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The time required to coat a metal surface of $80cm^2$ with $5\times 10^{-3}$cm thick layer of silver (density $10.5gcm^{-3})$ with the passage of 3A current through a silver nitrate solution is

$\begin{array}{1 1}1150s\\1250s\\1350s\\1450s\end{array} $

1 Answer

Answer : 1250 s
Using the expression $m=(Q/F)(M/|v_c|)$
We get
$(80\times 5\times 10^{-3}cm^3)(10.5 gcm^{-3})=\large\frac{(3A)t}{(96500Cmol^{-1})}\big(\large\frac{108gmol^{-1}}{1}\big)$
This gives $t=\large\frac{(80\times 5\times 10^{-3})(10.5)(96500)}{3(108)}=$1251 s
answered Jul 23, 2014 by sreemathi.v

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