Calculate the quantity of electricity in coulombs required to deposit all the silver metal from 2.50 mL of 1 M AgNO3 solution.
(Given molar mass of Ag = 108 g mol−1 and 1 F = 96500 C)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 241.25 C
Amount of \( \ce{Ag^{+}} \) in the solution:
\( n(\ce{Ag+}) = 1\ \text{mol L}^{-1} \times 2.50 \times 10^{-3}\ \text{L} = 2.50 \times 10^{-3}\ \text{mol} \)
Each \( \ce{Ag^{+}} \) ion takes one electron at the cathode, so one mole of Ag needs one faraday:
\( \ce{Ag+(aq) + e- -> Ag(s)} \)
\( Q = 2.50 \times 10^{-3} \times 96500 = 241.25\ \text{C} \)