Chemistry 3/4 2013 Thread

[Homer]

how can we experimentally determine the exact location of an metal/ion in an electrochemical series?

[thushan]

how can we experimentally determine the exact location of an metal/ion in an electrochemical series?

Construct half-cells at standard conditions with standard hydrogen electrode.

[Alwin]

how can we experimentally determine the exact location of an metal/ion in an electrochemical series?

exact location? That's a bit hard lol.. but the basic concept to finding the relative position under a certain condition can be achieved but setting up a series of half cells containing different metals and the sample metal/ion.

We then connect up these galvanic cells and note the reading on the voltmeter. From this, it is possible to list the metals in order of oxidation strength hence the position of the sample metal within these metals.
However, this method will only provide a relative position, but can be achieved in school labs with school equipment.

To find the EXACT position you need to set up half-cells at standard conditions with standard hydrogen electrode and measure the voltage etc. Then, compare it to the data book or similar electrochemical series to determine the exact position
This is a lot harder to achieve accurately at school (if this is for a prac haha)

Hope it helps!

EDIT: beaten, but its by Thushan himself so its okay

[Edward21]

With galvanic cells, there's examples in the (lovely purple) Heinamann textbook of this sort of equation, with potential volt difference calculations but a lot more complicated with -log involved and an equlibrium expression :O Do we have to know how to do this? It says extension, but I don't really know if that's completely off the course or? 

[lzxnl]

With galvanic cells, there's examples in the (lovely purple) Heinamann textbook of this sort of equation, with potential volt difference calculations but a lot more complicated with -log involved and an equlibrium expression :O Do we have to know how to do this? It says extension, but I don't really know if that's completely off the course or? 

Yay Nernst equation! Nah, VCE is too simple to make you use that.
It essentially allows you to deal with scenarios where you're not at standard states. AKA 99.999% of the time.

[09Ti08]

Use the electrochemical series to predict whether a reaction will occur when the reactions listed under normal laboratory conditions: Cu(s)/Pb(NO₃)₂

My answer is a reaction can occur and the overall equation is: 3Cu(s)+2NO₃^-(aq)+8H⁺(aq)->2NO(g)+3Cu²⁺(aq)+4H₂O(l). But the answer is no. Now I'm confused, do we need to consider NO₃^- in the electromical series?

[clıppy]

Can someone please give me a quick explanation as to how the energy released by a reaction is = calibration factor * temperature difference

[jgoudie]

Calibration factor is a measure of the amount of energy per degree celsius.  (j/oC) or (kJ/oC).  So in this case to get energy multiple (oC x kJ/oC).

Hope this makes sense.  It does to me.

Can someone please give me a quick explanation as to how the energy released by a reaction is = calibration factor * temperature difference

[lzxnl]

Use the electrochemical series to predict whether a reaction will occur when the reactions listed under normal laboratory conditions: Cu(s)/Pb(NO₃)₂

My answer is a reaction can occur and the overall equation is: 3Cu(s)+2NO₃^-(aq)+8H⁺(aq)->2NO(g)+3Cu²⁺(aq)+4H₂O(l). But the answer is no. Now I'm confused, do we need to consider NO₃^- in the electromical series?

If you look carefully, your cell just has copper, lead and nitrate.
Unfortunately, the reduction half-equation for nitrate requires a lot acid. Do you see the 8H+ there is? There's not enough H+ for your nitrate reaction to go through.

[SocialRhubarb]

When you calibrate a calorimeter, what you do is you put in a set amount of energy, and then measure the change in temperature of the calorimeter.

If you put in 100 kJ and the temperature of the calorimeter increases by 10 degrees, you can say, "Right, well, averaging it out, it looks like I need to put in 10 kJ of energy to raise the temperature by 1 degree Celsius."

Now, imagine we put some amount of methane into the calorimeter and burnt it, and, combustion of methane being an exothermic reaction, the temperature of your calorimeter went up 5 degrees Celsius.

Now, if it took 10 kJ of energy to raise the calorimeter's temperature by 1 degree Celsius, and your combustion of methane raised the temperature by 5 degrees Celsius, you, being a brilliant chemistry student, could work out that the reaction produced five times the change in heat and hence produced five times the energy, so the reaction must have released 50 kJ.

So the calibration factor tells you how much energy you need to raise the temperature of the calorimeter by 1 degree Celsius. You multiply your change in temperature by your calibration factor to work out how much energy you would need to produce the temperature change you observe.

[Edward21]

Hey, so I decided to read ahead (probably a poor decision haha) because now I'm confused. With the prediction of redox equations at SLC conditions for galvanic cells, you can do the whole top goes to the right thing (becomes reduces) bottom one/stronger reductant goes to the left (gets oxidised)... the backwards Z thing or whatever you want to call it  NOW let's say I have a 1M solution of Copper(II) Bromide undergoing electrolysis tell me why Br2 and Cu is produced, does it does the exact opposite to the spontaneous reaction of galvanic cells??? I'm a little confused with the questions

[SocialRhubarb]

Ok, so firstly, let's say we've got a beaker with the copper (II) bromide with two electrodes attached to a battery, with one electrode attached to the battery's negative end and one electrode attached to the battery's positive end.

We've got electrons coming out of the battery's negative terminal into the attached electrode. You know what likes electrons? Copper (II) ions. Copper (II) ions react with the electrons, and are reduced, to form copper metal. Cu²⁺ (aq) + 2e^----> Cu (s). Even though this happens at the negative electrode, it is where reduction occurs and hence is at the cathode.

But with all these copper (II) ions coming out of solution, we have a build up of negative charge. To balance the charge and complete the circuit, the bromide ions are oxidised. The bromide ions react to form bromine and two electrons. 2Br^- (aq)---> Br₂ (aq) + 2e^-. Although this happens at the positive electrode, it is where oxidation occurs and hence is at the anode.

The reason that we can go 'backwards' on the electrochemical series is because we apply a voltage through the battery. In a galvanic cell, we let an oxidant and a reductant react, and in the process we generate a voltage. In electrolysis, we apply a voltage to cause a reaction, as opposed to a reaction generating a voltage. It's like the reverse of a galvanic cell.

[Edward21]

Ok, so firstly, let's say we've got a beaker with the copper (II) bromide with two electrodes attached to a battery, with one electrode attached to the battery's negative end and one electrode attached to the battery's positive end.

We've got electrons coming out of the battery's negative terminal into the attached electrode. You know what likes electrons? Copper (II) ions. Copper (II) ions react with the electrons, and are reduced, to form copper metal. Cu²⁺ (aq) + 2e^----> Cu (s). Even though this happens at the negative electrode, it is where reduction occurs and hence is at the cathode.

But with all these copper (II) ions coming out of solution, we have a build up of negative charge. To balance the charge and complete the circuit, the bromide ions are oxidised. The bromide ions react to form bromine and two electrons. 2Br^- (aq)---> Br₂ (aq) + 2e^-. Although this happens at the positive electrode, it is where oxidation occurs and hence is at the anode.

The reason that we can go 'backwards' on the electrochemical series is because we apply a voltage through the battery. In a galvanic cell, we let an oxidant and a reductant react, and in the process we generate a voltage. In electrolysis, we apply a voltage to cause a reaction, as opposed to a reaction generating a voltage. It's like the reverse of a galvanic cell.

Omg, you are simply amazing, A+ answer, I am printing this out as reference, sometimes all you need is someone to translate textbook talk into forum talk and you get it 

[Edward21]

Can we definitively say, for VCE chemistry purposes, that electrons always flow from anode to cathode, where the species is oxidised at the anode (negative in galvanic, positive in electrolysis) producing electrons, that flow to the cathode (positive in galvanic, negative in electrolysis) to be reduced by the electrons? 

[thushan]

Can we definitively say, for VCE chemistry purposes, that electrons always flow from anode to cathode, where the species is oxidised at the anode (negative in galvanic, positive in electrolysis) producing electrons, that flow to the cathode (positive in galvanic, negative in electrolysis) to be reduced by the electrons? 

That's correct.