For the reaction between $M n O_{4}^{-}$ and $C_{2} O_{4}^{-}$ in basic solution, the unbalanced equation is
$M n O_{4}^{-} + C_{2} O_{4}^{2 -} ⟶ M n O_{2} (s) + C O_{3}^{2 -}$
In a balanced equation, the number of ${O H}^{-}$ ions is :

0

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4

on the right

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4

on the left

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2

on the left

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Solution

The correct option is A $4$ on the left
The balanced equation is as follows:
$4 M n O_{4}^{-} + 3 C_{2} O_{4}^{2 -} + 2 H_{2} O \to 4 M n O_{2} + 6 C O_{3}^{2 -} + 4 O H^{-}$
Hence, the number of $O H^{-}$ is $4$ on the left.

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Similar questions

M n O_{4}^{-}

C_{2} O_{4}^{-}

M n O_{4}^{-} + C_{2} O_{4}^{2 -} ⟶ M n O_{2} (s) + C O_{3}^{2 -}

{O H}^{-}

F e {(O H)}_{2} (s) + O_{2} (g) ⟶ F e {(O H)}_{3} (s)

{N O}_{3}^{-} + B i (s) \to {B i}^{3 +} + {N O}_{2}

F e {(O H)}_{3} (s) + H_{2} O_{2} \to F e {(O H)}_{2} (s) + H_{2} O + O_{2}

{C r}_{2} O_{7}^{2 -} + C_{2} H_{4} O \to C_{2} H_{4} O_{2} + {C r}^{3 +}

M n (O H)_{2} (s) + M n O_{4}^{-} (a q) \to M n O_{2} (s)

{[M n O_{4}]}^{2 -} + H^{+} ⟶

+ {[M n O_{4}]}^{-} + H_{2} O

A

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