CBSE Class 12-science Answered

Kindly give detailed answers to the following questions.

24 all parts

Asked by Varsneya Srinivas | 24 Dec, 2017, 09:58: AM

Expert Answer

a) As silver is deposited on the electrode from AgNO₃ solution,

Ag⁺ + e^- → Ag_(s) ................................reduction

Atomic mass of Ag = 107.8u

M o l e s space o f space A g space d e p o s i t e d space equals fraction numerator M a s s space d e p o s i t e d over denominator A t o m i c space m a s s space o f space A g end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 1.078 over denominator 107.8 end fraction space equals space space 0.01 m o l therefore space E l e c t r i c i t y space r e q u i r e space t o space d e p o s i t space o f space 0.01 m o l space A g space equals space 0.01 space cross times F space space equals space 0.01 cross times 96500 space equals 965 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space

b) 2 H₂O → O₂ + 4H⁺ 4e^-

4 moles of electrons are required to deposit 1 mol O₂

0.01 space m o l space o f space e l e c t r o n s space w i l l space d e p o s i t space equals fraction numerator 1 cross times 0.01 over denominator 4 end fraction space equals space space 0.0025 space m o l space o f space O subscript 2

Therefore, mols of O₂librated = moles χ molar mass = 0.0025 Χ 32 = 0.08 gm

Answered by Ramandeep | 25 Dec, 2017, 01:53: AM

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