Redox Reactions And Electrochemistry Ques 627
Question: 1.08 g of pure silver was converted into silver nitrate and its solution was taken in a beaker. It was electrolysed using platinum cathode and silver anode. 0.01 Faraday of electricity was passed using 0.15 volt above the decomposition potential of silver. The silver content of the beaker after the above shall be
Options:
A) 0 g
B) 0.108 g
C) 1.08 g
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ A{g^{+}}+\underset{1F}{\mathop{{e^{-}}}},\xrightarrow{{}}\underset{108g}{\mathop{Ag}}, $ 1 F = 1 mole of electrons = 96500 C 0.01F= 1.08 g Ag;Ag left = $ 1.08-1.08=0g $
