Kindly give detailed answers to the following questions.
24 all parts

Asked by Varsneya Srinivas | 24th 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 O2 librated = moles χ  molar mass  =  0.0025 Χ 32 = 0.08 gm
 

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