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

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JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.0@.Sam. @nincompoop

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.0I get A. Can anyone confirm, please?

JFraser
 one year ago
Best ResponseYou've already chosen the best response.0Do you have each halfcell potential?

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.0U need to have the potentials. E cell = Ecathose  Eanode

abb0t
 one year ago
Best ResponseYou've already chosen the best response.0Yes. To find the overall potential of this cell, you need to have the standard potentials of each. It goes from left to right.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I didn't get any of the answers. What half reactions did you choose? Cr > Cr+3 oxidation Mn+2 >Mn reduction

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.0I used those half reactions you did, but my chart said that the Mn one had a red potential of 1.04 V

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.0So, I technically got 1.78, which I figured was pretty close to A

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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

abb0t
 one year ago
Best ResponseYou've already chosen the best response.0Ok, so you can also separate it by left and right cell. and do E\(\sf _{total} = E_{left}E_{right}\)

abb0t
 one year ago
Best ResponseYou've already chosen the best response.0Where E = E\(\sf ^o\) + \(\frac{0.0831}{n}\)ln\(\sf \frac{prod}{react}\) n = numer of electrons being transferes, here it's 1.

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.0Wow, thanks so much everyone!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@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.
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