ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: VCAA Official on October 22, 2010, 03:44:46 pm
-
Is my understanding correct as per the diagram I made?
-
not entirely sure (sorry the diagram is a little all over the place), but I believe that this will be a two waters reating situation as you have the battery pack and electrolysis will occur.
EDIT: I should be more clear, looking at the electrochemical series, you see that there are 3 H2O reactions available and the Al (s) and Al3+, there will be a spontaeous reaction between water and Al(s) but I would focus on the electrolysis since we have a power supply.
-
If I understand this correctly:
electrons are flowing out of the right hand side, into the left hand side.
The cathode is on the left, the anode is on the right.
The solution is aqueous 1M Al2(SO4)3.
At the anode, Al(s) -> Al3+ (aq) + 3e
At the cathode, 2H2O(l) + 2e -> H2(g) + 2OH-(aq)
-
but what would you do about the electrolysis (i am assuming the thing on the external circuit is a power supply)
I have attached a electrochemical series with markings, There is both a possible spontaneous reaction and a electrolysis reaction.
-
Do NOT remember electrolysis as positive gradients on the electrochemical series.
For electrochemical cells (galvanic)
1. note all chemicals/compounds you have in the system.
2. pick the highest on the left, and lowest on the right
3. If it is a negative gradient, spontaneous and the reaction goes. Copy those equations down.
For electrolytic cells
1. Determine the direction of electrons, find the anode/cathode
2. At the anode, determine all chemical species available. Pick the lowest on the RHS. This is one of the half equation
3. At the cathode, determine all chemical species available, pick the higest on the LHS. This is the other half of the equation
4. REGARDLESS of the gradient, the reaction goes. Because electrons are being forced.
-
isnt the case with electrolytic cells, the closest reactants react first? not the two 'furtherest' away on electrochemical series.
i know this is the case with galvanic, as this is the biggest potential difference and the most reactive..
-
isnt the case with electrolytic cells, the closest reactants react first?
That's what I've learnt, and as you've already said the furtherest for galvanic cells,
-
An electrolytic reaction doesn't have to be endothermic. It can be a spontaneous AND driven reaction.
For all electrochemical and electrolytic reactions, you will always have the strongest oxidant reacting with the strongest reductant.
-
I'm confused now.
I just did a question involving zinc and copper aqueous ions.
I said the copper would plate, then the zinc would. This was apparently the correct answer...
I've always thought, because you were applying energy, the one which would require the least amount of energy would reduce or oxidise first..
-
I'm confused now.
I just did a question involving zinc and copper aqueous ions.
I said the copper would plate, then the zinc would. This was apparently the correct answer...
I've always thought, because you were applying energy, the one which would require the least amount of energy would reduce or oxidise first..
Exactly. The combination of Top Left (Cu2+) and Bottom Right (H2O) will give the equation which require the least amount of energy.
-
isnt the case with electrolytic cells, the closest reactants react first? not the two 'furtherest' away on electrochemical series.
i know this is the case with galvanic, as this is the biggest potential difference and the most reactive..
It's not always the closest. It's like Mao said, the strongest oxidant and strongest reductant. It's just that coincidentally most of the time you will get ones that are closer togehter as reactants inch closer and closer to the top (oxidants) and bottom (reductants) of the electrochemical series.
Would not bromine (l) and Iron (s) react under eletrolysis? Yet they are very far apart and are considered a spontaneous reaction. It's simply because we use electrolysis for objects that don't spontenously react so you start to think of the reactants as needing to be "close together" on the ES, but actually that's a shortcut method that doesn't consider the details. So in a tricky question as seen above you get caught out.
-
oh yeah sorry mao, i just got the good ol' oxidants and reductants mixed up ><. something that happens way too often hahaha
-
If I understand this correctly:
electrons are flowing out of the right hand side, into the left hand side.
The cathode is on the left, the anode is on the right.
The solution is aqueous 1M Al2(SO4)3.
At the cathode, Al(s) -> Al3+ (aq) + 3e
At the anode, 2H2O(l) + 2e -> H2(g) + 2OH-(aq)
"At the cathode, Al(s) -> Al3+ (aq) + 3e
At the anode, 2H2O(l) + 2e -> H2(g) + 2OH-(aq)"
Doesn't reduction occur at the cathode?
Do you mean:
At the anode, Al(s) -> Al3+ (aq) + 3e
At the cathode, 2H2O(l) + 2e -> H2(g) + 2OH-(aq)"
That. I was being silly and mixed them up. <.< Cheers man