Let’s first consider the ions present
H+ OH-
Na+ Cl-
what exactly happens in the solution when electricity is passed through
The negative electrode(cathode) attracts the Hydrogen ions are attracted to it. They accept electrons to form hydrogen gas(which does not produce any change in the indicator)
At the positive electrode(anode), chloride ions are attracted. Here they lose two electrons to form chlorine gas. The chlorine bleaches the indicator.
The ions remaining form sodium hydroxide but the color change to blue is observed at the other electrode only because at the anode the indicator is bleached.
Ask again if u don't understand
oh allright got it thanx a lot nid. really nice of u to help
one more thing so de hydrogen formed at cathode doesnt affect de indicator becoz its as gas nd not as H+ ions,is dat de reason why de hydrogen gas has no effect on de indicator???
also so de sodium hydroxide is formed everywhere however it only affects the cathode region, because at the anode the litmus paper is bleached.. is dat rite
so if de litmus paper had not been bleached den even at de anode the litmus paper would have turned blue??
nd before chlroine turns damp blue litmus paper white it first turn the damn blue litmus paper red rite?? y is dat
nd one more question will sulphur dioxide gas react weith sodium hydroxide.. if yes what are the products
+ rep for u. Thanks a lot for ur help. hope im not wasting too much of ur time
which country r u from??
hydrogen gas simply does not affect the indicator. H
+ions affect indiacators(since acids contain them)
if the indicator was not bleached it would have turned it blue throughout
It is in indicator in a solution(it's a universal indicator not litmus paper). *Chlorine does not turn litmus paper red anyways. The indicator is green in neutral solutions.
Sulphur dioxide will also react directly with bases such as sodium hydroxide solution. If sulphur dioxide is bubbled through sodium hydroxide solution, sodium sulphite solution is formed first followed by sodium hydrogensulphite solution when the sulphur dioxide is in excess. But y do u want to know this??
