1
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:
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Solution
The correct option is A 71
As given,The Redox changes during electrolysis of SnCl2 are:
Anode: 2Cl−→Cl2+2e
Cathode:Sn2++2e→Sn
Also Cl2 formed at anode reacts with SnCl2 to give SnCl4
SnCl2+2e→SnCl4
Now 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 = 2×10−3
Total 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=4×10−3
Initial 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
As given,
The Redox changes during electrolysis of SnCl2 are:
Anode: 2Cl−→Cl2+2e
Cathode:Sn2++2e→Sn
Also Cl2 formed at anode reacts with SnCl2 to give SnCl4
SnCl2+2e→SnCl4
Now 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 = 2×10−3
Total 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=4×10−3
Initial 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
Anode: 2Cl−→Cl2+2e
Cathode:Sn2++2e→Sn
Also Cl2 formed at anode reacts with SnCl2 to give SnCl4
SnCl2+2e→SnCl4
Now 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 = 2×10−3
Total 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=4×10−3
Initial 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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