Question

A DMFC generates 1A at 0.3V at STP. Methanol and air are supplied to the fuel cell at 0.003 mmol/s and 0.03 mmol/s respectively. Please calculate: i) the output flow rate of methanol; ii) consumption rate of oxygen; iii) output flow rate of air; and iv) water generation rate.

Answer 1

Power generated by DMFC = V*I = 0.3V*1A = 0.3 J/s

This means that the fuel cell is able to produce 0.3 J of work every second

$\therefore \Delta G^{o} = - 0.3 J (every second) \Rightarrow \Delta G^{o} = - nFE^{o} = -0.3$

E^o = 0.3V & F = 96500 C per mole of electron & n = moles of electron

$nF*0.3 = 0.3 \Rightarrow nF = 1$

$\therefore n = \frac{1}{F} \Rightarrow n = \frac{1}{96500}$

Now you must know the reactions (anode & cathode or overall) of DMFC to solve the question, The overall reaction is as follows;

$CH_{3}OH + \frac{3}{2}O_{2} \rightarrow 2H_{2}O + CO_{2}$

After this, you can either use Farday's Law to solve the question or use simple logic (unitary method you study in school mathematics). I prefer unitary method

As per the reaction, 6 mole electrons are required to oxidise 1 mole methanol; so

n mole of electron will oxidise n/6 mole of methanol

Methanol oxidised(consumed) every second = n/6 mole

$\frac{n}{6} = \frac{1}{96500*6} = 1.73 * 10^{-6} mol/s$ $\frac{n}{6} = \frac{1}{96500*6} = 1.73 * 10^{-6} mol/s = 0.00173 mmol/s$

i) Output flow rate of methanol = Input - Consumed = 0.003 - 0.00173 = 0.00127 mmol/s

For oxygen, similar analysis is applicable:

ii)

As per the reaction, 6 mole electrons are required to reduce 3/2 mole of oxygen; so

1 mole of electron will reduce 3/(2*6) mol of oxygen, i.e.1/4 mole of oxygen

n mole of electron will reduce n/4 mole of oxygen

Oxygen reduced(consumed) every second = n/4 mole

$\frac{n}{4} = \frac{1}{96500*4} = 2.59 * 10^{-6} mol/s = 0.00259 mmol/s$

(this is the consumption rate of oxygen)

iii) For third part composition of air must have been given,

Let consumption rate of air = p

Assuming, 21% oxygen in air (this should have been given in the question),

Consumption rate of oxygen = 21% of consumption rate of air

$0.00259 mmol/s = \frac{21}{100} * p$

$\therefore p = 0.0123 mmol/s$

Output of air = Input - Consumption = 0.03 - p

Therefore, output flow rate of air = 0.0177 mmol/s

iv) Water generation rate follows the same method;

As per the reaction, 6 mole electrons are required to produce 2 mole of water; so

1 mole of electron will produce 2/6 mol of water, i.e.1/3 mole of water

n mole of electron will produce n/3 mole of water

Water generated every seond = n/3 mole

$\frac{n}{3} = \frac{1}{96500*3} = 3.45 * 10^{-6} mol/s = 0.00345 mmol/s$

This is the answer for the fourth part.

Verify the answers & let me know if you face any more problems.