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Q.

In the refining of silver by electrolytic method, what will be the final mass (in gm) of 72.8 g silver anode (60% pure, by weight), if 9.65A current is passed for 1h? (Ag = 108)

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answer is 8.

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Detailed Solution

neq = Ag oxidized =QF⇒W108×1=9.65×1×360096500
∴W = 38.88 gm 
Mass of anode dissolved =38.88×10060=64.8gm
Final mass of anode = 72.8 – 64.8 = 8 gm

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In the refining of silver by electrolytic method, what will be the final mass (in gm) of 72.8 g silver anode (60% pure, by weight), if 9.65A current is passed for 1h? (Ag = 108)