On passing a current of 1.0 ampere for 16 min and 5 sec through one litre solution of CuCl 2, all copper of the solution was deposited at cathode. The strength of CuCl 2 solution was (Molar mass of Cu =63.5; Faraday constant =96,500 C mol -1 )

Question

On passing a current of 1.0 ampere for 16 min and 5 sec through one litre solution of CuCl 2, all copper of the solution was deposited at cathode. The strength of CuCl 2 solution was (Molar mass of Cu =63.5; Faraday constant =96,500 C mol -1 ) (A) 0.01 N (B) 0.01 M (C) 0.02 M (D) 0.2 N

Tardigrade Admin · Accepted Answer

By Faraday's Is t Law, (W/E)=(q/96500) (where q= it = charge of ion) =(W/E)=(i t/69500)=(1 × 965/96500)=(1/100) (where i=1 A, t=16 × 60+5=965 sec.) Since, we know that Normality =( text no- of equivalent / text Volume (in litre) ) =((1/100)/1)=0.01 N

Q. On passing a current of 1.0 ampere for 16min and 5sec through one litre solution of CuCl2​, all copper of the solution was deposited at cathode. The strength of CuCl2​ solution was (Molar mass of Cu=63.5; Faraday constant =96,500Cmol−1 )

0.01 N

0.01 M

0.02 M

0.2 N

By Faraday's Ist Law, EW​=96500q​ (where q= it = charge of ion) =EW​=69500it​=965001×965​=1001​ (where i=1A,t=16×60+5=965 sec.) Since, we know that Normality = Volume (in litre) no- of equivalent ​ =11001​​=0.01N

Q. On passing a current of $1.0$ ampere for $16 min$ and $5 sec$ through one litre solution of $C u C l_{2}$ , all copper of the solution was deposited at cathode. The strength of $C u C l_{2}$ solution was (Molar mass of $C u = 63.5$ ; Faraday constant $= 96, 500 C m o l^{- 1}$ )