Consider the voltaic cell

Zn(s) + Cu{2+}(aq)--> Zn{2+}(aq)+Cu(s)
{}=charge
Under standard conditions, what is the maximum electrical work, in Joules that can accomplish if 60 g of copper is plated out?

G = -nFE = J/mol

Zn(s)-->Zn^2+(aq)+2e^- 0.765 V
Cu^2+(aq)+2e^- -->Cu(s) 0.337 V

n = 2, F = 96,485, E =(0.765+0.337) = 1.102

G = -212652.94 J/mol

60g Cu * 1 mol Cu/63.546g Cu = 0.944 mol

Max. work = -212652.94 J/mol * 0.944 mol
= -200786

Well, I can certainly help you with that! Let's get electrified!

To determine the maximum electrical work, we first need to calculate the number of moles of copper that is being plated out. So, let's gather the necessary information:

The molar mass of copper (Cu) is approximately 63.55 g/mol.

Using the molar mass, we can calculate the number of moles of copper plated out:

Number of moles of copper = mass / molar mass
Number of moles of copper = 60 g / 63.55 g/mol

Calculating this gives us approximately 0.944 moles of copper.

Now, since the reaction is 1:1 between zinc and copper, we know that—for every mole of copper plated out—one mole of zinc is consumed. Therefore, we also have 0.944 moles of zinc consumed.

Now we need to find the maximum electrical work. The maximum electrical work is given by the formula:

W = -nFE

Where "W" is the maximum electrical work, "n" is the number of moles of electrons transferred (which is equal to the number of moles of zinc consumed), "F" is Faraday's constant (approximately 96,485 C/mol), and "E" is the cell potential.

Under standard conditions, the cell potential for this voltaic cell is approximately 1.10 V (given).

Substituting the values into the formula, we get:

W = - (0.944 mol) x (96,485 C/mol) x (1.10 V)

Doing the math, we find that the maximum electrical work is approximately -99,704 J.

Now, keep in mind that the negative sign indicates that work is being done on the system rather than being done by the system. So, if we want a positive value for work done by the system, we can take the absolute value.

In that case, the maximum electrical work is approximately 99,704 J (and that's a shocking amount of work!).

I hope this electrifying answer sparks some joy!

To calculate the maximum electrical work (ΔG) in Joules that can be accomplished during the plating of 60 g of copper, we need to use the equation:

ΔG = -nFΔE°

Where:
n = number of moles of electrons transferred during the reaction
F = Faraday's constant
ΔE° = standard cell potential

First, let's calculate the moles of copper (Cu) using its molar mass.

Molar mass of Cu = 63.55 g/mol

Number of moles of Cu = Mass / Molar mass
Number of moles of Cu = 60 g / 63.55 g/mol ≈ 0.944 moles

Since the stoichiometry of the reaction shows that 1 mole of copper corresponds to 2 moles of electrons transferred, we have:

n = 2 * moles of Cu
n = 2 * 0.944 moles = 1.888 moles

Next, we need to determine the standard cell potential, ΔE°. This information can be found in tables or online resources. For the given reaction:

Zn(s) + Cu{2+}(aq) → Zn{2+}(aq) + Cu(s)

The standard cell potential, ΔE°, is 1.10 V.

Now, we can substitute the values into the equation to calculate the maximum electrical work, ΔG.

ΔG = -nFΔE°

F = 96,485 C/mol (Faraday's constant)

ΔG = -1.888 * 96,485 C/mol * 1.10 V
ΔG ≈ -199,512C V = -199,512 J

Therefore, the maximum electrical work that can be accomplished if 60 g of copper is plated out is approximately -199,512 Joules.

To determine the maximum electrical work that can be accomplished in this voltaic cell reaction, we need to use the equation:

ΔG = -nFE

Where:
- ΔG is the change in Gibbs free energy
- n is the number of moles of electrons transferred in the balanced redox reaction
- F is the Faraday constant (approximately 96,485 Coulombs/mole)
- E is the cell potential in volts

First, let's determine the number of moles of electrons transferred in the balanced redox reaction:

Zn(s) + Cu{2+}(aq) --> Zn{2+}(aq) + Cu(s)

From the balanced equation, we can see that 2 moles of electrons are transferred for every 1 mole of copper formed. Therefore, for 60 g of copper, we can calculate the moles of copper (Cu) using its molar mass:

molar mass of Cu = 63.55 g/mol

moles of Cu = mass of Cu / molar mass of Cu
= 60 g / 63.55 g/mol
= 0.943 mol

Since 2 moles of electrons are transferred per mole of copper, the number of moles of electrons (n) is 2 times the moles of Cu:

n = 2 * 0.943 mol
= 1.886 mol

Next, we need to determine the cell potential (E). In this case, under standard conditions, the cell potential can be looked up from tables. The standard reduction potentials for the Zn{2+}/Zn and Cu{2+}/Cu half-reactions are:

Zn{2+}(aq) + 2e- --> Zn(s) E° = -0.76 V
Cu{2+}(aq) + 2e- --> Cu(s) E° = +0.34 V

To calculate the cell potential (E), we simply subtract the reduction potential of the anode (Zn{2+}/Zn) from the reduction potential of the cathode (Cu{2+}/Cu):

E = E°cathode - E°anode
= (+0.34 V) - (-0.76 V)
= +1.10 V

Now, we have all the values we need to calculate the maximum electrical work (ΔG). The equation becomes:

ΔG = -nFE

Substituting the values:

ΔG = -(1.886 mol) * (96,485 C/mol) * (+1.10 V)
= -196,259 J

Since the question asks for the absolute value of the work, the maximum electrical work that can be accomplished is 196,259 Joules.