To determine the composition of the material deposited on the cathode, we need to consider the reduction potentials and the Faraday's laws of electrolysis.
Given:
- X2+ ions have a reduction potential of E = 0.35 V.
- Y2+ ions have a reduction potential of E° = 0.40 V.
- Atomic weight of X is 63.
- Atomic weight of Y is 200.
According to Faraday's laws of electrolysis, the amount of substance deposited on the cathode is directly proportional to the quantity of electricity passed through the electrolytic cell.
Since X2+ and Y2+ ions do not interact with each other, we can consider their reduction reactions separately.
For X2+ ions:
2 mol of electrons are required to reduce 1 mol of X2+ ions.
For Y2+ ions:
2 mol of electrons are required to reduce 1 mol of Y2+ ions.
Since the same amount of electricity is passed through the cell for both X2+ and Y2+ ions, the ratio of the amount of substance deposited on the cathode will be equal to the ratio of their respective molar masses.
Let's calculate the ratio:
Ratio of X2+ ions deposited : Ratio of Y2+ ions deposited
= Atomic weight of X : Atomic weight of Y
= 63 : 200
= 0.315 : 1
The composition of the material deposited on the cathode is approximately 0.315 (or 31.5%) X and 1 (or 100%) Y by weight.
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Answer:
To determine the composition of the material deposited on the cathode, we need to consider the reduction potentials and the Faraday's laws of electrolysis.
Given:
- X2+ ions have a reduction potential of E = 0.35 V.
- Y2+ ions have a reduction potential of E° = 0.40 V.
- Atomic weight of X is 63.
- Atomic weight of Y is 200.
According to Faraday's laws of electrolysis, the amount of substance deposited on the cathode is directly proportional to the quantity of electricity passed through the electrolytic cell.
Since X2+ and Y2+ ions do not interact with each other, we can consider their reduction reactions separately.
For X2+ ions:
2 mol of electrons are required to reduce 1 mol of X2+ ions.
For Y2+ ions:
2 mol of electrons are required to reduce 1 mol of Y2+ ions.
Since the same amount of electricity is passed through the cell for both X2+ and Y2+ ions, the ratio of the amount of substance deposited on the cathode will be equal to the ratio of their respective molar masses.
Let's calculate the ratio:
Ratio of X2+ ions deposited : Ratio of Y2+ ions deposited
= Atomic weight of X : Atomic weight of Y
= 63 : 200
= 0.315 : 1
The composition of the material deposited on the cathode is approximately 0.315 (or 31.5%) X and 1 (or 100%) Y by weight.
Therefore, the correct answer is:
C. 98% X and 2% Y by weight.