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An electrolytic cell contains a solution of Ag2SO4 and have platinum electrodes. A current is passed until 1.6 gm of O2 has been liberated at anode. The amount of silver deposited at cathode would be
- a)107.88 gm
- b)1.6 gm
- c)0.8 gm
- d)21.60 gm
Correct answer is option 'D'. Can you explain this answer?
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An electrolytic cell contains a solution of Ag2SO4 and have platinum e...
Wt. of O2 = 1.6 g; Wt. of Ag deposited = ? ;
Eq. Wt. of oxygen = 8, Eq. Wt. of Ag = 108
We know :
Eq. Wt. of oxygen = 8, Eq. Wt. of Ag = 108
We know :
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An electrolytic cell contains a solution of Ag2SO4 and have platinum e...
Solution:
Given, the amount of oxygen liberated at the anode = 1.6 gms
We know that the reaction at the anode is:
2H2O → O2 + 4H+ + 4e-
From the above equation, we can see that 4 electrons are required to produce 1 molecule of O2.
Now, let's calculate the amount of charge required to produce 1.6 gms of O2.
Molar mass of O2 = 32 g/mol
Number of moles of O2 = 1.6/32 = 0.05
Number of electrons required = 4 × 0.05 = 0.2 moles
Charge required = 0.2 × 96500 (Faraday's constant) = 19300 C
Now, the amount of silver deposited at the cathode can be calculated using Faraday's laws of electrolysis.
According to Faraday's first law, the amount of substance deposited at the electrode is directly proportional to the amount of charge passed through the cell.
m = ZIt
where m is the mass of substance deposited, Z is the electrochemical equivalent, I is the current, and t is the time.
For silver deposition, Z = 0.001118 gm/C
As the same amount of charge is passed through the cell for both the reactions, the amount of silver deposited would be:
m = 0.001118 × 19300 = 21.60 gm
Hence, the correct option is (d) 21.60 gm.
Given, the amount of oxygen liberated at the anode = 1.6 gms
We know that the reaction at the anode is:
2H2O → O2 + 4H+ + 4e-
From the above equation, we can see that 4 electrons are required to produce 1 molecule of O2.
Now, let's calculate the amount of charge required to produce 1.6 gms of O2.
Molar mass of O2 = 32 g/mol
Number of moles of O2 = 1.6/32 = 0.05
Number of electrons required = 4 × 0.05 = 0.2 moles
Charge required = 0.2 × 96500 (Faraday's constant) = 19300 C
Now, the amount of silver deposited at the cathode can be calculated using Faraday's laws of electrolysis.
According to Faraday's first law, the amount of substance deposited at the electrode is directly proportional to the amount of charge passed through the cell.
m = ZIt
where m is the mass of substance deposited, Z is the electrochemical equivalent, I is the current, and t is the time.
For silver deposition, Z = 0.001118 gm/C
As the same amount of charge is passed through the cell for both the reactions, the amount of silver deposited would be:
m = 0.001118 × 19300 = 21.60 gm
Hence, the correct option is (d) 21.60 gm.
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An electrolytic cell contains a solution of Ag2SO4 and have platinum electrodes. A current is passed until 1.6 gm of O2 has been liberated at anode. The amount of silver deposited at cathode would bea)107.88 gmb)1.6 gmc)0.8 gmd)21.60 gmCorrect answer is option 'D'. Can you explain this answer?
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