Question

Help me understand. Determine the equilibrium constant for the following reaction at 398 K.
3Cu(s) + 2MnO4- (aq) + 8H+(aq) --> 3Cu2+(aq) + 2MnO2(s) + 4H2O (l)

Answer 1

hm have you used the nernst equation?

Answer 2

yes we have before but the instructor, she goes thru it so fast in her slides that I can hardly keep up an we have an exam on tuesday
@aaronq

Answer 3

@aaronq its that E cell = ECell standard potential -.0592V/n logQ

Answer 4

at equilibrium \((i.e.~E_{cell}=E^o_{cell})\), so you can solve for K.

\(0=E^o_{cell}-\dfrac{RT}{nF}lnK\rightarrow E^o_{cell}=\dfrac{RT}{nF}lnK\rightarrow lnK=\dfrac{nFE^0_{cell}}{RT}\)

Answer 5

wha?! haha but do i have k?

Answer 6

K is the equilibrium constant (what you're looking for).

Answer 7

remember that the reduced equation you wrote can only be used at 298 K, 1M concentration and at 1 atm (i.e. standard conditions) which is not the case here (398 K).

Answer 8

oh...okay haha so what do I do?

Answer 9

@aaronq im looking at my note paper with equtions there is a Ecell=.0592V/n * log K

and there is a
Ecell = Ecell- .0592V/n *logQ

Answer 10

use the original form: \(E_{cell}=E^0_{cell}-\dfrac{RT}{nF}lnK\) , which solves to what i wrote above

Answer 11

okay. so then... here hold on ..

the nF = hold on I have to do that reduction oxidation thingy right?
and for that oxidation would be anode so negative,
an cathode is reduction that is positive right? Im just making sure I have this part right first. @aaronq

Answer 12

I also found this chart just in case

http://bilbo.chm.uri.edu/CHM112/tables/redpottable.htm

Answer 13

UGH!!! IM SO LOST!

okay this is what I understand from your formula:
nF= the number of mol electrons
R= the constant, i think its 8.314 J/mol K
T= is the temperature=398K

@aaronq Is all of this right so far? okay I just dont remember how to find the \[E^0 cell\]

Answer 14

n = # of electrons exchanged in redox reaction:

3Cu(s) + 2MnO4- (aq) + 8H+(aq) --> 3Cu2+(aq) + 2MnO2(s) + 4H2O (l)

6 electrons

Answer 15

F = faradays constant

Answer 16

okay F = Faradays constant is 96485 C/mol e-

Answer 17

yep, thats right. now you just need to find the potential of the cell \(E^0_{cell}=E^{red}_{cathode}-E^{red}_{anode}\)

Answer 18

sometimes it's also written as \(E^0_{cell}=E^{red}_{cathode}+E^{oxd}_{anode}\)

Answer 19

oh....okay wait haha i always thought u would subtract it, but I guess I keep getting it wrong, so basically it becomes an addition

Answer 20

you subtract it if both of your values are reduction potentials. you add them if one is oxidation (at anode) and one is reduction potential (at cathode).

Answer 21

okay so @aaronq the anode would be the positive charge right? or am i mixed up.
I remember An-Ox for Anode=Oxidation= -
and Red-Cat for Cathode=Reduction=+

Answer 22

\(\huge \color{red}OI\color{red}L ~~\color{blue}RI\color{blue}G\)

Oxidation loss of electrons (at anode)
Reduction gain of electrons (at cathode)

Answer 23

okay let me see if I have this right then

Answer 24

SORRY @aaronq INTERNET CONNECTION IS being lame

Answer 25

no worries !

Answer 26

okay the Cathode, or reduction reaction is going to be the 2MnO4

Answer 27

yes

Answer 28

right? cuz it goes from a 2MnO4- to a 2MnO2; it ganed the e- to make it a positive and added Oxygen :D

Answer 29

ok, so then now that i know that

Answer 30

okay E cell.er....um..... okay hold on

Answer 31

sorry, i read that as "now what" lol

Answer 32

Would I use....half reactions for the Ecell(V) ?

Answer 33

yes

Answer 34

so find these reactions in a reduction potential chart and set them up

Answer 35

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch20/electro.php
Im using this website

Answer 36

Cu2+ + e- <-----> Cu+ 0.158 so I would flip it right? and

MnO4- + 2 H2O + 3 e- <-----> MnO2 + 4 OH- 0.588

Answer 37

@aaronq

Answer 38

you're using this one Cu2+ + 2 e- <-----> Cu 0.3402

Answer 39

but yeah flip that one

Answer 40

wait haha I just forgot, but why do I flip it?

Answer 41

and why do I use that one an not the others? @aaronq

Answer 42

@aaronq I also cant figure out which MnO4 to use:
MnO4- + 2 H2O + 3 e- <-----> MnO2 + 4 OH- 0.588
or
MnO2 + 4 H+ + 2 e- <-----> Mn2+ + 2 H2O 1.208
or
MnO4- + 8 H+ + 5 e- <-----> Mn2+ + 4 H2O 1.491

Answer 43

well in the main equation they have an 8H+ --> 4H+
would I use the bottom one that gives me 1.491?
@aaronq

Answer 44

you flip it if you're adding the values together, because as it's written it is a reduction equation and you're looking for the oxidation.

Answer 45

i would use the top one from those 3

Answer 46

ohh...okay haha @aaronq thank you
so then is this right, okay dang hold on gotta do it again but not use the bottom number

Answer 47

no i mean the bottom one. use the bottom one of the 3

Answer 48

OKAY, oh........haha okay let me do it again hold on haha

Answer 49

okay @aaronq I did the Ecell= C+A
1.491+-.34 = 1.151v

Answer 50

right?

Answer 51

right?

Answer 52

yep thats right

Answer 53

@aaronq sorry I keep loosing connection, is ur computer the same way?

Answer 54

no, i was just away from the computer.

so, \(\huge K=e^{\small(\dfrac{nFE^o_{cell}}{RT})}\)

Answer 55

OH! ahah u rearranged that one equation huh? haha

Answer 56

haha yeah, you should try it yourself on paper to make sure you can arrive at the same result

Answer 57

@aaronq Im going to need your help big time right now because. I dont remember the n, you said the n was the number of elections exchanged in redox reaction. u said that was 6 because there was 3 Cu and 2MnO4's so you times them and got 6 right?

Answer 58

yes

Answer 59

okay then so.. @aaronq

Answer 60

e=1.151v

Answer 61

\[1.151v (\frac{ 8.314*398K }{ 6*96485 })lnK\]

Answer 62

oops I think I did it wrong @aaronq

Answer 63

\[k=1.151v(\frac{ 6*96485*e^0cell }{ 8.314*398K })\]

Right ????@aaronq

Answer 64

@aaronq is the answer....Well I dont know what to put for the e^0 Cell :/

Answer 65

this dumb connection! @aaronq

Answer 66

so you had to end up with: \(\sf K=e^\dfrac{6*(9.65*10^4 C/mol)*1.151V}{(8.314 J/mol*K)*(398K)}\)

Answer 67

im back @aaronq

Answer 68

yes I did @aaronq but what do i put for the e??

Answer 69

@aaronq I get K=e 201.4

Answer 70

do i use that 10^?? to get.... WOW, haha I got an over flow error

Answer 71

"e" is euler's number. it's transcendental but you can write it as 2.718

so it's \(2.718^{2.93*10^{87}}\) i'm getting \(\infty\), which just means that the number is too big for my calculator lol

Answer 72

okay @aaronq i got a big number haha. I dont even think we were doing that part right. Here is what I am given on an equation sheet of mine

Answer 73

I will be back online in the morning.

Answer 74

i think it was late and i was doing something wrong.

now ii'm getting \(e^{201.4}=2.93*10^{87}\) which seems right.

Answer 75

ps i'm 100% sure were using the right equation.

Answer 76

haha @aaronq okay Im back! had to go out to store.

Answer 77

@aaronq but where okay hold on so this was the equation we were working on: well this part:

\[K=e(\frac{ 6*96485*1.151v }{ 8.314*398K })\]
giving us:
\[K=e(\frac{ 666325.4 }{ 8.314*398K }) \]

\[K=e201.4\]

@aaronq NOW, do we raise both to the e^, but how does it get rid of the K? I understand how u got\[K=e^{201.4} = 2.93x10^{87}\]

Answer 78

when you have: \(lnK=\dfrac{nFE}{RT}\) you have to raise both sides to the \(e^x\) to get rid of ln.

so you end up with \(\Huge K=e^{\small \frac{nFE}{RT}}\)

e^(nFE/RT)

Answer 79

oh. okay that makes sense now lol sorry

Answer 80

the formatting its kinda weird because there are so many symbols in the exponent, which makes like it look like a multiplication, but it's not.

Answer 81

no problem ! hope everything is good now

Answer 82

okay so now that we have the K=2.93x10^87 that will be our volts, right?

Answer 83

no that's the equilibrium constant (usually unitless)

Answer 84

really? so we just solved it? I thought there would need another equation

Answer 85

hah yep thats it

Answer 86

Dang....Longest problem ever

Answer 87

thank you so much. I have other problems too that I will post on here in a little bit.

Answer 88

no problem !