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# Acidified $K_2Cr_2O_7$ turns green by

(a) $SO_2$
(b) $CO$
(c) $SiO_2$
(d) $HCl$

Answer: $SO_2$
$K_2Cr_2O_7$ turns green by $SO_2$ due to its reduction.
$K_2Cr_2O_7 + H_2SO_4 + 3SO_2 \longrightarrow K_2SO_4 + Cr_2(SO_4)_3$(Green)$+ H_2O$