What is the quantity of electricity (in coulombs)required to deposit all the silver from 250 mL of 1 M AgNO3 solution ? (Ag = 108)

Question

What is the quantity of electricity (in coulombs)required to deposit all the silver from 250 mL of 1 M AgNO3 solution ? (Ag = 108) (A) 96500 (B) 24125 (C) 48250 (D) 12062.5

Tardigrade Admin · Accepted Answer

Amount of Ag in solution = M × V (mL) = 1 × 250 = 250 m mol = 250 × 10-3 mol = 0.25 mol underset1 molAg+ + underset96500 Ce- arrow Ag Reduction of 1 mol of Ag requires 96500 C Reduction of 0.25 mol of Ag requires = 96500 × 0.25 = 24125 C

Q. What is the quantity of electricity (in coulombs)required to deposit all the silver from $250 m L$ of $1 M A g N O_{3}$ solution ? ( $A g = 108$ )

96500

24125

48250

12062.5

Amount of $A g$ in solution $= M \times V (m L)$
$= 1 \times 250 = 250 m m o l$
$= 250 \times 1 0^{- 3} m o l$
$= 0.25 m o l$
$1 m o l A g^{+} + 96500 C e^{-} \to A g$
Reduction of $1 m o l$ of $A g$ requires $96500 C$
Reduction of $0.25 m o l$ of $A g$ requires
$= 96500 \times 0.25 = 24125 C$

Q. What is the quantity of electricity (in coulombs)required to deposit all the silver from 250mL of 1MAgNO3​ solution ? (Ag=108)