Redox Reactions And Electrochemistry Ques 318
Question: The amount of silver deposited by passing 241.25 coulomb of current through silver nitrate solution is [MHCET 2003]
Options:
A) 2.7 g
B) 2.7 mg
C) 0.27 g
D) 0.54 g
Show Answer
Answer:
Correct Answer: C
Solution:
Given, Current = 241.25 columb 1 coulomb current will deposite $ =1.118\times {10^{-3}}gm\ Ag $ .
$ \therefore $ 241.25 current will deposite = $ 1.118\times {10^{-3}}\times 241.25 $ $ =0.27\ gm $ silver.
