Electroylsis - half equations!! (1 Viewer)

[Pace_T]

Hey all
I'm really confused about this (btw it's shipwrecks topic, i posted in the forum but im still confused, and maybe this can still be figured out without shipwrecks topic?)
Umm im really lost at the selection process for the half equations:
i *think* i understand the selection process, however I am totally clueless when it comes to pooling the possible half equations.
here is an example question - i hope someone can please go through the thinking process for the solution:
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Use the table of Standard potentials to write the half equations and net cell equations for the electrolysis of : 1. water 2. sodium iodide solution, at inert platinum electrodes. Determine the voltage required to initiate electrolysis in each case

Answer:
Water is the only species present that can be reduced or oxidised.
Cathode reaction:
2H2O(l) + 2e --> H2(g) + 2OH- E -0.83
Anode reaction:
2H2O(l) -->O2(g) + 4H+ + 4e- E -1.23
**why is it those two reactions? There's 3 reactions that involve H and O and stuff i nthe reduction potential list.
ie 6H2O(l) -->2H2(g) + O2(g) + 4OH- + 4H+
4OH- + 4H+ -->4H2O(l) <----- **where did this come from?
The net cell equation is
2H2O(l) --> 2H2(g) + O2(g)
.'. 2.06V needed

Sodion iodide solution:
anode can be:
2H2O(l) -->O2(g) + 4H+ + 4e- EMF -1.23V
or
2I- --> I2(aq) + 2e- E = -0.62V
or
2I- --> I2(s) + 2e- E = -0.54

.'. it is 2I- --> I2(s) + 2e- (more negative reduction potential)
**how do we know its those 3 reactions?

cathode reaction is
2H2O(l) + 2e- --> H2(g) + 2OH- E = -0.83V <--- ** why is it this one?

.'. V needs to be 1.37V
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ANY help would be greatly appreciated. My teacher doesn't know this either and I would be extremely thankful if you could please just explain this for me.


Thank you so much!!!

 

[Jumbo Cactuar]

When we apply a voltage the potential relates who much the electrons want to get to through the circuit. Our electrolysis is set up in such a way that for electrons to get through the circuit they need to participate in redox chemistry. Now, sensibly, the electrons want to take the easiest route. So when choosing the correct half-cells you are looking for what has the highest net redox potential such that the electons are transported through the solution. From their you can work out the minimum voltage required.

The idea of two half-cell equations is so you can easily balance electrons with having to calculate oxidation numbers. Obviously electons and charges in the net equation need to be balanced. Hence we need an oxidation half-cell (loss of electrons) and a reduction half-cell (gain of electrons).

Electrolysis of water:

Relevant half-cell equations;

2H20 + 2e- <-> H2 + 2OH- ........ E0 = -0.83V - gain of e- = reduction
2H20 <-> 4e- + 4H+ + O2 ......... E0 = -1.23V - loss of e- = oxidation
2H20 <-> 2e- + H2+ + H2O2 .... E0 = -1.76V - loss of e- = oxidation
4OH- <-> 4e- + 2H20 +O2 ..........E0 = -0.40V - loss of e- = oxidation

consider the two redox pairs;

2H20 + 2e- <-> H2 + 2OH- ........ E0 = -0.83V - gain of e- = reduction
2H20 <-> 4e- + 4H+ + O2 ......... E0 = -1.23V - loss of e- = oxidation
...balanced...
4H20 + 4e- <-> 2H2 + 4OH- ........ E0 = -1.66V - gain of e- = reduction
2H20 <-> 4e- + 4H+ + O2 ............ E0 = -1.23V - loss of e- = oxidation
--------------------------------------------------------
2H20 <-> 2H2 + O2 ................. E0 = -2.89V
*Problem due to neutralisation; higher than expected*


2H20 + 2e- <-> H2 + 2OH- ........ E0 = -0.83V - gain of e- = reduction
2H20 <-> 2e- + 2H+ + H2O2 .... E0 = -1.76V - loss of e- = oxidation
--------------------------------------------------------
2H20 <-> H2O2 + H2 ................. E0 = -2.59V


4H20 + 4e- <-> 2H2 + 4OH- ...... E0 = -1.66V - gain of e- = reduction
4OH- <-> 4e- + 2H20 +O2 ......... E0 = -0.40V - loss of e- = oxidation
---------------------------------------------------------
2H20 <-> 2H2 + O2 ................... E0 = -2.06V


So;

The first one isn't valid since the acid/base reaction provides additional energy. The second requires 2.59V. The third requires 2.06V and is determined to be the appropriate reaction.


Similarly, sodium iodide;

Relevant half-cell equations;

2H20 + 2e- <-> H2 + 2OH- ........ E0 = -0.83V - gain of e- = reduction
2I- <-> I2(aq) + 2e- ................... E0 = -0.62V - loss of e- = oxidation
2I- <-> I2(s) + 2e- ...................... E0 = -0.54V - loss of e- = oxidation
3I- <-> I3-(aq) + 2e- .................. E0 = -0.54V - loss of e- = oxidation


The redox reaction with the lowest E0 is;

2H20 + 2e- <-> H2 + 2OH- ........ E0 = -0.83V - gain of e- = reduction
2I- <-> I2(s) + 2e- ...................... E0 = -0.54V - loss of e- = oxidation
--------------------------------------------------------------
2H20 + 2I- <-> I2(s) + H2 + 2OH- .. E0 = -1.37V

AND

2H20 + 2e- <-> H2 + 2OH- ........ E0 = -0.83V - gain of e- = reduction
3I- <-> I3-(aq) + 2e- .................. E0 = -0.54V - loss of e- = oxidation
-------------------------------------------------------------
2H20 + 3I- <-> I3-(aq) + H2 + 2OH- .. E0 = -1.37V

The answer is the first one and I think thats to do with entropy of the system, but you don't have to worry about that.

 

[Pace_T]

thank you very much!
just one thing to clarify: in the 1st equation you balanced, the E0 is doubled. shouldn't it stay the same?
thanks.

 

[Jumbo Cactuar]

good point ... Luckily I am not doing electrochem currently. You scratch my back I'll scratch yours!

 

[Ghost1788]

good point ... Luckily I am not doing electrochem currently. You scratch my back I'll scratch yours!

that....came out the wrong way....