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 need to use the formula:
G = -nFE
Where:
G is the Gibbs Free Energy (maximum electrical work done),
n is the number of moles of electrons transferred,
F is Faraday's constant (96485 C), and
E is the standard cell potential (Eo) in volts.
In this case, you correctly calculated:
n = 2 (since the balanced equation shows the transfer of two moles of electrons),
Eo = -0.37 V,
and F = 96485 C.
Now, let's substitute these values into the formula:
G = - (2) (96485) (-0.37)
G = 71713 J
To convert the energy from joules to kilojoules, you divide by 1000:
G = 71.713 kJ
Now, let's address the confusion about dividing the total value in half.
The reason you need to divide the value in half is because the cell potential given (-0.37 V) is the reduction potential (Eo) for the overall cell reaction. It represents the potential difference between the reactants and products in the cell.
However, when you use this potential in the Gibbs Free Energy equation, you need to consider that you're calculating the work done by the entire reaction, not just the reduction half-reaction.
In this particular cell, the balanced equation involves the transfer of two moles of electrons, so the total work done by the reaction is twice the work done by one mole of electrons (0.5 nFΔE). Therefore, you divide the total value you obtained by 2 to adjust for this.
Hence, after dividing the value by 2, the maximum electrical work done by the cell is 71.713 / 2 = 35.856 kJ.
Remember to round your answer to three significant figures, which gives you the final result of 35.9 kJ.