Question

A fuel cell designed to react grain alcohol with oxygen has the following net reaction:
C2H5OH(l) + 3 O2(g) –––> 2 CO2(g) + 3 H2O(l)
The maximum work that 1 mole of alcohol can yield by this process is 1320 kJ. What is the
theoretical maximum voltage that this cell can achieve?
NOTE: The first step in solving this problem is to determine n, the number of moles of electrons
transferred when this reaction takes place as written.
A) 1.14 V B) 2.28 V C) 2.01 V D) 0.760 V E) 13.7 V

Answer 1

no. of electron transferred = change in oxidation no. of oxygen * 6 = (0- (-2)) * 6 = 12

now for a cell reaction  

$\Delta$ G = - n F E_cell

-1320 * 1000 = - 12 * 96500 * E_cell

so E_cell = 1320000 / (12*96500) = 1.139

so maximum voltage of cell = 1.14 Volt (Answer is option A)