In the electrolysis of SnCl2 aqueous solution, 4.48 L of Cl2 was liberated at STP. Find the amount of tin deposed at the cathode if current efficiency was equal to 100%. Molecular mass: Sn=118.7g, SnCl2=189.7g
Answers
Answer: 23.74 g of Tin is deposited on cathode .
Explanation: By Electrochemical series we can easily determine the reactions occurring on both Cathode and Anode
ELECTROLYSIS OF SnCl2
ANODE: 2 Cl– ———>Cl2(gas) + 2 electron
CATHODE : Sn 2+ + 2electon -——> Sn
———————————————————————
NET: 2Sn2+ + 2Cl– ———> Cl2 (g) + 2Sn(s)
Now by Faradays 2nd law we all know that if same amount of current is passed in different electrolytes for same time then mass deposited or liberated at any electrode is directly proportional to Eqivalent weight of that substance
So Equivalent wieght of Cl2 gas is=
(2 × 35.5 )/2 = 71/2
Moles of Cl2 gas liberated at STP =
4.48L / 22.4L = 2/10 moles
Mass of Cl2 liberated = (71 × 2)/10 gram
Now Equivalent wieght of Sn = 118.7 /2 Mass of Sn deposited = W2
Mass of Cl2 liberated Mass of Sn deposited
——————————— = ———————————
Equivalent weight of Equivalent weight of
Cl2 Sn
71 × 2/10 W2
—————— = ——————
71/2 118.7/2
W2 = ( 4/10) × (118.7/2) = 23.74 gm
Hence 23.74 gm of Sn is deposited on the cathode