Four faradays of electricity were passed through AgNO 3 l- CdSO 4 l - AlCl 3 l - - PbCl 4 l kept in four vessels using inert electrodes- The ratio of moles of Ag - Cd - Al - - Pb deposited will beA- 12- 4- 6- 3B-12- 6- 4- 3

Ans. B) 12 : 6 : 4 : 3 is correct Ag(aq) + e —– gt; Ag(s) 1 mol = 1F 1 mol 3F 3 mo ...

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Standard XII

Chemistry

Froth Floatation

Four faradays...

Question

Four faradays of electricity were passed through AgNO₃ (l), CdSO₄(l), AlCl₃(l), & PbCl₄(l) kept in four vessels using inert electrodes. The ratio of moles of Ag, Cd, Al,& Pb deposited will be

A. 12 : 4 : 6 : 3

B.12 : 6 : 4 : 3

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Solution

Ans. B) 12 : 6 : 4 : 3 is correct

Ag(aq) + e^{- -----> Ag(s)
1 mol = 1F 1 mol
3F 3 mol}
Cd²⁺(aq)+ 2e^{- -----> Cd(s)
2 mol = 2F 1 mol
3F 3/2 mol}
Al³⁺(aq)+ 3e^{- -----> Al(s)
3 mol = 3F 1 mol}
Pb⁴⁺(aq)+ 4e^{- -----> Pb(s)
4 mol = 4F 1 mol
3F 3/4 mol

therefore ratio = 3 : 3/2 : 1 : 3/4 = 12 : 6: 4 : 3}

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Four faradays of electricity were passed through AgNO3 (l), CdSO4(l), AlCl3(l), & PbCl4(l) kept in four vessels using inert electrodes. The ratio of moles of Ag, Cd, Al,& Pb deposited will be A. 12 : 4 : 6 : 3 B.12 : 6 : 4 : 3

Ans. B) 12 : 6 : 4 : 3 is correct Ag(aq) + e- -----> Ag(s) 1 mol = 1F 1 mol 3F 3 mol Cd2+(aq) + 2e- -----> Cd(s) 2 mol = 2F 1 mol 3F 3/2 mol Al3+(aq) + 3e- -----> Al(s) 3 mol = 3F 1 mol Pb4+(aq) + 4e- -----> Pb(s) 4 mol = 4F 1 mol 3F 3/4 mol therefore ratio = 3 : 3/2 : 1 : 3/4 = 12 : 6: 4 : 3

Four faradays of electricity were passed through AgNO₃ (l), CdSO₄(l), AlCl₃(l), & PbCl₄(l) kept in four vessels using inert electrodes. The ratio of moles of Ag, Cd, Al,& Pb deposited will be

A. 12 : 4 : 6 : 3

B.12 : 6 : 4 : 3