Using the half reaction potentials calculate the E^o cell for the Ni-Al cell.

Ni (aq) + 2e- <-> Ni (s) -0.25v
Al3+ (aq) + 3e- <-> Al (s) -1.66v

Using that answer determine Ecell for Ni-Al cell when the concentration of Al3+ is 4.00M and the concentration of Ni2+ (aq) is 2.00*10^-5 M.

I am unsure about the nomenclature for this. I have set it up for the cell to be spontaneous; the Al is the anode and Ni is the cathode. If the term "Ni-Al cell" means that the cell is non-spontaneous, you will need to reverse everything. Hope this helps.

Al ==> Al^3+ + 3e Eo = 1.66v
Ni^2+ + 2e ==> Ni Eo = -0.25
----------------------------
2Al + 3Ni^2+ ==> 2Al^3+ + 3Ni
Eocell = EAl above + ENi above = 1.66+(-0.25) = ?.

For the second part:
Ecell = Eocell - (0.0592/6)log Q
where Q = (Al^3+)^2(Ni)^3/(Al)^2(Ni^2+)^3 and solve for Ecell.

To calculate the standard cell potential (E°cell) for the Ni-Al cell, you need to first calculate the standard reduction potentials for the half-reactions and then use the Nernst equation to calculate the cell potential under non-standard conditions.

1. Calculate the standard cell potential (E°cell):
E°cell = E°cathode - E°anode
E°cell = (-1.66V) - (-0.25V)
E°cell = -1.41V

2. Determine the cell potential (Ecell) under non-standard conditions:
Ecell = E°cell - (0.0592V/n) * log(Q)
where n is the number of electrons transferred and Q is the reaction quotient.

In this case, the balanced equation for the cell reaction is:
3Ni(s) + 2Al3+(aq) -> 3Ni2+(aq) + 2Al(s)

Since the number of electrons transferred is 6 (3 electrons per Ni2+ ion and 2 electrons per Al3+ ion), n = 6.

The reaction quotient (Q) can be determined using the concentrations of the species involved in the cell reaction:
Q = [Ni2+]^3 / [Al3+]^2 = (2.00*10^-5)^3 / (4.00)^2
Q = 1.25 * 10^-13

Using these values, plug in the numbers into the equation:
Ecell = -1.41V - (0.0592V/6) * log(1.25 * 10^-13)

Calculating this equation will give you the value of the cell potential under the given conditions.

To calculate the standard cell potential (E°cell) for the Ni-Al cell, we need to find the difference in potential between the two half-reactions.

1. Write the balanced equation for the overall cell reaction:
Ni (aq) + 2 Al3+ (aq) → Ni (s) + 2 Al (s)

2. Write the two half-reactions involved:
a) Ni (aq) + 2e- → Ni (s)
b) Al3+ (aq) + 3e- → Al (s)

3. Determine the standard potential for each half-reaction:
a) E°1 = -0.25 V (given)
b) E°2 = -1.66 V (given)

4. Add the two half-reaction potentials to get the cell potential:
E°cell = E°1 (cathode) - E°2 (anode)
= E°1 - E°2
= -0.25 V - (-1.66 V)
= -0.25 V + 1.66 V
= 1.41 V

The standard cell potential for the Ni-Al cell is 1.41 V.

To determine the cell potential (Ecell) for the Ni-Al cell under the given conditions:

1. Write out the balanced equation for the overall cell reaction with the species and their concentrations:
Ni (aq) + 2 Al3+ (aq) → Ni (s) + 2 Al (s)

2. Determine the number of electrons transferred in the reaction, which is the same as in the half-reactions (2e-).

3. Use the Nernst equation to calculate the cell potential:
Ecell = E°cell - (0.0592 V/n) * log(Q)

Where:
E°cell = standard cell potential (1.41 V)
n = number of moles of electrons transferred (2)
Q = reaction quotient = [Ni2+] /[Al3+]^2

4. Substitute the given values into the equation:
Q = ([Ni2+]) / ([Al3+]^2)
= (2.00*10^-5 M) / (4.00 M)^2
= 2.00*10^-5 / 16.00
= 1.25*10^-6

Ecell = 1.41 V - (0.0592 V/2) * log(1.25*10^-6)

5. Calculate the final answer:
Ecell = 1.41 V + (0.0296 V) * log(1.25*10^-6)

using the log function, you will find:
Ecell ≈ 1.41 V + 5.2443 * 10^(-8) V
Ecell ≈ 1.41 V

Therefore, the cell potential (Ecell) for the Ni-Al cell, when the concentration of Al3+ is 4.00 M and the concentration of Ni2+ is 2.00*10^-5 M, is approximately 1.41 V.