Thanks, what about Q13??
Ok remember when
concentrated NaCl is electrolyzed:
First the solution contains Na
+ Cl
- and H
+ and OH
- (from water). The positive inos go to the cathode, and negative ions to the anode.
At the CATHODE
the H+ ions accept electrons, since it is less reactive than Na
(always elements which is more reactive 'likes' to stay in the ionic form)2H
+ + 2e- ---> H2
At the ANODE
the Cl- ions give up the electrons more readily than the OH- ions do.
2Cl
- --> Cl
2 + 2e-
what left in the solution is Na+ and OH- which combine to form NaOH
in your ques ..is 'D' .. the indicator solution turns blue at -ve as NaOH is produced .. Chlorine doesnt show any change in presence of an indicator solution
jus understand the concept ..
see in this question u hav to act smart .. look at the options .. u now kno tht NaOH is produced so the indicator turns blue ..there are two options with 'bule/purple' A and B .. now see A .. at -ve 'red' is given which is not possible because no acid it produced during the reaction so u r left with D ..