An electrochemical cell consists of a nickel metal electrode immersed in a solution with [Ni2+] = 1.0 M separated by a porous disk from an aluminum metal electrode.

(a) What is the potential of this cell at 25°C if the aluminum electrode is placed in a solution in which [Al3+] = 7.3 10-3 M?
(b) When the aluminum electrode is placed in a certain solution in which [Al3+] is unknown, the measured cell potential at 25°C is 1.54 V. Calculate [Al3+] in the unknown solution. (Assume Al is oxidized.)

(a) Well, if you have a nickel metal electrode and aluminum metal electrode, and they're separated by a porous disk, then I guess you could say they've got a "charged" relationship. But seriously, to calculate the potential of the cell, you'll need to use the Nernst equation:

Ecell = Eºcell - (RT/nF) * ln(Q)

Where Ecell is the potential of the cell, Eºcell is the standard cell potential, R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin (25°C = 298 K), n is the number of electrons transferred in the cell reaction, F is Faraday's constant (96500 C/mol), and Q is the reaction quotient.

In this case, the overall cell reaction is:

Al(s) + Ni2+(aq) -> Al3+(aq) + Ni(s)

To simplify things, let's assume the reaction is balanced and that two electrons are transferred.

Now, you'll need to calculate the reaction quotient Q using the concentrations given. And then plug it all into the Nernst equation to find the potential of the cell.

(b) Oh, you've got an unknown solution with an unknown [Al3+], huh? Well, you'll still need to use the Nernst equation to find the potential of the cell. This time, however, you know the measured cell potential and you need to calculate the unknown [Al3+].

So, you'll set up the Nernst equation again, but this time, you'll rearrange it to solve for Q:

Q = exp((Eºcell - Ecell) * (nF/RT))

Once you have Q, you'll use it to calculate the unknown [Al3+] using the balanced cell reaction and the number of electrons transferred.

Now, go forth and calculate those cell potentials and concentrations. Just remember, even if the math gets a bit shocking, don't get too amped up!

To calculate the potential of the electrochemical cell, we need to use the Nernst equation, which relates the cell potential to the concentration of the species involved.

(a) To calculate the potential of the cell when the aluminum electrode is immersed in a solution with [Al3+] = 7.3 x 10^(-3) M, we need the standard reduction potentials for the half-reactions involved.

The standard reduction potential for the nickel half-reaction can be found in the table of standard reduction potentials and is given as:

Ni2+ + 2e- -> Ni (s) E° = -0.25 V

The half-reaction for the aluminum electrode can be considered as:

3e- + Al (s) -> Al3+ E° = ?

Since we assume that Al is oxidized, the sign of E° for the aluminum half-reaction will be positive.

Given that the potential is T = 25°C, we can use the Nernst equation:

E = E° - (0.0592 V/n) * log(Q),

where E is the cell potential, E° is the standard potential, n is the number of electrons transferred, and Q is the reaction quotient.

For the cell reaction:

Ni2+ + 2e- -> Ni (s) (reduction)
3e- + Al (s) -> Al3+ (oxidation)

The cell potential can be written as:

Ecell = E (reduction) - E (oxidation).

In this case, E (reduction) = E°(Ni) = -0.25 V (given)
and E (oxidation) = - E°(Al).

Substituting these values into the Nernst equation and solving for Ecell:

Ecell = (-0.25 V) - (- E°(Al))
= -0.25 V + E°(Al).

Now we need to find the value of E°(Al).

To do this, we can use the fact that the cell potential at standard conditions (1 M for all species involved) is zero. This means:

Ecell = E°(reduction) - E°(oxidation) = 0.

Since E°(reduction) = - E°(oxidation), we have:

E°(Ni) = - E°(Al),
-0.25 V = - E°(Al),
E°(Al) = 0.25 V.

Substituting this value in the expression for Ecell:

Ecell = -0.25 V + 0.25 V
= 0 V.

Therefore, the potential of the cell is 0 V when the aluminum electrode is placed in a solution with [Al3+] = 7.3 x 10^(-3) M.

(b) To calculate [Al3+] in the unknown solution when the measured cell potential is 1.54 V, we can use the Nernst equation again.

Since the standard potential for the nickel half-reaction remains the same, the standard potential for the aluminum half-reaction also stays the same: E°(Al) = 0.25 V.

Using the Nernst equation:

Ecell = E°cell - (0.0592 V/n) * log(Q),

where Ecell is the measured cell potential and E°cell is the standard cell potential.

Substituting the known values:

1.54 V = E°cell - (0.0592 V/3) * log(Q).

Rearranging the equation:

log(Q) = (E°cell - 1.54 V) / (0.0592 V/3).

Since the reaction quotient Q is given by:

Q = [Al3+]/[Ni2+],

we can rewrite the equation as:

log([Al3+]/[Ni2+]) = (E°cell - 1.54 V) / (0.0592 V/3).

Taking the antilog of both sides:

[Al3+]/[Ni2+] = 10^[(E°cell - 1.54 V) / (0.0592 V/3)].

Substituting the given values:

[Al3+]/(1.0 M) = 10^[(0.25 V - 1.54 V) / (0.0592 V/3)].

Simplifying the expression:

[Al3+]/(1.0 M) = 10^[-29.067].

Multiplying both sides by 1.0 M:

[Al3+] = 1.0 M * 10^[-29.067].

Calculating the value:

[Al3+] = 1.0 M * 10^[-29.067] ≈ 3.1 x 10^(-30) M.

Therefore, the concentration of Al3+ in the unknown solution is approximately 3.1 x 10^(-30) M.

To find the potential of an electrochemical cell, we need to use the Nernst equation, which relates the cell potential to the concentrations of the species involved. The Nernst equation is as follows:

Ecell = E°cell - (RT/nF) ln(Q)

Where:
Ecell is the cell potential,
E°cell is the standard cell potential (at standard conditions, usually 25°C and 1 atm),
R is the gas constant (8.314 J/(mol·K)),
T is the temperature in Kelvin,
n is the number of electrons transferred in the balanced cell reaction,
F is Faraday's constant (96485 C/mol),
and Q is the reaction quotient.

Let's solve part (a) first:

(a) What is the potential of this cell at 25°C if the aluminum electrode is placed in a solution in which [Al3+] = 7.3 × 10-3 M?

First, we need to identify the balanced cell reaction. From the information given, we know that nickel (Ni) is oxidized, and aluminum (Al) is reduced. Therefore, the balanced cell reaction is:

Al(s) + Ni2+(aq) -> Ni(s) + Al3+(aq)

In this reaction, 3 electrons are transferred.

Next, we need to calculate the value of Q for the reaction at the given concentrations. Q is calculated by multiplying the concentrations of the products divided by the concentrations of the reactants, each raised to the power of their stoichiometric coefficients.

Q = ([Ni] / [Al3+])^1

Plugging in the given concentrations:

Q = (1.0 M / 7.3 × 10-3 M)^1

Now, we can use the Nernst equation to calculate the cell potential:

Ecell = E°cell - (RT/nF) ln(Q)

The standard cell potential, E°cell, can be found in reference tables. For this reaction, let's assume E°cell = 0.00 V.

Now, we can substitute the values into the equation and solve for Ecell:

Ecell = 0.00 V - (8.314 J/(mol·K) × 298 K / (3 × 96485 C/mol)) ln(1.0 M / 7.3 × 10-3 M)

Calculating the right side of the equation:

Ecell = 0.00 V - (0.0257 V) ln(137)

Ecell ≈ -0.074 V

Therefore, the potential of the cell at 25°C, when the aluminum electrode is placed in a solution with [Al3+] = 7.3 × 10-3 M, is approximately -0.074 V.

Now let's move on to part (b):

(b) When the aluminum electrode is placed in a certain solution in which [Al3+] is unknown, the measured cell potential at 25°C is 1.54 V. Calculate [Al3+] in the unknown solution.

Here, we already know the measured cell potential (Ecell = 1.54 V) and the reaction stoichiometry (n = 3). We also know the standard potential (E°cell = 0.00 V).

We can rearrange the Nernst equation to solve for Q:

Q = 10^((Ecell - E°cell) × (nF/RT))

Substitute the values given:

Q = 10^((1.54 V - 0.00 V) × (3 × 96485 C/mol) / (8.314 J/(mol·K) × 298 K))

Calculating the right side of the equation:

Q = 10^(136.9)

Q ≈ 3.69 × 10^136

Now, we can write the Q expression using the unknown concentration [Al3+]:

Q = ([Ni] / [Al3+])^1

We can substitute Q into this expression and solve for [Al3+]:

3.69 × 10^136 = ([Ni] / [Al3+])^1

As the concentration of nickel [Ni2+] was given as 1.0 M, we can solve for [Al3+]:

[Al3+] = [Ni2+] / (3.69 × 10^136) = 1.0 M / (3.69 × 10^136)

This concentration of Al3+ is extremely small, practically zero.

a) Use the Nernst equation to calculate the Ni/Ni^+2 half cell potential. Do the same for the Al^+3/Al half cell potential.

Add the two to obtain Ecell. I get -1.45v if I didn't make an error. You need to confirm that. Also you need to write the balanced equation for the cell AS WRITTEN, keeping in mind that it is negative which means it will not proceed spontaneously as written and as drawn.
For part b, the directions tell you to make Al the oxidized electrode which means the cell is rearranged and will proceed spontaneously; you will need to adjust the signs of the numbers (as well as the equation) to fit.
Ecell = 1.54v
n = 6 electrons.
You can fill in the values for components in Q.
Ecell = Eocell -(0.0592/n)log Q