A voltaic cell is constructed that uses the following reaction and operates at 298 K: 2 Al(s) + 3 Mn^2+(aq) 2 Al^3+(aq) + 3 Mn(s). What is the potential, E, of this cell when [Mn^2+] = 0.10 M and [Al^3+] = 1.5 M?

See your other post. Same process but you must correct for concns that are not 1M. Use the Nernst equation for that. Post your work if you get stuck.

To determine the potential (E) of the voltaic cell using the given reaction, you can use the Nernst equation.

The Nernst equation relates the standard cell potential (E°) to the actual cell potential (E) under nonstandard conditions and is given by:

E = E° - (RT/nF) * ln(Q)

Where:
E = Cell potential under nonstandard conditions
E° = Standard cell potential
R = Gas constant (8.314 J/(mol·K))
T = Temperature in Kelvin (298 K)
n = Number of electrons transferred in the balanced equation
F = Faraday's constant (96485 C/mol)
Q = Reaction quotient (ratio of product concentrations to reactant concentrations)

First, we need to find the standard cell potential (E°) for the given reaction, which can be obtained from standard reduction potentials (E°red) of the involved species. The standard cell potential (E°) is calculated by taking the difference between the standard reduction potentials (E°red) of the reduction half-reaction and the oxidation half-reaction.

The standard reduction potentials (E°red) for the involved half-reactions are as follows:
Mn^2+(aq) + 2e- Mn(s) E°red = -1.18 V (reduction half-reaction)
Al^3+(aq) + 3e- Al(s) E°red = -1.66 V (oxidation half-reaction)

To get the standard cell potential (E°), we subtract the reduction potential of the oxidation half-reaction from the reduction potential of the reduction half-reaction:

E° = E°red(reduction) - E°red(oxidation)
E° = (-1.18 V) - (-1.66 V)
E° = 0.48 V

Next, we need to calculate the reaction quotient (Q) using the concentrations of the involved species. The reaction quotient (Q) is calculated by taking the ratio of the product concentrations raised to their stoichiometric coefficients divided by the reactant concentrations raised to their stoichiometric coefficients.

For the given reaction:
2 Al(s) + 3 Mn^2+(aq) 2 Al^3+(aq) + 3 Mn(s)

The reaction quotient (Q) is:
Q = ([Al^3+]^2 * [Mn(s)]^3) / ([Mn^2+]^3 * [Al(s)]^2)

Plugging in the given concentrations:
Q = ([1.5 M]^2 * [Mn(s)]^3) / ([0.10 M]^3 * [Al(s)]^2)

Now, we can substitute the values into the Nernst equation, along with the known values for the gas constant (R), Faraday's constant (F), temperature (T), and the number of electrons transferred (n = 3, as seen from the balanced equation) to calculate the cell potential (E):

E = E° - (RT/nF) * ln(Q)
E = 0.48 V - ((8.314 J/(mol·K)) * (298 K) / (3 * 96485 C/mol)) * ln(Q)

Finally, substitute the calculated value of Q:
E = 0.48 V - ((8.314 J/(mol·K)) * (298 K) / (3 * 96485 C/mol)) * ln({[1.5 M]^2 * [Mn(s)]^3} / {[0.10 M]^3 * [Al(s)]^2})

Evaluating this expression will give you the potential (E) of the cell under the given conditions.