In a hydrogen oxygen fuel cell- electricity is produced- In this process H 2 g is oxidised at anode and O 2 g reduced at cathode- Given - Cathode-O 2 g -2 H 2 O l -4 e - 4 OH -aq Anode - H 2 g -2 OH - aq - 2 H 2 O l -2 e - 4-48 L of H 2 at 1 atm and 273 K is oxidisedin 9650 sec-The mass of water produced is in grams-A- 7-2 gB- 1-8 g- 3-6 gD- 0-9 g

The correct option is C 3.6 gUsing Faraday’s Law: m_H_2O= MM/n× 96500× i× t Now, from the previous question: i= 4 A Putting the values: m_H_2O= 18× 4× 9650/2× ...

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Standard XII

Chemistry

Faraday's First Law

In a hydrogen...

Question

In a hydrogen oxygen fuel cell, electricity is produced. In this process $H_{2} (g)$ is oxidised at anode and $O_{2} (g)$ reduced at cathode. Given : Cathode: $O_{2} (g) + 2 H_{2} O (l) + 4 e^{-} \to 4 O H^{-} (a q)$ Anode : $H_{2} (g) + 2 O H^{-} (a q) \to 2 H_{2} O (l) + 2 e^{-}$ $4.48 L$ of $H_{2}$ at $1 atm$ and $273 K$ is oxidised in $9650 sec$ .
The mass of water produced is (in grams):

7.2 g

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1.8 g

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3.6 g

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0.9 g

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Solution

The correct option is C $3.6 g$
Using Faraday’s Law:
$m_{H_{2} O} = \frac{M M}{n \times 96500} \times i \times t$
Now, from the previous question:
$i = 4 A$
Putting the values:
$m_{H_{2} O} = \frac{18 \times 4 \times 9650}{2 \times 96500}$
$m = 3.6 g$

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Similar questions

Q. In a hydrogen oxygen fuel cell, electricity is produced. In this process

H_{2} (g)

is oxidised at anode and

O_{2} (g)

reduced at cathode.

Given: Cathode

O_{2} (g) + 2 H_{2} O (l) + 4 e^{-} \to 4 O H^{-} (a q)

Anode

H_{2} (g) + 2 O H^{-} (a q) \to 2 H_{2} O (l) + 2 e^{-}

4.48 litre

H_{2}

at 1 atm and 273K oxidised in 9650 sec.

The mass of water produced is:

Q. In a hydrogen oxygen fuel cell, electricity is produced. In this process

H_{2} (g)

is oxidised at anode and

O_{2} (g)

reduced at cathode.

Given: Cathode

O_{2} (g) + 2 H_{2} O (l) + 4 e^{-} \to 4 O H^{-} (a q)

Anode

H_{2} (g) + 2 O H^{-} (a q) \to 2 H_{2} O (l) + 2 e^{-}

4.48 litre

H_{2}

at 1 atm and 273K oxidised in 9650 sec.

The current produced is (in amp):

Q. In a hydrogen oxygen fuel cell, electricity is produced. In this process

H_{2} (g)

is oxidised at anode and

O_{2} (g)

reduced at cathode.
Given : Cathode:

O_{2} (g) + 2 H_{2} O (l) + 4 e^{-} \to 4 O H^{-} (a q)

Anode :

H_{2} (g) + 2 O H^{-} (a q) \to 2 H_{2} O (l) + 2 e^{-}

4.48 L

H_{2}

1 atm

and

273 K

is oxidised in

9650 sec

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The current produced is (in amp):

Q. In a hydrogen oxygen fuel cell, electricity is produced. In this process

H_{2} (g)

is oxidised at anode and

O_{2} (g)

reduced at cathode. Given : Cathode:

O_{2} (g) + 2 H_{2} O (l) + 4 e^{-} \to 4 O H^{-} (a q)

Anode :

H_{2} (g) + 2 O H^{-} (a q) \to 2 H_{2} O (l) + 2 e^{-}

4.48 L

H_{2}

1 atm

and

273 K

is oxidised in

9650 sec

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The mass of water produced is (in grams):

Q. In a hydrogen oxygen fuel cell, electricity is produced. In this process

H_{2} (g)

is oxidised at anode and

O_{2} (g)

reduced at cathode. Given : Cathode:

O_{2} (g) + 2 H_{2} O (l) + 4 e^{-} \to 4 O H^{-} (a q)

Anode :

H_{2} (g) + 2 O H^{-} (a q) \to 2 H_{2} O (l) + 2 e^{-}

4.48 L

H_{2}

1 atm

and

273 K

is oxidised in

9650 sec

.
If current produced in fuel cell, use for the deposition of

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The correct option is C 3.6 gUsing Faraday’s Law: mH2O=MMn×96500×i×t Now, from the previous question: i= 4 A Putting the values: mH2O= 18×4×96502×96500 m= 3.6 g

The correct option is C $3.6 g$
Using Faraday’s Law:
$m_{H_{2} O} = \frac{M M}{n \times 96500} \times i \times t$
Now, from the previous question:
$i = 4 A$
Putting the values:
$m_{H_{2} O} = \frac{18 \times 4 \times 9650}{2 \times 96500}$
$m = 3.6 g$