Chemistry, asked by cyb3r, 8 months ago

A sample of lead weighing 1.05 g was dissolved in a small quantity of nitric acid to produce
aqueous solution of Pb2+ and Ag* (which is present as impurity). The volume of the solution
was increased to 300 ml by adding water, a pure silver electrode was immersed in the
solution and the potential difference between this electrode and a standard hydrogen
electrode was found to be 0.503 V at 25°C. What was the % of Ag in the lead metal? Given
(Ag / Ag)
= 0.799 V. Neglect amount of Ag* converted to Ag.
E'Ag* I Ag)​

Answers

Answered by zuhairpagarkar
0

Answer:

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