Question

•A bar of silver is placed in a solution of sodium nitrate. A graphite rod is used as the other electrode. A constant current of 500 milliamps is passed for 6.00 hours through a liter of solution. Water is reduced at the graphite cathode.

2H2O +2e-→ H2 +2OH-

When the run is over, what is the concentration of silver ion?

Answer 1

Solution :-

Current = 500 milliamps * 1 amp / 1000 milliamp = 0.5 A

Time = 6.0 hr * (60 min/1 hr)*(60 sec / 1 min) = 21600 s

Charge = current x time

Charge = 0.5 A * 21600 s

             = 10800 J

Calculating the moles of electron using the charge

10800 J*1 mol e- / 96500 C = 0.112 mol e-

Using the moles of electron we can calculate the moles of Ag^+

Ag --- > Ag^+ + 1e-

0.112 mol e- * 1 mol Ag^+ / 1 mol e- = 0.112 mol Ag^+

Molarity = moles / volume in liter

                 = 0.112 mol Ag^+ / 1 L

                 = 0.112 M

Therefore the concentration of the silver ion is 0.112 M