The electrolysis of 0.100 l of 0.500 M AgNO3 (aq) using platiunum electrodes is

Question

The electrolysis of 0.100 l of 0.500 M AgNO3 (aq) using platiunum electrodes is carried out with a current of 1.05 A. What is the molarity of the AgNO3 (aq) 25.0 minutes after electrolysis is begun? Assume NO3- does not undergo electrolysis.

Explanation / Answer

Ag+ + 1e- => Ag 1.05 A = 1.05 Coulomb/sec 1 mol e- = 96485 Coulomb 1.05 C/s *25 min *(60s/1min) *1mole-/96485 C *(1mol Ag+/1mol e-) =0.0163 mol Ag+ consumed 0.10L *0.5mol/L = 0.05 mol Ag+ originally Afterwards 0.05 - 0.0163 = 0.0337 mol Ag+ Still in 0.1 Liters So M = 0.0337/0.1 = 0.337 M

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