Question

In the electrolysis of water shown below, a current of 2 amps is applied to 180 ml of h2o(l) for 6 hours and 42 minutes. how many grams of h2(g) are formed? (faraday's constant = 96,500 c/mol)

Answer 1

In electrolysis of water, water is decomposed into hydrogen and oxygen by passing a current.

Reaction at cathode :

`2 H_(2)O (l)+ 2e^(-)\rightarrow 2H_(2)(g) + 2OH^(-)(aq)`

Reaction at anode :

`2 H_(2)O (l)\rightarrow O_(2)(g) + 4H^(+)(aq)+ 4e^(-)`

The net electrolysis reaction can be written as

`2H_(2)O (l)\overset{electrolysis}{\rightarrow} 2H_(2)(g)+ O_(2)(g)`

Step 1 : Find the coulombs of charge passed through the solution.

We will first find how many coulombs of charge passed through water for the given amount of time.

The formula that relates current in ampere and charge in coulombs is

`Current (A)= (Charge(Coulombs))/(time(seconds))`

We have Current = 2 A

Time = 6 hrs 42 mins

Let us convert this to seconds.

`6 hrs * (60mins)/(1hr)=360 mins`

T = 360 mins + 42 mins = 402 mins

`402 mins * (60seconds)/(1min)=24120seconds`

T = 24120 seconds

Let us plug in the values in the charge (Q) formula

`2 A = (Q)/(24120s)`

`Q = 2 * 24120 coulombs`

Q = 48240 C

Step 2 : Find moles of electrons transferred during the reaction.

According Faradays constant, 1 mol of electrons carry a charge of 96500 C.

We have 48240 C. Let us find how many moles of e- would carry this amount of charge.

`48240 C * (1mol)/(96500C)= 0.500 mol`

0.5 mol of electrons are transferred during the reaction.

Step 3 : Find moles of H2 using reaction at cathode.

From the reduction reaction occurring at cathode, we can see that 1 mol of H₂ gas is formed when 2 moles are electrons are accepted by water.

Therefore the mole ratio of e⁻ to H₂ is 2 : 1 . Let us use this as a conversion factor to find moles of H₂ from the calculated moles of electrons.

`0.5 mol (e^(-))* (1mol H_(2))/(2mol (e^(-))) =0.25 mol H_(2)`

Moles of H₂ formed = 0.25 mol

Grams of H₂ =
`0.25 mol (H_(2))* (2.02g(H_(2)))/(1mol(H_(2)))`

Grams of H₂ formed = 0.505 g

0.505 grams of H₂ are formed during the electrolysis reaction