Question

The amount of silver deposited by passing one Faraday of current through silver nitrate solution is​

Answer 1

Answer:

According to Faraday's Law of Electrolysis:-

W=

nF

I×t×M

W= weight deposited

I= current F= Faraday's constant 96500

t= time M= Molar mass

n= number of electron

Given:-

It=9.65 coulomb

Molar mass of Silver, M=108g/mol n=1

W=

1×96500

9.65×108

W=0.0108g

⇒W=10.8mg

Answer 2

Answer:

answer in attachment