The electrochemical cell described by the balanced chemical equation has a standard emf (electromotive force) of -4.05 V. Calculate the maximum electrical work (kJ) the cell has done if 6.8500 mol of F2(g) (Molar Mass - 38.00 g/mol) forms. Round your answer to 3 significant figures.
2F-(aq) + Mn2+(aq) �¨ F2(g) + Mn(s)
How do we relate work with the equation for electrochemistry?
Is volts x coulombs = Joules?
In electrochemistry, work is related to the equation through the concept of electrical potential energy. The standard emf (electromotive force) of an electrochemical cell is a measure of the difference in electrical potential energy between the reactants and the products.
To calculate the maximum electrical work done by the cell, you need to consider the moles of electrons transferred in the reaction and the standard emf. The equation shows that 2 moles of F- ions are involved in the reaction. Since 1 mole of F2(g) forms, this implies the transfer of 2 moles of electrons.
The electrical work done (W) can be calculated using the equation:
W = -nFΔE
Where:
- W is the electrical work done (in joules)
- n is the number of moles of electrons transferred in the reaction
- F is the Faraday constant (96,485 C/mol)
- ΔE is the change in electrical potential energy, which is given by the standard emf of the cell (E)
To obtain the maximum electrical work done in kJ, you can follow these steps:
1. Convert the moles of F2(g) (given in the question) to moles of electrons by multiplying by the stoichiometric coefficient (2 moles of electrons/mol of F2(g)).
2. Calculate the electrical work done (W) in joules using the equation W = -nFΔE.
3. Convert the electrical work from joules to kilojoules by dividing by 1000.
Let's solve the problem using the given data:
Given:
- Standard emf (E) = -4.05 V
- Moles of F2(g) = 6.8500 mol
- Molar mass of F2(g) = 38.00 g/mol
Step 1: Convert moles of F2(g) to moles of electrons
Moles of electrons = 6.8500 mol x (2 mol of electrons / 1 mol of F2(g)) = 13.7000 mol
Step 2: Calculate the electrical work done (W) in joules
W = -nFΔE
W = -(13.7000 mol) x (96,485 C/mol) x (-4.05 V)
W = 5,050,592.65 J
Step 3: Convert the electrical work from joules to kilojoules
W (in kJ) = 5,050,592.65 J / 1000 = 5,050.59 kJ (rounded to 3 significant figures)
Therefore, the maximum electrical work done by the cell is 5,050.59 kJ.