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Question

# 19 g fused SnCl2 was electrolysed using inert electrodes. 0.119g Sn was deposited at cathode. If nothing was given out during electrolysis, calculate the ratio of mass of SnCl2 and SnCl4 in fused state after electrolysis is:

A
71
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B
101
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C
201
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D
None of these
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Solution

## The correct option is A 71As given,The Redox changes during electrolysis of SnCl2 are:Anode: 2Cl−→Cl2+2eCathode:Sn2++2e→SnAlso Cl2 formed at anode reacts with SnCl2 to give SnCl4SnCl2+2e→SnCl4Now Eq. of SnCl2 lost during elecrolysis = Eq. of Cl2 formed during electrolysis=Eq. of Sn formed during electrolysis=0.1190.119/2=2×10−3∴ Eq. of SnCl4 formed = 2×10−3Total loss in eq. of SnCl2 during complete course=Eq. of SnCl2 lost during electorlysis + Eq. of SnCl2 lost during reaction with Cl2 = 2×10−3+2×10−3=4×10−3Initial eq. of SnCl2 =\dfrac { 19 }{ 190/2 } =2\times { 10 }^{ -1 }∴ Eq.of SnCl2 left in solution =2×10−1−4×10−3 =0.196.Also Eq. of SnCl4 formed= 2×10−3=0.002∴ =MassofSnCl2leftMassofSnCl4formed=0.196×(190/2)0.002×(261/2)=71.34

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