$H_2 O_2$ acts as a reducing agent in
(1) $2 NaOCl+H_2 O_2 \to 2 NaCl+H_2 O+O_2$
(2) $2 Fe^{2+}+2 H^{+}+H_2 O_2 \to 2 Fe^{3+}+2 H_2 O$
(3) $Mn^{2+}+2 H_2 O_2 \to MnO_2+2 H_2 O$
(4) $Na_2 S+4 H_2 O_2 \to Na_2 SO_4+4 H_2 O$
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