A) Perchromic acid
B) Dichromate paper
C) Lead acetate paper
D) Tailing of mercury
Correct Answer: B
Solution :
\[{{H}_{2}}{{O}_{2}}\] acts as both reducing as well as oxidizing agent. With strong oxidizing agents (like \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}\]), it acts as reducing agent and itself is oxidized to \[{{O}_{2}}\]. Hence, it turns yellow dichromate paper, to green. \[C{{r}_{2}}O_{7}^{2-}+{{H}_{2}}{{O}_{2}}+8{{H}^{+}}\xrightarrow{{}}\underset{\text{green}}{\mathop{2C{{r}^{3+}}}}\,+5{{H}_{2}}O+2{{O}_{2}}\]
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