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 cell, you need to use the formula:
G = -nFE
Where:
G is the maximum electrical work done by the cell
n is the number of moles of electrons transferred in the balanced equation
F is Faraday's constant (96485 C)
E is the standard potential difference in volts (V)
In this case, the balanced equation for the electrochemical cell is:
Cu(s) → Cu+(aq) + e-
Sn4+(aq) + 2e- → Sn2+(aq)
From the balanced equation, we can see that two moles of electrons are transferred, so n = 2.
The standard potential difference (Eo) described in the cell notation is -0.37 V.
Now, substitute the values into the formula:
G = -nFE
G = -2 * 96485 * (-0.37)
Calculate the result:
G = 36162.62 J
To convert from joules to kilojoules, divide the result by 1000:
G = 36.16262 kJ
Now, round your answer to three significant figures:
G ≈ 36.2 kJ
Therefore, the maximum electrical work done by the cell is approximately 36.2 kJ.
It's important to note that the reason you multiply the value by 2 is because the standard potential obtained from the cell notation is the net potential difference for the overall reaction. Since the reaction involves two moles of electrons, you need to consider that in the calculation. If you divide the result by 2, it means you are considering the potential per mole of electrons transferred. In this case, you are calculating the overall maximum electrical work of the entire reaction, not just the potential per mole.