Sol :-
Net reaction is given :
C2H5OH (l) + 3O2 (g) ---------------> 2CO2 (g) + 3H2O (l)
Oxidation half cell reaction is :
C2H5OH (l) + 3H2O (l) -------------------> 2CO2 (g) + 12H+ (aq) + 12e-
Reduction half cell reaction is :
3O2 (g) + 12H+ (aq) + 12e- -------------------> 6H2O (l)
Number of electron exchange (n) in this redox reaction is = 12
The relationship between ΔG and Ecell is :
ΔG = -nFEcell
F= Farady constant = 96500 C
So,
- 1320 x 103 J = - 12 x 96500 C x Ecell
Ecell = - 1320 x 103 J / - 12 x 96500 C
= 1.14 V
| Hence, Potential of the cell = Ecell = 1.14 V |
QUESTION 3 A fuel cell designed to react grain alcohol with oxygen has the following net...
3. A fuel cell designed to react grain alcohol with oxygen has the following net reaction: C2H5OH(1) + 302(g) → 2CO2(g) + 3H2O(1) The maximum work one mole of alcohol can yield by this process is 1320 kJ. What is the theoretical maximum voltage this cell can achieve?
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...
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