Unit 4 Questions MEGATHREAD!

[Mao]

What do Ka and Kw actually do? (Does it indicate the strength of an acid or the mole? :S) I know that it's the Acidity Constant and Self Ionisation of Water.

They are equilibrium constants. By themselves they don't mean too much. You can however
-calculate concentration of one species if the concentration of other species are known
-compare strengths/extent of ionization given the same starting conditions, if equilibrium constants for similar reactions are known

[generalkorn12]

Could anyone explain the question:

An aqueous solution containing a mixture of 1.0 M KI and 1.0 M CaBr2 was electrolysed using unreactive
electrodes.
Which one of the following reactions is most likely to occur at the anode?
A. 2H2O(l) + 2e– → H2(g) + 2OH–(aq)
B. 2Br–(aq) → Br2(aq) + 2e–
C. Ca2+(aq) + 2e– → Ca(s)
D. 2I–(aq) → I2(aq) + 2e–

Since it's occuring at the Anode, I narrowed options to B or D, yet have no idea how to differentiate the two. When looking at the answers, VCAA says that B is being reduced, how?

[b^3]

Im guessing the answer is D?
So list what we have as reactants
H₂O, Br^-, I^-, Na⁺,Ca²⁺,K⁺
Next list the possible reactions
Red is what we have
O₂(g)+4H⁺(aq)+4e^--->2H₂O(l)
Br₂(l)+2e^--->2Br^-(aq)
I₂(s)+2e^--->2I^-(aq)
O₂+2H₂O(l)+4e^--->4OH^-(aq) (no oxygen present)
2H₂O(l)+2e^--->H₂(g)+2OH^-(aq)
Na+[/sup](aq)+e^--->Na(s)
Ca²⁺(aq)+2e^--->Ca(s)
K⁺(aq)+e^--->K(s)

So the solution is being electrolysed
So we are looking for a positive gradient (as it is electrolysis) and the shortest distance between them
that gives at the cathode - reduction
2H₂O(l)+2e^--->H₂(g)+2OH^-(aq)
and at the anode
2I^-(aq)--> I₂(s)+2e^-

What exam on what year and what question was it? VCAA 2009, MC Q19

Yeh D is right, they are just saying that a lot of students picked B, but I think it is a mistake as the forward reaction on the series is a reduction equation but the one given is and oxidation equation.

Just note since this is a MC question, don't write all this out somewhere, just circle the relevant species on the electrochemical series and then draw the gradient line or however you do it.

[generalkorn12]

Just forwarding from that, am I right to believe that Galvanic cells require a strong oxidant and a strong reductant to react? While Electrolysis requires a strong reductant and a strong oxidant?

[b^3]

Just forwarding from that, am I right to believe that Galvanic cells require a strong oxidant and a strong reductant to react? While Electrolysis requires a strong reductant and a strong oxidant?

You are partly right, in electrolysis we are helping the reaction along (well more like forcing it) in the opposite direction using an external power source. Think of it this way.
In galvanic cells you pick the two that are furthest away from each other and form a negative gradient. The higher reaction will go forward. The lower reaction will go backward,.
In electrolytic cells you pick the two that are closest to each other and form a positive gradient. The lower reaction will go forward. The higher reaction will go backward
 
I hope that makes sense.

[Christiano]

Say I have a reaction between HCl and NaOH in a calorimeter - would the heat of neutralisation be different if a different acid/base is used? i.e HNO3 with NaOH? I'm confused by conflicting explanations =\

[nbhindi]

Say I have a reaction between HCl and NaOH in a calorimeter - would the heat of neutralisation be different if a different acid/base is used? i.e HNO3 with NaOH? I'm confused by conflicting explanations =\

I'm not too sure but I think it would be different. The heat or enthalpy of a reaction is the difference b/w the enthalpy of the products from the enthalpy of the reactants. The use of a different acid or base will affect the total initial energy of the thermodynamic system (i.e. position of reactants in an energy profile) and also the total energy of the products formed, thus resulting in a different delta H value.

That's my logic...which could be completely flawed

[HarveyD]

A Student wishes to copper plate her iron locker key
She designs the appropriate cell but connects it to the wrong terminals of the power supply. What will happen to the locker key?

Will it undergo oxidation and then eventually disintegrate?

also does saying it is being oxidised the same as saying it undergoes oxidation?

[Mao]

A Student wishes to copper plate her iron locker key
She designs the appropriate cell but connects it to the wrong terminals of the power supply. What will happen to the locker key?

Will it undergo oxidation and then eventually disintegrate?

also does saying it is being oxidised the same as saying it undergoes oxidation?

yes.

yes.

[SamiJ]

This may be a really really really stupid question... Just a warning.
Can things be carbon plated?

[HarveyD]

What would the half equations be at each electrode (for the half cells attached) when the current commences to flow

[generalkorn12]

Anyone can correct me on this,
But as this is Electrolysis:

Anode will be the Positive Side - Oxidation occurs (electrons are on the right)
Cathode will be the Negative Side - Reduction occurs (electrons are on the left)

Hope that helps! If not, PM me and I'll give you the worked solutions.

[Vincezor]

What would the half equations be at each electrode (for the half cells attached) when the current commences to flow

Electrodes A+B
A = positive anode (oxidation)
B = negative cathode (reduction)

Species in solution:

A:

B:

Electrodes C+D
C = positive anode (oxidation)
D = negative cathode (reduction)

Species in solution:

C:

D:

Hope I didn't make a mistake :S Looks like I did.


You may be wondering why Fe(s) doesn't react with in reaction between electodes A+B. This is because the Fe electrode is the negative cathode,  where reduction will occur, meaning that it will not be oxidised and thus not be in the reaction.

It is a similar situation for electrodes C+D. Fe in this case is also the negative cathode, so Cu(s) will react with water instead...


So, in summary here's what I do for these kind of questions:

1. everything present in each reaction (I circle them with pencil in the electrochemical series). Remember in a solution there is H2O present!
2. Any metals in the cathode will not react so you can omit this species.
3. Bottom left -> top right and closest together will be the expected reactions
4. The reaction at cathode will be the bottom one from left to right
5. The reaction at anode will be higher up from right to left


I apologise for any mistakes I've made, but this is just how I remember how all this stuff works. Maybe Mao or someone else can give a more correct? answer?

[HarveyD]

A and B are right, but C and D are:
Cu(s) -----> Cu2+(aq) + 2e–
D: 2H2O(l) + 2e– -----> H2(g) + 2OH–(aq)

Does that mean the sulfuric acid does not react?



also when explaining why a spontaneous reaction has occurred between two galvanic half cells
What else would we need to say besides the fact that they are connected by a negative gradient on the electrochemical series?
and the same question for electrolysis reactions

[ggxxx]

Hi, I was just wondering would anyone be able to help me with this question?

In an experiment, a solution of 0.50M MgCl2 (aq) was electrolysed using graphite electrodes for 6.75min with a current of 2.00A.
No magnesium metal was deposited at the cathode, but gas evolved at both electrodes.
i) Calc the volume of the dried gas produced at the anode if it was collected at 100kPa and 20C.

ii) Which of the following changes to the experiment could produce a different gas at the anode in this experiment?
- changing the electrode surface area
- increasing the concentration of the electrolyte
- Decreasing the duration of the electrolysis
- Using platinum electrodes instead of graphite

Thanks heaps in advance!