Question

the electricity required for the preparation of 0.01 GM equivalent of AG + ions would be​

Answer 1

Answer: 8.94 Coloumb

Explanation:

$Q=I\times t$

where Q= quantity of electricity in coloumbs

I = current in amperes

t= time in seconds

$Ag^{+}+e^{-1}\rightarrow Ag$

1 mole of electrons carry charge=$\frac{1.6\times 10^{-19}}{1}\times 6.023\times 10^{23}=96500C$

96500 Coloumb of electricity deposits 1 mole of Ag

Thus 108 gram of Ag is deposited by 96500 C

0.01 gram of Ag is deposited by=$\frac{96500}{108}\times 0.01=8.94C$