Chemistry 3/4 2013 Thread

[Alwin]

When you say that after connecting the positive terminal to the positive electrode, the two are therefore at the same potential, what do you mean? Why therefore? Would you be able to explain further?

I just means that its an open circuit so both electrodes are at the same potential. It's like a torch switched off (open circuit) will not glow because at both globe connections there is the same potential so no electron flow.

I wouldn't be too fussed about this, more physics than chemistry

[clıppy]

Just doing some preparation for my SAC on calorimetry and enthalpy and I've got a few questions hopefully you lovely folks can answer:

• We've calibrated our calorimeter using 100mL of water and found our calibration constant. Would there be an effect on the constant if we used say, 50mL of water or even 100mL of ethanol instead of water?
• We're reacting HCl and NaOH and finding the enthalpy of that, and also reacting HNO3 and KOH and finding the enthalpy of that. Would there be any similarities between the enthalpies given that both their ionic equations are H⁺ + OH^- --> H₂O
• What are some main sources of error that may have affected the calibration of our calorimeter?
• We're also trying to find the enthalpy of the dissolution of potassium nitrate, in the equation for the reaction is H2O a reactant, or does it go above the arrow?

Huge thanks in advance

[barydos]

Erm...last time I checked...positive charges attract negative charges. I don't see how your logic works there. For a galvanic cell, the positive electrode is the site of reduction, i.e. where the electrons are headed.

Don't think of them as "positive" and "negative" electrodes. Think of them as electrodes with higher and lower electric potential, so the positive electrode has higher electric potential. Generally, positive charges move from higher to lower electric potential spontaneously; negatives do the opposite. Makes sense; electrons leave the negative terminal of the battery to go to the positive. So, negative electrode = electrode at lower potential. ETC.

The way I think about it is like this. To create an electrolytic cell, you need to meet two requirements.
1. Voltage of the power source is higher than the voltage of the galvanic cell that would otherwise be created.
2. Positive terminal of battery (assuming we have a battery) is connected to the positive electrode, negative terminal connected to negative electrode.

So what happens? Let's assume that initially the positive electrode of the battery and the positive terminal were connected first. Therefore they are at the same potential. Then, we connect the negative terminal to the negative electrode. As the potential drop across the battery is larger than the potential difference from the positive electrode to the negative by definition, the negative terminal of the battery is actually at a lower potential than the negative electrode. Therefore, electrons flow from the negative terminal to the negative electrode, i.e. the negative electrode is the cathode.
Now, as the voltage across the battery is constant, electrons are drawn from the positive electrode to the positive terminal of the battery to replace the electrons lost at the negative terminal. This way, a current flows where the electrons come from the positive electrode and go to the negative.

Alternatively, if the negative terminal is connected first to the negative electrode, when the positives are connected, the positive terminal of the battery is at higher potential than the positive electrode as the battery's voltage is bigger than the cell's voltage. This means the electrons will flow from the positive electrode to the positive battery terminal. Et cetera.

This is what the books really mean by "external power sources drive a non-spontaneous reaction". You get a larger power source to override the potential difference.

Thanks, you're a legend!

[zvezda]

I just means that its an open circuit so both electrodes are at the same potential. It's like a torch switched off (open circuit) will not glow because at both globe connections there is the same potential so no electron flow.

I wouldn't be too fussed about this, more physics than chemistry

I sort of see now.

Thats true but a little bit of extra knowledge never hurt anybody heh

[Alwin]

Just doing some preparation for my SAC tomorrow on calorimetry and enthalpy and I've got a few questions hopefully you lovely folks can answer:

• We've calibrated our calorimeter using 100mL of water and found our calibration constant. Would there be an effect on the constant if we used say, 50mL of water or even 100mL of ethanol instead of water?
• We're reacting HCl and NaOH and finding the enthalpy of that, and also reacting HNO3 and KOH and finding the enthalpy of that. Would there be any similarities between the enthalpies given that both their ionic equations are H⁺ + OH^- --> H₂O
• What are some main sources of error that may have affected the calibration of our calorimeter?
• We're also trying to find the enthalpy of the dissolution of potassium nitrate, in the equation for the reaction is H2O a reactant, or does it go above the arrow?

Huge thanks in advance

1. Using 50ml of water means that it will heat of more quickly (twice as much). Remember the units for specific heat capacity are joules, per gram per degree C.
The specific heat capacity of water is 4.18 J/g/K but ethanol is only 2.44 J/g/K. HOWEVER ethanol is less dense than water so the mass of ethanol is not the same (numerical) value as the volume. The point is, that the ethanol heats up quicker (about twice as fast accounting for the density).
So the calibration factor will be around half if either of the changes are made.

2. Hmm, yes i think they would be similar so long as you add both in aq solution. That way you don't need to account for temperature change when a solid ionises in water. Also, they are all strong acid or bases so adding them (ie increasing the overall volume) doesn't favour forward or back reaction so there'd be no change in temp due to shift of POE. HOWEVER, chem can do weird things so I wouldn't be surprised if they are not similar enthalpy values

3. Incorrectly reading the voltmeter (esp if analogue not digital). losing water because you over zealously stirred the water. losing temperature due to the surrounding (inefficient insulation) butt this last one should be accounted for my the calibration factor. Although, for more detailed analysis one should account for the energy lost but doesn't really matter for pracs and stuff

4. NO NO NO KNO₃ --> K⁺ + NO₃^-

Good luck

[Emily C]

We did a reaction on fuel cells where we formed our reactants of hydrogen gas and oxygen through electrolysis , we used 1M KOH as the  electrolyte. We than disconnected the power supply and connected it to a voltmeter, why would our recordings be about 0.04 volts , not the predicted 1.23 volts ? What are limitations of these results?

[Alwin]

We did a reaction on fuel cells where we formed our reactants of hydrogen gas and oxygen through electrolysis , we used 1M KOH as the  electrolyte. We than disconnected the power supply and connected it to a voltmeter, why would our recordings be about 0.04 volts , not the predicted 1.23 volts ? What are limitations of these results?

I'm assuming that because you have the same setup for electrolysis and galvanic cells, its something like a beaker with two electrodes (graphite?) that are covered by testtubes filled with KOH? Then when connected to the power supply gas is formed in each of the test tubes?

Sorry for the preamble, its coz the setup is quite important!

Okay, so now the bubbles form around the electrodes during electrolysis as that is where the reaction occurs. These bubbles rise up and collect at the top (bottom of the upside down test tube) ABOVE the electrode. This means that when you reverse the reaction, there may be too little gas on the surface of the electrode to react - the voltage will probably drop off after a while.

HOWEVER, this does not really account for the massive difference..
Were you by any change using an analogue voltmeter with 3 positive connections and 1 negative connection?
Each +ve connection has a different scale.. you have to make sure you read the right one

[Emily C]

I'm assuming that because you have the same setup for electrolysis and galvanic cells, its something like a beaker with two electrodes (graphite?) that are covered by testtubes filled with KOH? Then when connected to the power supply gas is formed in each of the test tubes?

Sorry for the preamble, its coz the setup is quite important!

Okay, so now the bubbles form around the electrodes during electrolysis as that is where the reaction occurs. These bubbles rise up and collect at the top (bottom of the upside down test tube) ABOVE the electrode. This means that when you reverse the reaction, there may be too little gas on the surface of the electrode to react - the voltage will probably drop off after a while.

HOWEVER, this does not really account for the massive difference..
Were you by any change using an analogue voltmeter with 3 positive connections and 1 negative connection?
Each +ve connection has a different scale.. you have to make sure you read the right one

Yes that's exactly how we set it up.
So the gas collects at the top or bottom of the test tube? and this effects the reaction ? we also didn't wait till all gas was produced before we switched off the power supply , so could that have anything to do with it ?
Also the voltage dropped off rapidly and we were told to record the voltage that it hovered at for a while ...

[lzxnl]

How much gas do you think you got? Anywhere near 101.3 kPa?
That's the pressure at which the electrochemical series was devised. Only if your hydrogen and oxygen gas pressures are at 101.3 kPa will you get the 1.23 V.

[Edward21]

With spontaneous redox reactions, how does the distance between the two reactions you've located determine the equilibrium constant for that overall reaction? Is there an equilibrium expression?? Does the varying distance between the reactions do anything? I can't seem to locate this in the textbook 

[Emily C]

How much gas do you think you got? Anywhere near 101.3 kPa?
That's the pressure at which the electrochemical series was devised. Only if your hydrogen and oxygen gas pressures are at 101.3 kPa will you get the 1.23 V.

That could be why , we just need to know some general limitations of our results on fuel cells , could concentration of hydrogen and oxygen have anything to do with or , or are concentrations not  ? I don't know ...

[lzxnl]

With spontaneous redox reactions, how does the distance between the two reactions you've located determine the equilibrium constant for that overall reaction? Is there an equilibrium expression?? Does the varying distance between the reactions do anything? I can't seem to locate this in the textbook 

Not part of the course, but here it goes.
Take a redox cell in which the difference between the standard reduction potential of the cathode and the reduction potential of the anode is E. Then, this E is equal to where R is the ideal gas constant, T is the temperature in K, n is the number of electrons per mole of reaction (so in the case of fluorine gas reacting with lithium, two electrons are transferred) and K is the equilibrium constant.

That could be why , we just need to know some general limitations of our results on fuel cells , could concentration of hydrogen and oxygen have anything to do with or , or are concentrations not  ? I don't know ...

The concentrations of hydrogen and oxygen and their partial pressures are directly proportional, so yeah. Using the ideal gas law, PV = nRT => n/V = P/RT

[lzxnl]

Can someone please clarify this for me?
Find the ionic equation of dilute sulfuric acid on solid copper(II) carbonate.
Is the answer Cu^2+(aq) + SO4^2-(aq) -> CuSO4(aq) or 2H+ + CO3 > H2O + Co2?

It's the second one. Nothing happens in the first reaction; CuSO4(aq) is the same as Cu 2+ and SO4 2-
Besides, sulfuric acid doesn't actually contain that much sulfate. Most of it is in the form of HSO4-.

[lzxnl]

Of dilute sulfuric acid, it's H2SO4(l) => H+(aq) + HSO4-(aq)
Only the first dissociation is complete, the second isn't.
If it's concentrated, then something different occurs which you don't need to worry about.

[9_7]

Of dilute sulfuric acid, it's H2SO4(l) => H+(aq) + HSO4-(aq)
Only the first dissociation is complete, the second isn't.
If it's concentrated, then something different occurs which you don't need to worry about.

Yep its dilute! You legend. Im assuming that there is no ionic equation?