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A 100.0 mL dillute solution of Ag+ is electrolysed for 15.0 minutes with a current of 1.25 mA and the silver is removed completely. What was the initial [Ag+]?
  • a)
    2.32 × 10–1
  • b)
    2.32 × 10–4
  • c)
    2.32 × 10–3
  • d)
    1.16 × 10–5
Correct answer is option 'D'. Can you explain this answer?
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A 100.0 mL dillute solution of Ag+ is electrolysed for 15.0 minutes wi...
No. of moles of
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A 100.0 mL dillute solution of Ag+ is electrolysed for 15.0 minutes wi...
The first step is to calculate the total charge passed through the solution during the electrolysis.

Current = 1.25 mA = 1.25 x 10^-3 A
Time = 15.0 minutes = 15.0 x 60 seconds = 900 seconds

Charge = Current x Time
Charge = 1.25 x 10^-3 A x 900 s
Charge = 1.125 C

Since each Ag+ ion requires one electron to form Ag, the total number of moles of electrons passed through the solution is equal to the total number of moles of Ag ions present initially in the solution.

1 mole of electrons = 1 Faraday = 96485 C

Number of moles of Ag ions = Charge / (1 x 96485)
Number of moles of Ag ions = 1.125 C / 96485 C/mol
Number of moles of Ag ions = 1.166 x 10^-5 mol

Initial volume of solution = 100.0 mL = 0.100 L

Initial [Ag] = Number of moles of Ag ions / Initial volume
Initial [Ag] = 1.166 x 10^-5 mol / 0.100 L
Initial [Ag] = 0.0001166 mol/L

Initial [Ag] = 2.32 x 10^-4 mol/L

Therefore, the initial [Ag] was 2.32 x 10^-4 mol/L.
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A 100.0 mL dillute solution of Ag+ is electrolysed for 15.0 minutes with a current of 1.25 mA and the silver is removed completely. What was the initial [Ag+]?a)2.32 × 10–1b)2.32 × 10–4c)2.32 × 10–3d)1.16 × 10–5Correct answer is option 'D'. Can you explain this answer?
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