Answer : The correct option is, (B) Oxygen diffuses through the anode.
Explanation :
Hydrogen-oxygen fuel cell : As we know that the a fuel cell is a type of device that converts the chemical energy into electrical energy. In the hydrogen-oxygen fuel cell, the proton exchange membrane cell uses hydrogen and oxygen gas as a fuel.
In this cell reaction, hydrogen and oxygen gas are the reactants and water, electricity and heat are the products.
The half-cell electrode reaction are:
At anode :
At cathode :
That means the hydrogen undergoes reduction at anode and oxygen undergoes oxidation at cathode.
As per question, we can say that oxygen diffuses towards anode and hydrogen diffuses towards cathode.
Hence, the correct option is, (B) Oxygen diffuses through the anode.
Answer:
b) The volume of stock solution used is less than 3 liters.
Explanation:
Given:
Concentration of stock solution of NaCl, M1 = 10 M
Concentration of diluted solution of NaCl, M2 = 1 M
Volume of diluted NaCl solution, V2 = 3 L
Let the volume of stock NaCl solution = V1
Formula:
Use the dilution relation:
Based on the above equation:
Therefore, the volume of stock solution required is 0.3 L which is less than 3 L
Answer:
a) molar composition of this gas on both a wet and a dry basis are
5.76 moles and 5.20 moles respectively.
Ratio of moles of water to the moles of dry gas =0.108 moles
b) Total air required = 68.51 kmoles/h
So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.
Explanation:
Let assume we have 100 g of mixture of gas:
Given that :
Mass of methane =75 g
Mass of ethane = 10 g
Mass of ethylene = 5 g
∴ Mass of the balanced water: 100 g - (75 g + 10 g + 5 g)
Their molar composition can be calculated as follows:
Molar mass of methane
Molar mass of ethane
Molar mass of ethylene
Molar mass of water
number of moles =
Their molar composition can be calculated as follows:
4.69 moles
0.33 moles
0.18 moles
0.56 moles
Total moles of gases for wet basis = (4.69 + 0.33 + 0.18 + 0.56) moles
= 5.76 moles
Total moles of gas for dry basis = (5.76 - 0.56)moles
= 5.20 moles
Ratio of moles of water to the moles of dry gas =
=
= 0.108 moles
b) If 100 kg/h of this fuel is burned with 30% excess air(combustion); then we have the following equations:
4.69 2× 4.69
moles moles
0.33 3.5 × 0.33
moles moles
0.18 3× 0.18
moles moles
Mass flow rate = 100 kg/h
Their Molar Flow rate is as follows;
Total moles of required = (2 × 4.69) + (3.5 × 0.33) + (3 × 0.18) k moles
= 11.075 k moles.
In 1 mole air = 0.21 moles
Thus, moles of air required =
= 52.7 k mole
30% excess air = 0.3 × 52.7 k moles
= 15.81 k moles
Total air required = (52.7 + 15.81 ) k moles/h
= 68.51 k moles/h
So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.