[kim21]

hey guys, I don't really understand this questions. any help? cheers

[jyodesh.com]

What is it about the question that you dont understand?
convert minutes to seconds then use
Q=It to answer part a
in part b you have to use the electrode polarity to deduce whether oxidation or reduction is occuring at each electrode and then use the reagents present to determine the half equation

[kim21]

What is it about the question that you dont understand?
convert minutes to seconds then use
Q=It to answer part a
in part b you have to use the electrode polarity to deduce whether oxidation or reduction is occuring at each electrode and then use the reagents present to determine the half equation

How do you know which H20 from the electrochemical series to use? Part 1 is fine, just part B. I've done a few more similar questions but sometimes they use 2H20 => O2 + 4e- + 4H+ whereas other times they'll use 2H20 + 2e- => H2 + 2OH-
i had originally thought it was whatever was closest to the metal but its not.

[jyodesh.com]

How do you know which H20 from the electrochemical series to use? Part 1 is fine, just part B. I've done a few more similar questions but sometimes they use 2H20 => O2 + 4e- + 4H+ whereas other times they'll use 2H20 + 2e- => H2 + 2OH-
i had originally thought it was whatever was closest to the metal but its not.

The difference is that in the first equation, water is being oxidised, in the second equation, water is being reduced. So water will be oxidised at the anode if it is the strongest reductant present in the solution. Conversely, water will be reduced at the cathode if it is the strongest oxidant in the solution.

[kim21]

The difference is that in the first equation, water is being oxidised, in the second equation, water is being reduced. So water will be oxidised at the anode if it is the strongest reductant present in the solution. Conversely, water will be reduced at the cathode if it is the strongest oxidant in the solution.

In the first beaker there is, water, Fe, and Al3+, in order of the electrochemical series, it goes O2 being reduced to water, Fe, H20 reduced to H2, and Al3+
I don't see a spontaneous reaction, thanks for your reply

[jyodesh.com]

In the first beaker there is, water, Fe, and Al3+, in order of the electrochemical series, it goes O2 being reduced to water, Fe, H20 reduced to H2, and Al3+
I don't see a spontaneous reaction, thanks for your reply

Electrode B
The reduction of oxygen to water isnt one of the half equations you should consider in these kinds of questions because oxygen is poorly soluble in water and its not being bubbled in. Additionally the solution is neutral not acidic.
Fe cannot be reduced because its already at its lowest oxidation state.
So this leaves the reduction of water and the reduction of aluminium. Since water is the strongest oxidant it will be reduced to H2 and OH-.
If Al3+ was reduced to Al you would have a reaction between water and the aluminium metal

[jyodesh.com]

Electrode A
This should be quite simple as long as you're aware that the Fe cannot be oxidised at this electrode
Theres only one reactant that is in contact with the carbon electrode that can be oxidised

[jyodesh.com]

Electrode C
This just relies on you being aware that the two cells are connected and this electrode is an anode. From there it should be pretty easy to figure out whats being oxidised.

[jyodesh.com]

Electrode D
Don't forget about the sulfuric acid!

[kim21]

Electrode D
Don't forget about the sulfuric acid!

Thanks man, youre a real life save I owe you big time !

[jyodesh.com]

Aaalll GGG