In a Ag voltameter,2.68 , g of silver is depo | vedclass

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(A) According to Faraday's first law of electrolysis,the mass $m$ deposited is given by $m = ZIt$,where $Z$ is the electrochemical equivalent of silve

Vedclass · Accepted Answer

$192 \, kJ$

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(A) According to Faraday's first law of electrolysis,the mass $m$ deposited is given by $m = ZIt$,where $Z$ is the electrochemical equivalent of silver $(Ag)$.For silver,the atomic mass is $108 \, g/mol$ and its valency is $1$. Thus,$Z = \frac{108}{96500} \, g/C$.Given: $m = 2.68 \, g$,$t = 10 \, \min = 600 \, s$.Calculating the current $I$:$I = \frac{m}{Zt} = \frac{2.68}{\left(\frac{108}{96500}ight) 	imes 600} = \frac{2.68 	imes 96500}{108 	imes 600} \approx 4 \, A$.Now,the heat $H$ developed in the resistor $R = 20 \, \Omega$ is given by Joule's law of heating:$H = I^2 Rt = (4)^2 	imes 20 	imes 600 = 16 	imes 20 	imes 600 = 192000 \, J = 192 \, kJ$.

In a $Ag$ voltameter,$2.68 \, g$ of silver is deposited in $10 \, \min$. The heat developed in a $20 \, \Omega$ resistor during the same period will be:

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