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3 A current was passed through an aqueous solution of an unknown salt of Pd for 1 hour. 2.977 g of Pdn+ was deposited at the cathode. Find nn. (Atomic weight of Pd=106.4)
Correct answer is '4'. Can you explain this answer?
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3 A current was passed through an aqueous solution of an unknown sal...
Pdn+ + ne− → Pd
For
n = 4
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3 A current was passed through an aqueous solution of an unknown sal...
Given data:
- The atomic weight of Pd = 106.4
- Mass of Pd deposited at the cathode = 2.977 g
Calculations:
- First, calculate the moles of Pd deposited at the cathode:
Moles of Pd = Mass of Pd / Molar mass of Pd
= 2.977 g / 106.4 g/mol
≈ 0.0279 mol
- Since the unknown salt of Pd is Pdn+, the moles of electrons transferred can be calculated using Faraday's law:
Moles of electrons = Moles of Pd
= 0.0279 mol
- The charge passed can be calculated using the equation:
Charge (Coulombs) = Moles of electrons * Faraday's constant
= 0.0279 * 96500 C/mol
≈ 2693.5 C
- Knowing that 1 hour = 3600 seconds, we can calculate the current:
Current (Amperes) = Charge / Time
= 2693.5 C / 3600 s
≈ 0.748 A
- The formula for current is given as:
Current (I) = nF
where n is the number of moles of electrons and F is Faraday's constant.
Substituting the values:
0.748 A = n * 96500 C/mol
n ≈ 0.0077 mol
- Since the formula for the salt of Pd is Pdn+, n = 4.
Therefore, the correct answer is n = 4.
- The atomic weight of Pd = 106.4
- Mass of Pd deposited at the cathode = 2.977 g
Calculations:
- First, calculate the moles of Pd deposited at the cathode:
Moles of Pd = Mass of Pd / Molar mass of Pd
= 2.977 g / 106.4 g/mol
≈ 0.0279 mol
- Since the unknown salt of Pd is Pdn+, the moles of electrons transferred can be calculated using Faraday's law:
Moles of electrons = Moles of Pd
= 0.0279 mol
- The charge passed can be calculated using the equation:
Charge (Coulombs) = Moles of electrons * Faraday's constant
= 0.0279 * 96500 C/mol
≈ 2693.5 C
- Knowing that 1 hour = 3600 seconds, we can calculate the current:
Current (Amperes) = Charge / Time
= 2693.5 C / 3600 s
≈ 0.748 A
- The formula for current is given as:
Current (I) = nF
where n is the number of moles of electrons and F is Faraday's constant.
Substituting the values:
0.748 A = n * 96500 C/mol
n ≈ 0.0077 mol
- Since the formula for the salt of Pd is Pdn+, n = 4.
Therefore, the correct answer is n = 4.
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3 A current was passed through an aqueous solution of an unknown salt of Pd for 1 hour. 2.977 g of Pdn+ was deposited at the cathode. Find nn. (Atomic weight of Pd=106.4)Correct answer is '4'. Can you explain this answer?
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3 A current was passed through an aqueous solution of an unknown salt of Pd for 1 hour. 2.977 g of Pdn+ was deposited at the cathode. Find nn. (Atomic weight of Pd=106.4)Correct answer is '4'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 3 A current was passed through an aqueous solution of an unknown salt of Pd for 1 hour. 2.977 g of Pdn+ was deposited at the cathode. Find nn. (Atomic weight of Pd=106.4)Correct answer is '4'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 3 A current was passed through an aqueous solution of an unknown salt of Pd for 1 hour. 2.977 g of Pdn+ was deposited at the cathode. Find nn. (Atomic weight of Pd=106.4)Correct answer is '4'. Can you explain this answer?.
3 A current was passed through an aqueous solution of an unknown salt of Pd for 1 hour. 2.977 g of Pdn+ was deposited at the cathode. Find nn. (Atomic weight of Pd=106.4)Correct answer is '4'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 3 A current was passed through an aqueous solution of an unknown salt of Pd for 1 hour. 2.977 g of Pdn+ was deposited at the cathode. Find nn. (Atomic weight of Pd=106.4)Correct answer is '4'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 3 A current was passed through an aqueous solution of an unknown salt of Pd for 1 hour. 2.977 g of Pdn+ was deposited at the cathode. Find nn. (Atomic weight of Pd=106.4)Correct answer is '4'. Can you explain this answer?.
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