October 17, 2019, 01:41:30 AM
Forum Rules: Read This Before Posting Topic: Generalized Redox Potential  (Read 725 times)

0 Members and 1 Guest are viewing this topic. Therapsid

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• Mole Snacks: +0/-0 Generalized Redox Potential
« on: February 02, 2019, 08:51:08 AM »
Hi,

Does (the amount of energy released) per (unit charge transferred as electrons) in a general redox reaction have a special name, or is it simply called the redox potential of that reaction?

I think that this quantity (I'll just call it redox voltage for now) is a very important property of redox reactions in general, not just ones happening in solution, because it tells us how "strong" the redox reaction is. Is it true that the redox voltage also allows us to predict whether a given redox reaction will take place? (The advantage of the redox voltage over the enthalpy change of reaction would be that it doesn't depend on how much of the reaction takes place, so the part reactions don't have to be balanced with each other for the prediction to be made.)

Example:

Mg (s) + 1/2 O2 (g) MgO (s)
ΔH = -601.6 kJ mol-1 (4 s.f.)
redox voltage = (-ΔH/NA) / (2e) = (-(-601.6 kJ mol-1) / (6.022×1023 mol-1)) / (2 × 1.602×10-19 C) = 3.118 V (4 s.f.)

2 Na (s) + 1/2 O2 (g) Na2O (s)
ΔH = -418.0 kJ mol-1 (4 s.f.)
redox voltage = (-ΔH/NA) / (2e) = (-(-418.0 kJ mol-1) / (6.022×1023 mol-1)) / (2 × 1.602×10-19 C) = 2.166 V (4 s.f.)

3.118 > 2.166, therefore the reaction

Mg (s) + Na2O (s) MgO (s) + 2 Na (s)

is exothermic (a little surprising, I know).
In addition, the redox voltage of this third reaction = redox voltage of 1st reaction - redox voltage of 2nd reaction = 3.118 V - 2.166 V = 0.952 V (3 d.p.).

If any of my assumptions, calculations and/or conclusions are incorrect, please tell me.

Therapsid mjc123

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• Mole Snacks: +230/-11 Re: Generalized Redox Potential
« Reply #1 on: February 04, 2019, 04:34:33 AM »
The redox potential is not related to the enthalpy change, but to the Gibbs energy change (which takes into account enthalpy and entropy, by the relation ΔG = ΔH - TΔS).
E is related to ΔG by the equation ΔG = -nFE.
Therefore your calculated example is invalid.
ΔG°f for many compounds is tabulated. (For elements in standard states it is zero.) I find values of -569.4 kJ/mol for MgO and -379.1 kJ/mol for Na2O. Try your calculation again. Therapsid

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« Reply #2 on: February 04, 2019, 09:04:13 AM »

Applying your correction ΔG = -nFE = -nNAeE to my calculation:

Mg (s) + 1/2 O2 (g) MgO (s)                                  (1)
ΔGr1 = -569.4 kJ mol-1 (4 s.f.)
n1 = 2
Er1 = -ΔGr1 / (n1 NA e) = -(-569.4 kJ/mol) / (2 × 6.022...×1023 mol-1 × 1.602...×10-19 C) = 2.950707... V = 2.951 V (4 s.f.)

2 Na (s) + 1/2 O2 (g) Na2O (s)                               (2)
ΔGr2 = -379.1 kJ mol-1 (4 s.f.)
n2 = 2
Er2 = -ΔGr2 / (n2 NA e) = -(-379.1 kJ/mol) / (2 × 6.022...×1023 mol-1 × 1.602...×10-19 C) = 1.964547... V = 1.965 V (4 s.f.)

Mg (s) + Na2O (s) MgO (s) + 2 Na (s)                      (3)
ΔGr3 = ΔGr1 − ΔGr2 = (-569.4 kJ mol-1) − (-379.1 kJ mol-1) = -190.3 kJ mol-1 (4 s.f.)
n3 = n1 = n2 = 2 =: n
Er3 = -ΔGr3 / (n3 NA e) = -(-190.3 kJ/mol) / (2 × 6.022...×1023 mol-1 × 1.602...×10-19 C) = 0.986160... V
= -(ΔGr1 − ΔGr2) / (n NA e) = (-ΔGr1 / (n NA e)) − (-ΔGr2 / (n NA e)) = Er1Er2 = 2.950707... V − 1.964547... V = 0.986160... V
= 0.986 V (3 d.p.)

We see that, as expected, the reaction is indeed thermodynamically favorable. mjc123

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• Mole Snacks: +230/-11 Re: Generalized Redox Potential
« Reply #3 on: February 06, 2019, 07:17:06 AM »
That is correct; the only quibble I have with your presentation is writing e.g. "569.4 kJ/mol" in your calculations, rather than "569,400 J/mol". If you multiply out the numbers you have written, the answer will be wrong. Obviously you have remembered to make the conversion, but to avoid mistakes it is helpful to write it out explicitly, and also in an exam will make your working clearer to the examiner. Therapsid

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• Mole Snacks: +0/-0 Re: Generalized Redox Potential
« Reply #4 on: February 10, 2019, 06:16:09 AM »
Thanks!