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1 Faraday of electricity is passed through the solution containing 1 mole each of AgNO3, CuSO4, AlCland SiCl4. Elements are discharged at the cathode.Number of moles of Ag, Cu,Al and Si formed will be in the ratio of  
  • a)
    1 : 1 : 1 : 1
  • b)
    1 : 2 : 3 : 4
  • c)
    6 : 4 : 3 : 1
  • d)
    12 : 6 : 4 : 3
Correct answer is option 'D'. Can you explain this answer?
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Explanation:

To solve this problem, we need to determine the moles of each element formed at the cathode when 1 Faraday of electricity is passed through the solution containing 1 mole each of AgNO3, CuSO4, AlCl3, and SiCl4.

Step 1: Determine the number of moles of each compound

Given that there is 1 mole of each compound in the solution, the number of moles of AgNO3, CuSO4, AlCl3, and SiCl4 is 1 mole each.

Step 2: Calculate the number of moles of each element

We need to determine the moles of Ag, Cu, Al, and Si formed at the cathode.

Since AgNO3 contains 1 mole of Ag, CuSO4 contains 1 mole of Cu, AlCl3 contains 1 mole of Al, and SiCl4 contains 1 mole of Si, the number of moles of Ag, Cu, Al, and Si formed will be the same as the number of moles of the corresponding compounds.

Therefore, the ratio of moles of Ag, Cu, Al, and Si formed will be 1:1:1:1.

Step 3: Simplify the ratio

The given options are in different ratios, so we need to simplify the ratio of 1:1:1:1.

The simplified ratio of 1:1:1:1 is 12:6:4:3.

Therefore, the correct answer is option D) 12:6:4:3.
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1 Faraday of electricity is passedthrough the solution containing 1 mole each of AgNO3, CuSO4, AlCl3and SiCl4. Elements are discharged at the cathode.Number of moles of Ag, Cu,Al and Si formed will be in the ratio of a)1 : 1 : 1 : 1b)1 : 2: 3: 4c)6 : 4 : 3 :1d)12 : 6 : 4 :3Correct answer is option 'D'. Can you explain this answer?
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