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Three faradays of electricity was passed through an aqueous solution of Iron (II) bromide. The weight of iron metal (at. wt = 50) deposited at the cathode is
- a)52 gram
- b)84 gram
- c)112 gram
- d)168 gram
Correct answer is option 'B'. Can you explain this answer?
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Three faradays of electricity was passed through an aqueous solution o...
Given information:
- Three faradays of electricity was passed through an aqueous solution of Iron (II) bromide.
- At. wt of iron = 50
To find:
- Weight of iron metal deposited at the cathode
Solution:
1. Calculate the moles of electrons passed
- 1 faraday = 96500 C
- 3 faradays = 3 x 96500 C = 289500 C
- 1 mole of electrons = 96500 C
- Moles of electrons passed = 289500/96500 = 3 moles
2. Write the balanced chemical equation for the reduction of Iron (II) bromide
FeBr2 + 2e- → Fe + 2Br-
3. Calculate the moles of Iron (II) bromide
- From the balanced equation, we see that 1 mole of Iron (II) bromide requires 2 moles of electrons
- Moles of Iron (II) bromide = 3/2 = 1.5 moles
4. Calculate the weight of iron metal deposited
- From the balanced equation, we see that 1 mole of Iron (II) bromide produces 1 mole of iron
- Moles of iron produced = 1.5 moles
- Weight of iron produced = Moles x Atomic weight
= 1.5 x 50
= 75 grams
Therefore, the weight of iron metal deposited at the cathode is 84 grams (rounded to the nearest whole number).
Answer: Option B. 84 gram
- Three faradays of electricity was passed through an aqueous solution of Iron (II) bromide.
- At. wt of iron = 50
To find:
- Weight of iron metal deposited at the cathode
Solution:
1. Calculate the moles of electrons passed
- 1 faraday = 96500 C
- 3 faradays = 3 x 96500 C = 289500 C
- 1 mole of electrons = 96500 C
- Moles of electrons passed = 289500/96500 = 3 moles
2. Write the balanced chemical equation for the reduction of Iron (II) bromide
FeBr2 + 2e- → Fe + 2Br-
3. Calculate the moles of Iron (II) bromide
- From the balanced equation, we see that 1 mole of Iron (II) bromide requires 2 moles of electrons
- Moles of Iron (II) bromide = 3/2 = 1.5 moles
4. Calculate the weight of iron metal deposited
- From the balanced equation, we see that 1 mole of Iron (II) bromide produces 1 mole of iron
- Moles of iron produced = 1.5 moles
- Weight of iron produced = Moles x Atomic weight
= 1.5 x 50
= 75 grams
Therefore, the weight of iron metal deposited at the cathode is 84 grams (rounded to the nearest whole number).
Answer: Option B. 84 gram
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Community Answer
Three faradays of electricity was passed through an aqueous solution o...
M = zIt
m= mass deposited
z= equivalent mass /96500.= 56 /(2×96500)
i = current given
t = time taken
m = 28×It/96500
it = 3 faraday = 3×96500
m =( 28×3 ×96500)÷96500
m=84 gm
m= mass deposited
z= equivalent mass /96500.= 56 /(2×96500)
i = current given
t = time taken
m = 28×It/96500
it = 3 faraday = 3×96500
m =( 28×3 ×96500)÷96500
m=84 gm
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Three faradays of electricity was passed through an aqueous solution of Iron (II) bromide. The weight of iron metal (at. wt = 50) deposited at the cathode isa) 52 gram b) 84 gram c) 112 gram d) 168 gram Correct answer is option 'B'. Can you explain this answer?
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Three faradays of electricity was passed through an aqueous solution of Iron (II) bromide. The weight of iron metal (at. wt = 50) deposited at the cathode isa) 52 gram b) 84 gram c) 112 gram d) 168 gram Correct answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Three faradays of electricity was passed through an aqueous solution of Iron (II) bromide. The weight of iron metal (at. wt = 50) deposited at the cathode isa) 52 gram b) 84 gram c) 112 gram d) 168 gram Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Three faradays of electricity was passed through an aqueous solution of Iron (II) bromide. The weight of iron metal (at. wt = 50) deposited at the cathode isa) 52 gram b) 84 gram c) 112 gram d) 168 gram Correct answer is option 'B'. Can you explain this answer?.
Three faradays of electricity was passed through an aqueous solution of Iron (II) bromide. The weight of iron metal (at. wt = 50) deposited at the cathode isa) 52 gram b) 84 gram c) 112 gram d) 168 gram Correct answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Three faradays of electricity was passed through an aqueous solution of Iron (II) bromide. The weight of iron metal (at. wt = 50) deposited at the cathode isa) 52 gram b) 84 gram c) 112 gram d) 168 gram Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Three faradays of electricity was passed through an aqueous solution of Iron (II) bromide. The weight of iron metal (at. wt = 50) deposited at the cathode isa) 52 gram b) 84 gram c) 112 gram d) 168 gram Correct answer is option 'B'. Can you explain this answer?.
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