Question

Please explain this concept, the answer is an actual value.

Answer 1

@aaronq I don't understand how to get the Voltage by changing the concentration of moles in the anode (left) or the cathode (right) I just know the half reactions

\[5(Cu_{s} \rightarrow Cu^{2+} + 2e)\\2(MnO_{4} + 8H^{+} + 5e \rightarrow Mn^{2+} + 4H_{2}O)\]

Answer 2

I know the reaction \[E = E° -\frac{ RT }{ nF }lnQ\]

Answer 3

I just understand how to get the voltage in general what calculations were involved.

Answer 4

@thomaster

Answer 5

you would change the concentrations on the lnK

\(K=\dfrac{[products]}{[reactants]}\)

Answer 6

but it says by a factor of 100 and I don't know the concentrations...

Answer 7

so i'm guessing that the concentrations are equal

then by decreasing it by a factor of a hundred you have 1/100

Answer 8

does it mean like 1M and K = (1M/0.01))

I got ln(100)

I changed that to (2.303)(log(100))

so I got

E = 1.17 - (0.05915)/10)*(2)
= 1.17- 0.01183
@aaronq

The answer for a) is 0.059 V and
b) = -0.19 V my answers are no where close

Answer 9

i think n = 10, there are 10 electrons being exchanged

Answer 10

oh you do have 10, nvm

Answer 11

We alsor reviewd the \[\Delta G = -nFE\] but that has nothing to do with this.

Answer 12

or doe it O.o?

Answer 13

no, it doesn't. hm so E = 1.17 - (0.05915)/10)*(2)

why *2? shouldn't be E = 1.17 - (0.05915)/10)*(2.303)(log(100)) ?

Answer 14

no... because if you change ln to log its by a factor of 2.303

so ex log100 = 2
ln 100 = 4.605

but ln100 / 2.303 = 2
so 2.303log(x) =ln(x)

@aaronq

Answer 15

hmm makes sense. sorry, i don't really see what you're doing wrong. Maybe double check \(E^o\)?

Answer 16

on the table I see -0.34 for Cu and 1.51 MnO4- = 1.19


= 1.17..=(

Answer 17

hm no.. -0.34 + 1.19 = 0.85

Answer 18

what where did you get -1.19 on the table....

I meant that


CU = -0.34

MnO4- = 1.51
______
1.19

Answer 19

lol I'm sorry. I've been really busy all day i think i'm all tired out.

are both of those reduction potentials? did you reverse the oxidation one?

Answer 20

yes cu is reversed it was 0.34 to -0.34

Answer 21

In the picture there is also a species of H2SO4 but since MnO4 is a stronger oxidizing agent I wrote the reaction...

Answer 22

OMG I forgot to add the powers to the equation...

Answer 23

from the balanced equation

Answer 24

! there you go