## JoannaBlackwelder one year ago Half cell problem help: number 15 please

1. JoannaBlackwelder

@.Sam. @nincompoop

2. JoannaBlackwelder

I get A. Can anyone confirm, please?

3. JFraser

Do you have each half-cell potential?

4. Rushwr

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

5. abb0t

Yes. To find the overall potential of this cell, you need to have the standard potentials of each. It goes from left to right.

6. anonymous

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

7. JoannaBlackwelder

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

8. JoannaBlackwelder

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

9. anonymous

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

10. abb0t

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

11. abb0t

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

12. JoannaBlackwelder

Wow, thanks so much everyone!

13. anonymous

@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.