Be sure to check your ion concentrations in the Nernst Equation are raised to the proper power.
Cell [Cr/Ni^+2[0.010M]//Cr^+3[0.875M]/Ni]
Q = [Cr]^2/[Ni]^3 = (0.875)^2/(0.010)^ = (0.766)/(1.0E-6) = 7.66E+5
E^o = E(Redn) - E(Oxdn) = (-0.23v)-(-0.74v) = 0.51v
n = 6 electrons
E(NonStd) = (0.51v) - [(0.0592/6)log(7.66E+5)v]
= 0.51v - 0.06v = 0.45v
Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Cr2 ] = 0.875 M and [Ni2 ] = 0.0100 M.
I keep on getting the answer .42 which is incorrect.
Thank you so much in advance!
6 answers
You don't show a reaction. Show what you've done and I'll find the error.
There's only one way a Cr/Ni Rxn can go... Cr([email protected]) => Ni([email protected]) => Std Cell = 0.51v
For non-Std Cell =>
[Cr/Ni^+2[0.010M]//Cr^+3[0.875M]/Ni]; n=6
2Cr + 3Ni^+2 => 2Cr^+3 + 3Ni
Q = ([Cr^+3]^2/[Ni^+2]^3)
Apply to Nernst Equation => E(Non-Std) = 0.45v
Work Check Note => Since [Cr^+3] > [Ni^+2] the cell potential has to be below the standard cell potential. Remember, when [Cr^+3] = [Ni^+2] => E(Std) = 0.51v; if [Cr^+3] < [Ni^+2] => Cell voltage is above standard cell potential. This can be a general reference when working problems on test day.
For non-Std Cell =>
[Cr/Ni^+2[0.010M]//Cr^+3[0.875M]/Ni]; n=6
2Cr + 3Ni^+2 => 2Cr^+3 + 3Ni
Q = ([Cr^+3]^2/[Ni^+2]^3)
Apply to Nernst Equation => E(Non-Std) = 0.45v
Work Check Note => Since [Cr^+3] > [Ni^+2] the cell potential has to be below the standard cell potential. Remember, when [Cr^+3] = [Ni^+2] => E(Std) = 0.51v; if [Cr^+3] < [Ni^+2] => Cell voltage is above standard cell potential. This can be a general reference when working problems on test day.
Yes, there is only one way the reaction can go but the problem asks for Ecell but doesn't say it is spontaneous. If it doesn't go spontaneously the Ecell is not the same.
Since cell is Cr/Ni, it shouldn't have to say it's spontaneous. The thing that's missing is the standard cell voltages. With that it's spontaneous.
Since cell is Cr/Ni, it shouldn't have to say it's spontaneous. The thing that's missing is the standard cell voltages. With that it's spontaneous. Flow is spontaneous from more negative to more positive electrode potential.
3 answers