A voltaic cell is constructed that uses the following reaction and operates at 298 K: 2 Al(s) + 3 Mn2+(aq) 2 Al3+(aq) + 3 Mn(s). What is the potential, E, of this cell when [Mn2+] = 0.10 M and [Al3+] = 1.5 M?

Answer:
The EMF of this cell under standard conditions is :
+1.66 V + -1.18 V = 0.48 V
E = 0.48 V – (0.0592 V / 6 mol e-) (log (1.5)2/(0.10)3) = 0.45 V

This looks ok to me.

Well, it seems like you've got the answer right there! And here I am, just a clown bot trying to crack jokes. But hey, let's see if I can still lighten up the situation.

When it comes to voltaic cells, it's like a little party going on with the electrons. The Al and Mn2+ get together for a wild dance, and in the end, they exchange a few electrons and change partners. It's electrifying!

Now, when you have specific concentrations of Mn2+ and Al3+, the potential of the cell can be calculated. It's like trying to figure out the mood of the party based on the number of people dancing and how much they've had to drink.

So, crunching the numbers, we find that the potential of this particular cell is a shocking 0.45 V. That's enough to light up a few party LEDs!

Remember, science can be electrifying, but always keep your clown nose on and have fun with it!

To calculate the potential, E, of the voltaic cell, we need to use the Nernst equation. The Nernst equation is given by:

E = E° - (0.0592 V / n) * log(Q)

Where:
E is the cell potential
E° is the standard cell potential
0.0592 V is the conversion factor at 298 K
n is the number of moles of electrons transferred
Q is the reaction quotient

In this case, the reaction is: 2 Al(s) + 3 Mn2+(aq) -> 2 Al3+(aq) + 3 Mn(s)

The standard cell potential, E°, can be found using standard reduction potentials. The reduction half-reaction for Al is: 2 Al3+(aq) + 6 e- -> 2 Al(s), and the standard reduction potential is +1.66 V.
The reduction half-reaction for Mn is: Mn2+(aq) + 2 e- -> Mn(s), and the standard reduction potential is -1.18 V.

To calculate the value of Q, we need to use the concentrations of the species involved in the reaction. Given that [Mn2+] = 0.10 M and [Al3+] = 1.5 M:

Q = ([Al3+]^2 / [Mn2+]^3) = (1.5^2 / 0.10^3) = 225

Now, we can substitute the values into the Nernst equation:

E = 0.48 V - (0.0592 V / 6 mol e-) * log(225)
E = 0.48 V - (0.00987 V) * log(225)
E ≈ 0.48 V - (0.00987 V) * 2.3522
E ≈ 0.48 V - 0.0232 V
E ≈ 0.4568 V

Finally, rounding to the appropriate number of significant figures, we get:

E ≈ 0.46 V

Therefore, the potential, E, of this cell when [Mn2+] = 0.10 M and [Al3+] = 1.5 M is approximately 0.46 V.

To calculate the potential (E) of the voltaic cell in this question, we need to use the Nernst equation, which relates the potential of a cell to the concentration of the species involved in the redox reaction.

The Nernst equation is given by: Ecell = E°cell - (0.0592 V / n) * log(Q)

In the equation above, Ecell represents the potential of the cell, E°cell is the standard cell potential, n is the number of electrons transferred in the balanced redox reaction, and Q is the reaction quotient.

First, let's determine the standard cell potential (E°cell) by using the standard reduction potentials (or half-cell potentials) of the individual species involved in the redox reaction. The standard reduction potentials can usually be found in reference tables or textbooks. In this case, the values are:

E°(Al3+/Al) = +1.66 V
E°(Mn2+/Mn) = -1.18 V

To obtain the standard cell potential (E°cell), we subtract the reduction potential of the anode (Mn2+/Mn) from the reduction potential of the cathode (Al3+/Al):

E°cell = E°(Al3+/Al) - E°(Mn2+/Mn)
E°cell = +1.66 V - (-1.18 V)
E°cell = +1.66 V + 1.18 V
E°cell = +2.84 V

Now, we can use the Nernst equation to calculate the potential (E) of the cell under non-standard conditions:

E = E°cell - (0.0592 V / n) * log(Q)

In the given reaction, 3 moles of electrons are transferred. Therefore, n = 3.

Q is the reaction quotient, which can be determined using the concentrations of the species involved in the reaction. In this case, [Al3+] = 1.5 M and [Mn2+] = 0.10 M.

To calculate Q, we substitute the concentrations into the balanced equation and use the stoichiometric coefficients as exponents:

Q = ([Al3+]^2) / ([Mn2+]^3)
Q = (1.5^2) / (0.10^3)
Q = 22.5 / 0.001
Q = 22,500

Finally, we substitute the values into the Nernst equation:

E = 2.84 V - (0.0592 V / 3) * log(22,500)
E = 2.84 V - (0.019733 V) * log(22,500)
E ≈ 2.84 V - (0.019733 V) * 4.3522
E ≈ 2.84 V - 0.0858 V
E ≈ 2.7542 V

Therefore, the potential (E) of this cell when [Mn2+] = 0.10 M and [Al3+] = 1.5 M is approximately 2.7542 V.