Question

I'm trying to calculate the Cell Potential of a given balanced chemical reaction from looking up values in appendices, but I'm having some trouble. More info below.

Answer 1

The balanced chemical equation itself is

\[\text{CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)}\]

What I'm having trouble figuring out is that I can't find values for CH4 in my Appendix, and I'm assuming there's a good reason for that; why is that? @Hoslos

Answer 2

@abb0t , could you help me at all on this?

Answer 3

You can use gibbs free energy.

Answer 4

I see that, but I don't see a compound in my appendix-which should have it, it says to utilize the Appendix to solve the problem-I don't see a compound CH4, which leads me to thinking I'm making some more fundamental error.

Answer 5

\[\triangle G_{f}^{} \ \ \rm H_{2}O= -237.2 \ kJ/mol,\]\[\triangle G_{f} \ \rm CO_{2}= -394.4 \ kJ/mol,\]\[\triangle G_{f} \rm \ O_{2} = 0 \ kJ/mol.\]

What about CH4? What should I do about that compound, since I can't find a value in the appendix; should I sum some other values or something like that?

Answer 6

@abb0t

Answer 7

Let us try to solve like the Standard Enthalphy change of Combustion method= reactants-products. Being -[-394.4 + 2(-237.2)]= 868.8kj/mol. Is this the answer ?

Answer 8

your appendix might be divided into organic and inorganic sections. Check around, I'm sure there's a \(\Delta G\) for \(CH_4\)

Answer 9

I'm looking for a value in volts, not kilojoules per mole. Seeking the Standard Cell Potential.

I think I first have to find Delta G in Kilojoules, as you have, and then use a formula to convert that value to the Standard Cell Potential.

Answer 10

\[\triangle G_{E}^{\circ} = -nFE^{\circ}\]

Answer 11

that's exactly what i'm suggesting

Answer 12

n can be found based on the number of moles of electrons, I believe, we can calculate delta G? I dunno, I'm having a hard time figuring this out, but being sleepy as heck probably has to deal with that moreso.

Answer 13

I checked both the listing under Carbon and Hydrogen to see if either had CH4 listed; I could find neither, will screencap to make sure I'm not just missing it.

(Actually, nevermind, I'm on a public PC and I don't have screencap software on here, can't do that.)

Answer 14

Yeah, neither element has it, although it should be under the Carbon heading.

Answer 15

I checked one of my books. \(CH_4\) has a \(\Delta G\) of -51kJ/mol

Answer 16

Alright, cool. So putting this altogether,

Answer 17

\[\triangle G^{\circ}=[2(-237.2)+(-394.4)]-[(-51)]\]

Answer 18

Does that setup look correct for solving for Gibbs Free Energy of the reaction?

Answer 19

*Change in Free Energy

Answer 20

looks good to me

Answer 21

Alright, cool. One moment.

Answer 22

\[\triangle G^{\circ}=-817.8=-nFE^{\circ}\]

Answer 23

(In KiloJoules, need to remember.)

Now I just need to find the number of moles of electrons involved.

Answer 24

I need to setup the half-reaction (at least one, I think only one) to find out the number of moles of electrons involved, but I'm not sure how to setup that half-reaction for this problem.

Answer 25

one half-reaction will be the methane - CO2 half-reaction, the other will be O2 to H2O

Answer 26

Thank you, sorry, had to go deal with something IRL really quick. One sec.

Answer 27

The thing that confuses me with these half-reactions is that I don't see how I can really separate them, since both terms are either lacking or needing something on one hand side provided by a separate compound, e.g., I'd like to setup the methane-CO2 half-reaction as such:

Answer 28

\[\rm CH_4 \rightarrow CO_{2}+4H,\]but the left hand is obviously lacking a necessary term with oxygen.

Answer 29

when you split a reaction like this, the half-reactions won't start out being balanced. we have to add pieces to both sides to see where the electrons come in

Answer 30

I balance half-reactions like this:

Answer 31

Is this sort of where I'm allowed to add Hydrogen and Water molecules at will?

Answer 32

yes, but not at will.

Answer 33

\[CH_4 \rightarrow CO_2\]

Answer 34

balance the oxygen by adding water\[2H_2O + CH_4 \rightarrow CO_2\]

Answer 35

balance the hydrogen by adding H+(aq)

Answer 36

\[2H_2O + CH_4 \rightarrow CO_2 + 8H^+\]

Answer 37

now this half-reaction is balanced for mass, but not charge

Answer 38

\[2H_2O + CH_4 \rightarrow CO_2 + 8H^{+1} + 8e^{-1}\]

Answer 39

(Taking a look at what was just written, thank you very much for your help so far.)

Answer 40

Alright, and since the other half-reaction will, by default, be required to have the exact same number of moles of electrons, we can stop here and use 8 as n, correct?

(Doesn't double just because additional values on other hand of eqn, equals the total number of electrons on any given side)

Answer 41

its very likely that the number of electrons in the reduction half-reaction will also be 8, or some factor of 8, but let's check to see

Answer 42

\[4e^{-1} + 4H^{+1} + O_2 \rightarrow 2H_2O\]

Answer 43

Alright, so does that mean we need to.....well, no, I don't see how that would work, maybe?

Does that 4e mean we need to collectively double everything in the reduction half-reaction to get the electrons to cancel?

Answer 44

exactly. also see how the H+ ions will cancel when we do that?

Answer 45

Yeah, I do. I'm assuming it's always like that, but that's still just something that's too convenient that I have to ask; is it always like that,?

Answer 46

(At least with any real, non-imaginary unconventional chemical equations or something/)

Answer 47

\[2H_2O + CH_4 \rightarrow CO_2 + 8H^{+1} + 8e^{-1}\]PLUS
\[2*(4e^{-1} + 4H^{+1} + O_2 \rightarrow 2H_2O)\]

Answer 48

the only thing that HAS TO cancel when you add half-reactions together is the electrons

If some other pieces cancel, great, but it's not required

Answer 49

Alright, so you can have extra (not sure about the proper term for free, radical Hydrogen atoms without an electron attached) Hydrogen not in the diatomic H_2 form, and you can have extra water, and that's cool.

Answer 50

Alright, well, on that note, e = 8, right?

Answer 51

you can, but in this reaction all the H+ ions cancel

Answer 52

yes, n = 8 for the electrons in the Gibbs' equation

Answer 53

Alright. Taking a shot at this now.

\[-817.8 kJ = \triangle G^{\circ}=-nFE^{\circ};\]
\[817,800=(8)(96,500)E^{\circ}\]\[E^{\circ}=\frac{817,800}{(8)(96,500)}=1.05 \ \rm V\]

Answer 54

looks good to me

Answer 55

Alright, gonna see if it likes the answer, heh. One sec.

Answer 56

(Hooray!)

Answer 57

Hah, thank you so much, I spent a while looking through my textbook and it never directly made the connection, at least in the chapter itself, between the two equations formally, it just vaguelly referred to Gibbs Free Energy earlier, and I wasn't sure if those two relations were equal.

Answer 58

you can find \(\Delta G\) about 3 different ways, depending on the information you're given. glad i could help