Question

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\(^{-1}\))

• A. 16.2
• B. 21.2
• C. 10.8
• D. 6.4

Hint:

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.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
This is an electrolysis problem where we need to calculate the mass of silver metal deposited by a specific amount of electric charge.

Step 2: Key Formula or Approach:
We will use Faraday's Laws of Electrolysis.
1. Write the reduction half-reaction for silver ions to determine the number of electrons involved per ion.
2. Calculate the moles of electrons corresponding to the given charge using the Faraday constant.
3. Use the stoichiometry from the half-reaction to find the moles of silver deposited.
4. Convert the moles of silver to mass, and then convert the units to milligrams.

Step 3: Detailed Explanation:
The reduction half-reaction for silver ions at the cathode is:
\[ \text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s) \]
This shows that 1 mole of electrons (e\(^-\)) deposits 1 mole of silver (Ag).
First, calculate the moles of electrons for the given charge Q = 9.65 C.
\[ \text{Moles of electrons} = \frac{\text{Total Charge (Q)}}{\text{Faraday Constant (F)}} = \frac{9.65 \text{ C}}{96500 \text{ C/mol}} = 0.0001 \text{ mol} = 10^{-4} \text{ mol} \]
From the 1:1 stoichiometry, the moles of Ag deposited equals the moles of electrons.
\[ \text{Moles of Ag} = 10^{-4} \text{ mol} \]
Next, convert the moles of silver to mass in grams using the molar mass (108 g/mol).
\[ \text{Mass of Ag (g)} = \text{moles} \times \text{molar mass} = 10^{-4} \text{ mol} \times 108 \text{ g/mol} = 0.0108 \text{ g} \]
Finally, convert the mass from grams (g) to milligrams (mg) by multiplying by 1000.
\[ \text{Mass of Ag (mg)} = 0.0108 \text{ g} \times 1000 \text{ mg/g} = 10.8 \text{ mg} \]

Step 4: Final Answer:
The amount of silver deposited is 10.8 mg.