Hydrogen sulphide truns an acidified `K_(2)Cr_(2)O_(7)`

Options

A. colorless
B. green
C. pink
D. orange

Answer

2

Explanation

Acidified `K_(2)Cr_(2)O_(7)` (orange solution ) is reduced froming green chromine sulphate:
`{:(K_(2)Cr_(2)O_(7)+4H_(2)SO_(4) rarr K_(2)SO_(4)+Cr_(2)(SO_(4))_(3)+4H_(2)O+3(O)),(H_(2)S+O rarr H_(2)O+S"]"xx3),(bar(K_(2)Cr_(2)O_(7)+4H_(2)SO_(4)+3H_(2)S rarr K_(2)SO_(4)+Cr_(2)(SO_(4))_(3)+7H_(2)O+3S)):}`
On the other hand acidified `KMnO_(4)` is decolourised.
`{:(2 KMnO_(4)+3H_(2)SO_(4) rarr K_(2)SO_(4) +2MnSO_(4)+3H_(2)O+5[O]),([H_(2)S+O rarr H_(2)O+S]xx5),(bar(2 KMnO_(4)+3H_(2)SO_(4)+5H_(2)S rarr K_(2)SO_(4)+2MnSO_(4)+8H_(2)O+5S)):}`