Question

Half cell problem help: number 15 please

Answer 1

@.Sam. @nincompoop

Answer 2

I get A. Can anyone confirm, please?

Answer 3

Do you have each half-cell potential?

Answer 4

U need to have the potentials.
E cell = Ecathose - Eanode

Answer 5

Yes. To find the overall potential of this cell, you need to have the standard potentials of each.

It goes from left to right.

Answer 6

I didn't get any of the answers. What half reactions did you choose?
Cr -> Cr+3 oxidation
Mn+2 ->Mn reduction

Answer 7

I used those half reactions you did, but my chart said that the Mn one had a red potential of -1.04 V

Answer 8

So, I technically got 1.78, which I figured was pretty close to A

Answer 9

I don't know where did you got your cell potentials This is my calculation.

Cr(s) -> Cr+3(aq) + 2 e- oxidation 0.740 V
Mn+2 (aq) +2e- ->Mn (s) reduction -1.180V

E= -0.44 V it is not spontaneous

Answer 10

Ok, so you can also separate it by left and right cell. and do E\(\sf _{total} = E_{left}-E_{right}\)

Answer 11

Where E = E\(\sf ^o\) + \(\frac{0.0831}{n}\)ln\(\sf \frac{prod}{react}\)
n = numer of electrons being transferes, here it's 1.

Answer 12

Wow, thanks so much everyone!

Answer 13

@abb0t , sorry to tell you but the formula is Ecell=Eright−Eleft is ALLWAYS final minus initial, and the number of electrons being transferred in these reactions are 2e-.
Respect to the other equation E = Eo + (0.0831/n) x ln ([prod]/[react]) it is not necessary .In this case the two concentrations are the same and the ln 1 =0 then E=Eo and besides the question in the problem is not asking for E but Eo.