Using Oxidation-number method- find the net ionic equation for the following reaction in basic medium-Mn OH 2 s - MnO 4- aq - MnO 2 s A- 6 Mn OH 2 s -2 MnO 4- aq - 5 MnO 2 s -2 H 2 O - OH -B- 4 Mn OH 2 s -2 MnO 4-aq - 4 MnO 2 s -2 H 2 O -2 OH -C- 2 Mn OH 2 s -2 MnO 4-aq - 5 MnO 2 s -2 H 2 O -2 OH -D- 3 Mn OH 2 s -2 MnO 4-aq - 5 MnO 2 s -2 H 2 O -2 OH -

The correct option is D 3Mn(OH)_2(s) + 2MnO^ _4(aq) → 5MnO_2(s) + 2H_2O + 2OH^ Mn(OH)_2(s) + MnO^ _4 (aq) → MnO_2(s) Mn oxidation state in Mn(OH)_2=+2 Mn oxi ...

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Byju's Answer

Standard XII

Chemistry

Oxidation Number Method

Using Oxidati...

Question

Using Oxidation-number method, find the net ionic equation for the following reaction in basic medium:
$M n (O H)_{2} (s) + M n O_{4}^{-} (a q) \to M n O_{2} (s)$

6 M n (O H)_{2} (s) + 2 M n O_{4}^{-} (a q) \to 5 M n O_{2} (s) + 2 H_{2} O + O H^{-}

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

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2 M n (O H)_{2} (s) + 2 M n O_{4}^{-} (a q) \to 5 M n O_{2} (s) + 2 H_{2} O + 2 O H^{-}

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3 M n (O H)_{2} (s) + 2 M n O_{4}^{-} (a q) \to 5 M n O_{2} (s) + 2 H_{2} O + 2 O H^{-}

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Solution

The correct option is D $3 M n (O H)_{2} (s) + 2 M n O_{4}^{-} (a q) \to 5 M n O_{2} (s) + 2 H_{2} O + 2 O H^{-}$
$M n (O H)_{2} (s) + M n O_{4}^{-} (a q) \to M n O_{2} (s)$

Mn oxidation state in $M n (O H)_{2} = + 2$
Mn oxidation state in $M n O_{4}^{-} = + 7$
Mn oxidation state in $M n O_{2} = + 4$

Cleary, $M n (O H)_{2}$ is undergoing oxidation and $M n O_{4}^{-}$ is undergoing reduction.
Formula used for the n-factor calculation,
$n_{f} = (| O . S ._{P r o d u c t} - O . S ._{R e a c t a n t} | \times number of atoms$
using the formula of n-factor given above,
$n_{f}$ of $M n (O H)_{2} = 2$
$n_{f}$ of $M n O_{4}^{-} = 3$
Cross multiplying these with $n_{f}$ of each other.
we get,
$3 M n (O H)_{2} (s) + 2 M n O_{4}^{-} (a q) \to M n O_{2} (s)$

Balancing the Mn element on both sides,
$3 M n (O H)_{2} (s) + 2 M n O_{4}^{-} (a q) \to 5 M n O_{2} (s)$

Adding the $H_{2} O$ to balance the oxygen,
The LHS of the equation contains 14 oxygens while RHS contains only 10 oxygens, so adding 4 $H_{2} O$ to RHS.
$3 M n (O H)_{2} (s) + 2 M n O_{4}^{-} (a q) \to 5 M n O_{2} (s) + 4 H_{2} O$

Adding $H^{+}$ to balance hydrogen,

As LHS of the equation contains 6 hydrogens while RHS contains 8 hydrogens, so adding 2 $H^{+}$ to LHS.
$3 M n (O H)_{2} (s) + 2 M n O_{4}^{-} (a q) + 2 H^{+} \to 5 M n O_{2} (s) + 2 H_{2} O$

Now adding $O H^{-}$ to both sides to combine with $H^{+}$ and make it $H_{2} O$ ,
$3 M n (O H)_{2} (s) + 2 M n O_{4}^{-} (a q) + 2 H^{+} + 2 O H^{-} \to 5 M n O_{2} (s) + 4 H_{2} O + 2 O H^{-}$
$3 M n (O H)_{2} (s) + 2 M n O_{4}^{-} (a q) + 2 H_{2} O \to 5 M n O_{2} (s) + 4 H_{2} O + 2 O H^{-}$

Eliminating the unrequired common $H_{2} O$ which is present on the both sides,

$3 M n (O H)_{2} (s) + 2 M n O_{4}^{-} (a q) \to 5 M n O_{2} (s) + 2 H_{2} O + 2 O H^{-}$

This is the final balanced equation.

We can also see that charge on both sides is -2. which also indicates that the equation is balanced now.

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Similar questions

Q. Write balanced net ionic equation for the following reaction in basic solution:

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

Q. Balance the following chemical equation in basic medium:

M n O_{4}^{-} (a q) + B r^{-} (a q) \to M n O_{2} (s) + B r O_{3}^{-} (a q)

Q. For a given redox reaction:

3 M n (O H)_{2} (s) + 2 M n O_{4}^{-} (a q) \to 5 M n O_{2} (s) + 2 H_{2} O + 2 O H^{-}

Choose the correct option(s):

Q. Which of the following reactions are disproportionation reactions?
a)

2 C u^{+}

\to

C u^{2 +}

C u^{0}

3 M n O_{4}^{2 -}

4 H^{+}

\to

2 M n O_{4}^{-}

M n O_{2}

2 H_{2} O

2 K M n O_{4}

△ - \to

K_{2} M n O_{4}

M n O_{2}

O_{2}

2 M n O_{4}^{-}

3 M n^{2 +}

2 H_{2} O

⟶

5 M n O_{2}

4 H^{+}

Q. Which of the following reactions are disproportion reactions?
(a)

C u^{+} \to C u^{2 +} + C u

(b)

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

(c)

2 K M n O_{4} \to K_{2} M n O_{4} + M n O_{2} + O_{2}

(d)

2 M n O_{4}^{-} + 3 M n^{+ 2} + 2 H_{2} O \to 5 M n O_{2} + 4 H^{+}

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Using Oxidation-number method, find the net ionic equation for the following reaction in basic medium: Mn(OH)2(s)+MnO−4(aq)→MnO2(s)

Using Oxidation-number method, find the net ionic equation for the following reaction in basic medium:
$M n (O H)_{2} (s) + M n O_{4}^{-} (a q) \to M n O_{2} (s)$