The total number of $H_{2} O$ molecules present in the balanced chemical reaction is :
$M n O_{4}^{⊖} + {F e}^{2 +} ⟶ {M n}^{2 +} + {F e}^{3 +}$ (in alkaline medium)

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Solution

The balanced equation is as follows:
$M n O_{4}^{⊖} + 4 H_{2} O + 5 {F e}^{2 +} ⟶ {M n}^{2 +} + 5 {F e}^{3 +} + 8 ⊖ O H$
Hence, the total number of water molecules present $= 4$ .

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Similar questions

M n O_{4} + F e^{2 +} O H^{-} \to M n^{2 +} + F e^{3 +}

{M n O_{4}}^{-} + F e^{2 +} + H^{+} \to M n^{2 +} + F e^{3 +} + H_{2} O

. . . H^{+} + . . . M n O_{4}^{-} + . . . F e^{2 +} \to

. . . M n^{2 +} + . . . F e^{3 +} + . . . H_{2} O

5 {F e}^{2 +} + M n O_{4}^{-} + 8 H^{+} ⇌ {M n}^{2 +} + 5 {F e}^{3 +} + 4 H_{2} O

{F e}^{3 +} + e^{-} ⟶ {F e}^{2 +}, {E_{1}}^{o}

M n O_{4}^{-} + 8 H^{+} + 5 e^{-} ⟶ {M n}^{2 +} + 4 H_{2} O, {E_{2}}^{o}

F e^{2 +} \to F e^{3 +} + e^{-}; M n O_{4}^{-} + 5 e^{-} \to M n^{2 +}

F e^{2 +}

M n O_{4}^{-}

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