The electrochemical cell described by the cell notation has an Eo of -0.37 V. Calculate the maximum electrical work (kJ) the cell has done if 331.73 g of Cu(s) (Molar Mass - 63.55 g/mol) reacts. Round your answer to 3 significant figures.
Cu(s) l Cu+(aq) ll Sn4+(aq), Sn2+(aq) l Pt(s)
St. Red. Pot. (V)
Cu+/Cu +0.52
Sn4+/Sn2+ +0.15
Faraday's Constant
F = 96485 C
n=2
convert 331g to mole= 5.21
G=-nFE=2(5.21)(96485)(-.37)=372kJ, why do i have to divide it in half to get 186kJ. Why do i have to times it by 5.21 moles as well
To calculate the maximum electrical work done by the electrochemical cell, you need to understand the concept of standard cell potential (E°) and Faraday's law of electrolysis.
The standard cell potential (E°) is the potential difference between the two electrodes of an electrochemical cell under standard conditions, which means at 298 K (25°C), 1 atm pressure, and 1 M concentration for all species involved in the electrochemical reaction.
In your given cell notation, Cu(s) is the anode (oxidation half-reaction), and Sn4+(aq), Sn2+(aq) is the cathode (reduction half-reaction). The standard reduction potentials (E°) of Cu+/Cu and Sn4+/Sn2+ are given as +0.52 V and +0.15 V, respectively.
Now, to calculate the maximum electrical work done by the cell, you need to use the formula:
ΔG = -nFE
where ΔG is the change in Gibbs free energy, n is the number of moles of electrons transferred, F is Faraday's constant (96485 C/mol), and E is the standard cell potential.
First, you need to determine the number of moles of electrons transferred (n). Considering your given balanced cell reaction:
Cu(s) → Cu+(aq) + e-
(Sn4+(aq) + 2e- → Sn2+(aq))
You can see that for every 2 mol of Cu(s) reacting, 2 mol of electrons are transferred. Therefore, n = 2.
Next, you need to convert the mass of Cu(s) given to moles. Since the molar mass of Cu is 63.55 g/mol, you divide the given mass (331.73 g) by the molar mass:
n = (331.73 g) / (63.55 g/mol) ≈ 5.21 mol
Now, you can calculate the maximum electrical work:
ΔG = -nFE
ΔG = -2(5.21)(96485)(-0.37)
ΔG ≈ 372 kJ
To answer your question, you don't need to divide it in half or multiply it by the number of moles used. The reason you may see the result expressed as 186 kJ is that one mole of Cu(s) would produce half of the calculated ΔG value, as indicated by stoichiometry. Therefore, if the question asks for the maximum electrical work done when 331.73 g of Cu(s) reacts, the answer should be 372 kJ.