In the refining of silver by electrolytic method what will be the weight of `100g Ag` anode if 5 ampere curretn is passed for 2 hours? Purity of silver is `95%` by weight?

Answer

`57.5894g`

Explanation

`w = (108 xx 5 xx 2 xx 3600)/(96500) = 40.29`
but since anode is `95%`pure with Ag so along with silver some impurities will alo come out so actual
wt. coming out of anode is `(40.29)/(0.95) = 42.41g`
weight of anode now ` = 100 - 42.41 = 57.58g`