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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?
Most Upvoted Answer
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
Free Test
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
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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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