Asked by Gabriella
Find the equilibrium constant for the reaction Cr(s) + Zn2+(aq)→Cr2+(aq) + Zn(s) if the standard cell emf is −0.76 V at the cathode and −0.91 V at the anode.
RT
-
F= 0.025693 V
RT
-
F= 0.025693 V
Answers
Answered by
DrBob222
Cr(s) + Zn^2+(aq) ==> Cr^2+(aq) + Zn(s)
Cr(s) ==> Cr^2+ + 2e E = 0.91
Zn^2+ + 2e ==> -0.76
---------------------
Cr(s) + Zn^2+(aq) ==> Cr^2+(aq) + Zn(s)
Ecell = 0.91 + (-0.76) = ?
I don't get the significance of the RT/F = 0.025693 v.
dG = -nEF = -RTlnK
Substitute the numbers and calculate K.
Cr(s) ==> Cr^2+ + 2e E = 0.91
Zn^2+ + 2e ==> -0.76
---------------------
Cr(s) + Zn^2+(aq) ==> Cr^2+(aq) + Zn(s)
Ecell = 0.91 + (-0.76) = ?
I don't get the significance of the RT/F = 0.025693 v.
dG = -nEF = -RTlnK
Substitute the numbers and calculate K.
Answered by
yo
the problem doesn't give u the temperature. but instead they give u the RT/F which you can use instead. you can change the equation to: nE=(RT/F)lnK. Then plug and chug ->
(2)(.15) = (.025693)lnK
then solve for K: i got roughly 117752.538
but as for if that's right idk, cus it seems like an abnormally large answer for a hw question
(2)(.15) = (.025693)lnK
then solve for K: i got roughly 117752.538
but as for if that's right idk, cus it seems like an abnormally large answer for a hw question
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