Unit 4 questions

[roly182]

if your saying that temperature is doubled, considering CF is J/Temp change, wouldnt doubling temp change actually halve CF?

[Collin Li]

Not quite. The CF is calculated by running the calibration step.

If you already know the CF, and it was the reaction that made the temperature double, then you use that temperature change in conjunction with the CF to calculate the energy output.

[bec]

I've never really thought about that. I don't think it matters. Sulfuric acid is covalently bonded, so I think you technically should include the whole species.

In a proper ionic equation, you should include the calcium ion too, because the calcium ion has changed state (solid to aqueous).

That's what I thought - thanks!

[bec]

I don't think this question is that hard but I just don't know how to approach it:

A 100g lump of heated gold was dropped into 100g of water, which was initially at 20 degrees Celcius. The final temperature of the mixture was 30 degrees. Assuming that the heat lost by the gold equals the heat gained by the water, what was the inital temperature of the gold?
(Given that C(H20)=4.184J/g^oC and C(Au)=0.13J/g^oC)

Any help would be appreciated!

[Mao]

1 degree change in 1 gram of water = 4.184/0.13 = 32.2 degrees change in 1 gram of gold.

10 degrees change in 100 grams of water = 322 degrees change in 100 grams of gold

temperature of Gold [final]: 20 degrees

temperature of Gold [initial]: 340 degrees (2 sig fig)

[bec]

You make it so straight forward, thanks a heap!

[bec]

I get confused with the different ways energy is expressed in the calorimetry topic.
When a question asks to calculate the "heat of reaction", is that the delta H value?
Or is it the product of specific heat capacity, mass and delta T?
Or, if it's in a calorimeter, is it the product of CF and deltaT?

[Mao]

heat of reaction is a rate, Energy per amount

sometimes, it means (kJ/mol or J/mol).
sometimes, it means the energy content (kJ/g or J/g).

the two latter cases you have listed both calculated the energy, not per amount (which is what heat of reaction is)

as for which one the heat of reaction actually refers to, that depends on the question. if food was burnt and no chemical formula was given, then you can safely assume kJ/g or J/g. If the chemical formula is given, it may be asking for either (specified by the question). There are cases, however, where a consequential question asks for , hence you'd calculate the kJ/g or J/g first...

to the best of my knowledge, there's no clear cut distinction.

[bec]

I don't have the solution to these questions, so could anyone tell me if this is right or where I've gone wrong? (I have a feeling it's going to be the latter- we haven't started electrolysis at school and I'm not too confident...)

Predict what changes will occur around each electrode, and the products formed after electrolysis proceeds for a short time. What differences, if any, might be observed after a long time?

1. A solution of 1M Na₂SO₄ is electrolysed using copper electrodes.

Cathode:
Na⁺(aq)+2e^- --> Na(s)
-Initially, sodium plating will form on the copper electrode.
-After a long time, the Na+ will be "used up"

Anode: Cu(s) --> Cu ²⁺(aq) + 2e^-
-Initially, copper electrode will dissolve
-After a long time, electrode will be "used up"
(***Question - the water doesn't react afterwards, does it? I'm thinking that it won't because once the Cu is used up there's no more cathode...or does this reaction: 2H₂O(l) --> O₂(g) + 4H⁺(aq) + 4e^- occur anyway, once the copper has fully reacted?


2. A solution of 0.1m CuCl₂ and 0.01M AlCl₃ is electrolysed using inert electrodes.

Cathode: Cu²⁺(aq) + 2e^- --> Cu(s)
and then: Al³⁺+3e^- -->Al(s)
- Initially, solution will lose blue-green colouring and copper plating will form on cathode
- Eventually (when copper ions are used up), aluminium plating will form on the cathode

Anode: 2H₂O(l) --> O₂(g) + 4H⁺ + 4e^-
Then: 2Cl^-(aq) --> Cl₂(g) + 2e^-
- Initially, pH will decrease due to H+ production, and O2 gas will form bubbles at the anode
- Eventually, bubbles of Cl2 gas will form at the anode


3. A solution of 0.05M FeCl₂ using Ag electrodes

Cathode: 2H₂O(l) + 2e^- --> H₂(g) + 2OH^-(aq)
Fe³⁺ + e^---> Fe²⁺
Initially, pH increases and H2 gas is produced around cathode
Eventually, iron (III) ions react...

Anode: Fe²⁺ --> Fe³⁺ + e^-)
Ag(s) --> Ag⁺ + e^-
Initially, ion (III) ions are produced
Eventually, ion (III) ions are used up in the reaction at the cathode; Ag anode dissolves

(I know this is probably wrong, having Fe2+ acting as an oxidant and a reductant....how do you work it out?)

Thanks, sorry this is so long!

[Collin Li]

1. A solution of 1M Na₂SO₄ is electrolysed using copper electrodes.

Cathode:
Na⁺(aq)+2e^- --> Na(s)
-Initially, sodium plating will form on the copper electrode.
-After a long time, the Na+ will be "used up"

Anode: Cu(s) --> Cu ²⁺(aq) + 2e^-
-Initially, copper electrode will dissolve
-After a long time, electrode will be "used up"
(***Question - the water doesn't react afterwards, does it? I'm thinking that it won't because once the Cu is used up there's no more cathode...or does this reaction: 2H₂O(l) --> O₂(g) + 4H⁺(aq) + 4e^- occur anyway, once the copper has fully reacted?

Haven't looked at it all, but the part in red is wrong. Think about what else is in a solution of sodium sulfate.

[Collin Li]

2. A solution of 0.1m CuCl₂ and 0.01M AlCl₃ is electrolysed using inert electrodes.

Cathode: Cu²⁺(aq) + 2e^- --> Cu(s)
and then: Al³⁺+3e^- -->Al(s)
- Initially, solution will lose blue-green colouring and copper plating will form on cathode
- Eventually (when copper ions are used up), aluminium plating will form on the cathode

Anode: 2H₂O(l) --> O₂(g) + 4H⁺ + 4e^-
Then: 2Cl^-(aq) --> Cl₂(g) + 2e^-
- Initially, pH will decrease due to H+ production, and O2 gas will form bubbles at the anode
- Eventually, bubbles of Cl2 gas will form at the anode

Not really sure what would happen in reality, since these two reactions are so close to each other on the series, but if we assumed the order stayed the same, would exist after the has all gone?

It would precipitate, but more importantly, it would lose contact with the electrode!

[bec]

1. A solution of 1M Na₂SO₄ is electrolysed using copper electrodes.

Cathode:
Na⁺(aq)+2e^- --> Na(s)
-Initially, sodium plating will form on the copper electrode.
-After a long time, the Na+ will be "used up"

Anode: Cu(s) --> Cu ²⁺(aq) + 2e^-
-Initially, copper electrode will dissolve
-After a long time, electrode will be "used up"
(***Question - the water doesn't react afterwards, does it? I'm thinking that it won't because once the Cu is used up there's no more cathode...or does this reaction: 2H₂O(l) --> O₂(g) + 4H⁺(aq) + 4e^- occur anyway, once the copper has fully reacted?

Haven't looked at it all, but the part in red is wrong. Think about what else is in a solution of sodium sulfate.

ahhh...
so does this reaction:
2H2O(l) +2e^- --> H₂(g) +2OH^-(aq)
happen simultaneously with the Na+ reaction?

[Collin Li]

ahhh...
so does this reaction:
2H2O(l) +2e^- --> H₂(g) +2OH^-(aq)
happen simultaneously with the Na+ reaction?

Only the reaction with water would occur. After the water is gone, sodium ions can come next (it's a sequential preference-based system - in ideality, at least), but see my post just above this for why it won't happen (it'd precipitate or lose contact with the electrodes once the water level goes low enough)

[Collin Li]

3. A solution of 0.05M FeCl₂ using Ag electrodes

Cathode: 2H₂O(l) + 2e^- --> H₂(g) + 2OH^-(aq)
Fe³⁺ + e^---> Fe²⁺
Initially, pH increases and H2 gas is produced around cathode
Eventually, iron (III) ions react...

Anode: Fe²⁺ --> Fe³⁺ + e^-)
Ag(s) --> Ag⁺ + e^-
Initially, ion (III) ions are produced
Eventually, ion (III) ions are used up in the reaction at the cathode; Ag anode dissolves

(I know this is probably wrong, having Fe2+ acting as an oxidant and a reductant....how do you work it out?)

Thanks, sorry this is so long!

You don't have , so you can't write that reaction. You can only write the reaction backwards (because we have iron (II) ions as a reactant).

[bec]

Right...I think the underlying concept kind of eluded me but I've got it now:

The strongest reductant and strongest oxidant both react, and "second-strongest" etc of each can react afterwards, but only if they are still in contact with an electrode - so if water reacts first, in an aqueous solution, the solute will NOT react.

Is that right?