A fuel cell designed to react grain alcohol with oxygen has the following net reaction:
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
The maximum work that 1 mol of alcohol can yield by this process is 1320 kJ. What is the theoretical
maximum voltage this cell can achieve?
To find the theoretical maximum voltage of the fuel cell, we need to look at the energy change for the reaction. The maximum work that can be obtained from a reaction is equal to the product of the maximum electrical potential difference (voltage) and the number of moles of electrons involved in the reaction.
In this case, the net reaction is given as:
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
From the balanced reaction, we can see that 6 moles of electrons are involved, as the reaction involves the reduction of 6 moles of O2 to 6 moles of H2O.
Given that the maximum work is 1320 kJ and 1 mole of electrons is equivalent to 96,485 Coulombs (C), we can calculate the maximum electrical potential difference using the formula:
Work = Voltage × Charge
Rearranging the formula, we have:
Voltage = Work / Charge
So, the voltage (V) is equal to:
1320 kJ / (1 mole × 96,485 C/mole)
Converting kJ to Joules:
1320 kJ × 1000 J/kJ = 1,320,000 J
Therefore, the voltage (V) is:
1,320,000 J / (1 mole × 96,485 C/mole)
Simplifying:
V = 13.65 volts
Hence, the theoretical maximum voltage this cell can achieve is 13.65 volts.