(H\_{2}O\_{2})reduces(K\_{4}Fe(CN)\_{6}). | CHEMISTRY - REDOX REACTIONS AND ELECTRODE PROCESSES | SolveeIt

$H_{2}O_{2}$ reduces $K_{4}Fe(CN)_{6}$ .

In neutral solution

In acidic solution

In non-polar solvent

In alkaline solution

Answer

In acidic solution

Explanation

When $H_{2}O_{2}$ reduces with $K_{4}\lbrack Fe(CN)_{6}\rbrack$ .

It is present in acidic solution.

$2K_{4}\lbrack Fe(CN)_{6} + H_{2}SO_{4} + H_{2}O_{2} \rightarrow$

$2K_{3}\lbrack Fe(CN)_{6}\rbrack + K_{2}SO_{4} + 2H_{2}O$

Question: \(H_{2}O_{2}\)reduces\(K_{4}Fe(CN)_{6}\)....