Home
Class 12
CHEMISTRY
Exactly 0.4 faraday electric charge is p...

Exactly 0.4 faraday electric charge is passed through three electrolytic cells in series, first containing `AgNO_(3)`, second `CuSO_(4)` and third `FeCl_(3)` solution. How many grams of each metal will be deposited assuming only cathodic reaction in each cell ?

Text Solution

Verified by Experts

The cathodic reaction in the cells are respectively.
`underset(("1 mole"),(108 g))(Ag^(+))+underset(("1 mole"),("1 F"))(e^(-)) rarr Ag`
`underset(("1 mole"),(63.5 g))(Cu^(2+))+underset(("2 mole"),("2 F"))(2e^(-)) rarr Cu`
and `underset(("1 mole"),(56 g))(Fe^(3+))+underset(("3 mole"),("3 F"))(3e^(-)) rarr Fe`
Hence, Ag deposited `=108xx0.4=43.3 g`
Cu deposited `=63.5/2xx0.4=12.7 g`
and Fe deposited `=56/3 xx0.4 =7.47 g`
Promotional Banner

Topper's Solved these Questions

  • ELECTROCHEMISTRY

    OP TANDON|Exercise ILLUSTRATIONS|28 Videos

  • ELECTROCHEMISTRY

    OP TANDON|Exercise PRACTICE PROBLEMS|94 Videos

  • COORDINATION COMPOUNDS

    OP TANDON|Exercise SBJECTIVE TYPE|96 Videos

  • HALOALKANES AND HALOARENES

    OP TANDON|Exercise Single Interger answer type questions|14 Videos

Similar Questions

Explore conceptually related problems

Exactly 0.4 faraday of charge is passed through three electrolytic cells connected in series, first containing AgNO_(3) , second CuSO_(4) while third containing FeCl_(3) solution. How many grams of each metal will be deposited assuming only cathodic reaction in each cell ?

Exactly 0.2 mole electrons are passed through two electrolytic cells in series containing CuSO_(4) and ZnSO_(4) respectively. How many grams of each metal will be deposited on the respective cathodes in the two cells ?

Three mole of electrons are passed through three solutions in succession containing AgNO_(3), CuSO_(4) and AuCl_(3) respectively. The ratio of amounts of cations reduced at cathode will be :

If same quantity of electricity is passed through three electrolytic cells containing FeSO_(4),Fe_(2)(SO_(4))_(3) , and Fe(NO_(3))_(3) , then

Three moles of electrons are passed through three solutions in succession containing AgNO_(3),CuSO_(4) and AuCl_(3) respectively. The molar ratio of amounts of cations reduced at cathode will be

Three moles of electrons are passed through three solutions in succession containing AgNO_(3),CuSO_(4) and AuCI_(3) respectively. The molar ratio of amounts of cations reduced at cathode will be:-