Question:medium

The amount of silver (in mg) deposited when 9.65 coulombs of electricity is passed through an aqueous solution of silver nitrate is (Ag=108 u) (1F=96500 C mol\(^{-1}\))

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A useful proportion to remember for electrolysis is:
\( \frac{\text{Mass deposited}}{\text{Molar mass}} = \frac{\text{Charge passed}}{n \times F} \)
where 'n' is the number of moles of electrons per mole of substance from the half-reaction.
  • 16.2
  • 21.2
  • 10.8
  • 6.4
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Question:
9.65C passed through Ag⁺ solution. Find mass of Ag deposited.

Step 2: Key Formula (Alternate):
Ag⁺+e⁻→Ag. 1F=96500C deposits 108g. Use unitary method.

Step 3: Detailed Explanation:
96500C → 108g. 9.65C → (108×9.65/96500) = 0.0108g = 10.8mg.

Step 4: Final Answer:
10.8 mg of silver deposited.
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