Question:
The amount of silver in mg deposited when \(9.65\) coulombs of electricity is passed through an aqueous solution of silver nitrate is \(\left(Ag=108u,\ 1F=96500\text{ C mol}^{-1}\right)\)
The amount of silver in mg deposited when \(9.65\) coulombs of electricity is passed through an aqueous solution of silver nitrate is \(\left(Ag=108u,\ 1F=96500\text{ C mol}^{-1}\right)\)
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Use Faraday's law: \(m=\frac{EQ}{F}\). For silver, equivalent weight is \(108\).
Updated On: May 7, 2026
- \(16.2\)
- \(21.2\)
- \(10.8\)
- \(6.4\)
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The Correct Option is C
Solution and Explanation
For silver deposition:
\[
Ag^+ + e^- \rightarrow Ag.
\]
One mole of electrons deposits one mole of silver.
One Faraday charge:
\[
96500\text{ C}
\]
deposits:
\[
108\text{ g}
\]
of silver.
Given charge:
\[
Q=9.65\text{ C}.
\]
Mass deposited is:
\[
m=\frac{108\times 9.65}{96500}.
\]
Since:
\[
\frac{9.65}{96500}=0.0001,
\]
we get:
\[
m=108\times0.0001.
\]
\[
m=0.0108\text{ g}.
\]
Convert gram to milligram:
\[
1\text{ g}=1000\text{ mg}.
\]
\[
0.0108\text{ g}=10.8\text{ mg}.
\]
Hence, the amount of silver deposited is:
\[
10.8\text{ mg}.
\]
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