Two electrolytic cells A and B contain gold salt and copper sulphate solution respectively . Two cells are combined in series and current is passed for 5 h . If the oxidation state of gold in gold salt solution is +3 and the amount of gold accumulated in cell A is 9.85 g , then find the amount of copper discharged in cell B ? [ Au = 197 , Cu = 63.5]

Explanation

Equivalent mass of `Au=(197)/(3)=65.66.`
Hence, `65.66g Au-=96500C`
`therefore" 9.85 g Au "-=14476.47C.`
But, `Cu^(2+)+2erarrCu.` So, `2xx96500C-=63.5gCu`
`therefore" "14476.47C-=4.7629gCu.`