JoannaBlackwelder
  • JoannaBlackwelder
Half cell problem help: number 15 please
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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JoannaBlackwelder
  • JoannaBlackwelder
JoannaBlackwelder
  • JoannaBlackwelder
I get A. Can anyone confirm, please?
JFraser
  • JFraser
Do you have each half-cell potential?

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Rushwr
  • Rushwr
U need to have the potentials. E cell = Ecathose - Eanode
abb0t
  • 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.
Cuanchi
  • Cuanchi
I didn't get any of the answers. What half reactions did you choose? Cr -> Cr+3 oxidation Mn+2 ->Mn reduction
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JoannaBlackwelder
  • JoannaBlackwelder
I used those half reactions you did, but my chart said that the Mn one had a red potential of -1.04 V
JoannaBlackwelder
  • JoannaBlackwelder
So, I technically got 1.78, which I figured was pretty close to A
Cuanchi
  • Cuanchi
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
abb0t
  • abb0t
Ok, so you can also separate it by left and right cell. and do E\(\sf _{total} = E_{left}-E_{right}\)
abb0t
  • 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.
JoannaBlackwelder
  • JoannaBlackwelder
Wow, thanks so much everyone!
Cuanchi
  • Cuanchi
@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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