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!
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.
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.
To calculate the cell potential for the given reaction, you need to use the Nernst equation. The Nernst equation relates the cell potential (Ecell) to the standard cell potential (E°cell), the reaction quotient (Q), the number of electrons transferred in the reaction (n), the gas constant (R), the temperature (T), and the Faraday constant (F).
The Nernst equation is given by:
Ecell = E°cell - (RT/nF) * ln(Q)
Given:
[Cr2+] = 0.875 M
[Ni2+] = 0.0100 M
The balanced equation for this reaction is:
Cr2+(aq) + 2e- → 2Cr(s)
Ni2+(aq) + 2e- → Ni(s)
The standard cell potential, E°cell, can be found by looking up the reduction potentials for the half-reactions in a standard reduction potential table. However, since the standard cell potential is not given in the question, we cannot provide the value.
To calculate the reaction quotient, Q, you need to express the concentrations of the products and reactants raised to their stoichiometric coefficients:
Q = ([Cr]^2 / [Ni]^2)
Substituting the given concentration values:
Q = ([Cr2+]^2) / ([Ni2+]^2)
= (0.875^2) / (0.0100^2)
= 61.06
Now, you can substitute the values of E°cell, R, T, n, F, and Q into the Nernst equation to calculate the cell potential (Ecell). However, since the value of E°cell is not provided in the question, we cannot compute the exact cell potential at this point.
Make sure you have the correct value of E°cell for the given reaction. If you still get the incorrect answer after rechecking the calculations, please double-check the given information and make sure there are no errors in the concentrations or the stoichiometry of the reaction.
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