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
To calculate the maximum electrical work done by the electrochemical cell, you first need to use the equation:
G = -nFE
Where:
G is the maximum electrical work done by the cell (in joules),
n is the number of moles of electrons transferred in the balanced equation (2 moles in this case),
F is Faraday's constant (96485 C/mol),
E is the standard cell potential (Eo) given (-0.37 V).
First, convert the given mass of Cu(s) to moles:
331.73 g Cu x (1 mol Cu/63.55 g Cu) = 5.21 mol Cu
Next, insert the values into the equation:
G = - (2 mol)(96485 C/mol)(-0.37 V)
Calculate:
G = 2(5.21 mol)(96485 C/mol)(-0.37 V)
G ≈ - 372 kJ
The negative sign indicates that the electrical work is done by the system. However, the question asks for the maximum electrical work, which is defined as the amount of work done on the system. To obtain the maximum work done, you need to take the positive value:
Maximum work done = -G = 372 kJ
So, the maximum electrical work done by the cell is 372 kJ.
There is no need to divide the value in half in this case. The "n" value already accounts for the number of moles in the balanced equation, so you do not need to divide it again. The 186 kJ value you mentioned may be a result of a different calculation or context.