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Q.
$H_{2}O_{2}$ reduces $K_{4}Fe(CN)_{6}$
Redox Reactions
Solution:
When $H_{2}O_{2}$ reduces with $K_{4}[Fe(CN)_{6}]$ It is present in acidic solution
$2K_{4}[Fe(CN)_{6}+H_{2}SO_{4}+H_{2}O_{2} \to 2K_{3}[Fe(CN)_{6}]+K_{2}SO_{4}+2H_{2}O$