ClO 2- SbO 2- ClO 2- Sb OH 6-The balanced form of the above reaction in basic medium using oxidation number method is-A- 2 ClO 2- SbO 2-2 OH - 2 ClO 2- Sb OH 6-2 H 2 OB- ClO 2- SbO 2-2 H 2 O - OH - ClO 2- Sb OH 6- 2 ClO 2- SbO 2-2 H 2 O -2 OH - 2 ClO 2- Sb OH 6-D- 2 ClO 2-4 SbO 2-2 OH - ClO 2-4 Sb OH 6-3 H 2 O

The correct option is C 2ClO_2 + SbO_2^ + 2H_2O + 2OH^ → 2ClO_2^ + Sb(OH)_6^ +4ClO_2 + +3SbO_2^ →+3ClO_2^ + +5Sb(OH)_6^ ClO_2 is an oxidising agent. SbO_ ...

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Byju's Answer

Standard XII

Chemistry

Common Oxidation States

ClO 2+ SbO 2-...

Question

$C l O_{2} + S b O_{2}^{-} \to C l O_{2}^{-} + S b (O H)_{6}^{-}$
The balanced form of the above reaction in basic medium using oxidation number method is:

2 C l O_{2} + S b O_{2}^{-} + 2 O H^{-} \to 2 C l O_{2}^{-} + S b (O H)_{6}^{-} + 2 H_{2} O

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C l O_{2} + S b O_{2}^{-} + 2 H_{2} O + O H^{-} \to C l O_{2}^{-} + S b (O H)_{6}^{-}

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2 C l O_{2} + S b O_{2}^{-} + 2 H_{2} O + 2 O H^{-} \to 2 C l O_{2}^{-} + S b (O H)_{6}^{-}

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2 C l O_{2} + 4 S b O_{2}^{-} + 2 O H^{-} \to C l O_{2}^{-} + 4 S b (O H)_{6}^{-} + 3 H_{2} O

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Solution

The correct option is C $2 C l O_{2} + S b O_{2}^{-} + 2 H_{2} O + 2 O H^{-} \to 2 C l O_{2}^{-} + S b (O H)_{6}^{-}$
$+ 4 C l O_{2} + + 3 S b O_{2}^{-} \to + 3 C l O_{2}^{-} + + 5 S b (O H)_{6}^{-}$

$C l O_{2}$ is an oxidising agent.
$S b O_{2}^{-}$ is a reducing agent.

$n_{f} = (| O . S ._{P r o d u c t} - O . S ._{R e a c t a n t} |) \times number of atom$

$+ 3 S b O_{2}^{-} \to + 5 S b (O H)_{6}^{-}$ oxidation
$n_{f} = (| + 5 - 3 |) \times 1 = 2$

$+ 4 C l O_{2} \to + 3 C l O_{2}^{-}$ reduction
$n_{f} = (| + 3 - 4 |) \times 1 = 1$

Ratio of $n_{f}$ for oxidation to reduction is $2 : 1$ .

Cross mutiplying the oxidising and reducing agents with ratio of n-factors.
$2 C l O_{2} + S b O_{2}^{-} \to C l O_{2}^{-} + S b (O H)_{6}^{-}$

Balancing elements except $O$ and $H$
$2 C l O_{2} + S b O_{2}^{-} \to 2 C l O_{2}^{-} + S b (O H)_{6}^{-}$

Balancing oxygen by adding $H_{2} O$
$2 C l O_{2} + S b O_{2}^{-} + 4 H_{2} O \to 2 C l O_{2}^{-} + S b (O H)_{6}^{-}$

Balancing hydrogen by adding $H^{+}$
$2 C l O_{2} + S b O_{2}^{-} + 4 H_{2} O \to 2 C l O_{2}^{-} + S b (O H)_{6}^{-} + 2 H^{+}$

Adding $2 O H^{-}$ on both sides to balance in basic medium
$2 C l O_{2} + S b O_{2}^{-} + 4 H_{2} O + 2 O H^{-} \to 2 C l O_{2}^{-} + S b (O H)_{6}^{-} + 2 H^{+} + 2 O H^{-}$

$2 C l O_{2} + S b O_{2}^{-} + 4 H_{2} O + 2 O H^{-} \to 2 C l O_{2}^{-} + S b (O H)_{6}^{-} + 2 H_{2} O$

Canceling $H_{2} O$
$2 C l O_{2} + S b O_{2}^{-} + 2 H_{2} O + 2 O H^{-} \to 2 C l O_{2}^{-} + S b (O H)_{6}^{-}$

Balancing charge:

charge in reactant side $= - 1 - 2 = - 3$
charge in product side $= - 2 - 1 = - 3$

So the balanced equation is
$2 C l O_{2} + S b O_{2}^{-} + 2 H_{2} O + 2 O H^{-} \to 2 C l O_{2}^{-} + S b (O H)_{6}^{-}$

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Similar questions

C l O_{2} + S b O_{2}^{-} \to C l O_{2}^{-} + S b (O H)_{6}^{-}

The balanced form of the above reaction in basic medium using oxidation number method is:

The balanced equation in basic medium is :

C l O_{2} + {S b O}_{2}^{-} \to {C l O}_{2}^{-} + {S b (O H)}_{6}^{-}

Q. Which pairs represent conjugate pair of oxidant-reductant or reductant-oxidant in

2 C l O_{2} + 5 H_{2} O_{2} + 2 O H^{-} \to 2 C l^{-} + 5 O_{2} + 6 H_{2} O

.
I.

C l O_{2}

and

C l^{-}

II.

H_{2} O_{2}

and

O_{2}

III.

C l^{4 +}

and

C l^{-}

IV.

C l^{5 +}

and

C l^{-}

Q. In alkaline medium,

C l O_{2}

oxidizes

H_{2} O_{2}

O_{2}

and is itself reduced to

{C l O}_{2}^{-}

. Number of moles of

H_{2} O_{2}

oxidized by 1 mole of

C l O_{2}

is :

Q. 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)

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Common Oxidation States

Standard XII Chemistry

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ClO2+SbO−2→ClO−2+Sb(OH)−6 The balanced form of the above reaction in basic medium using oxidation number method is:

$C l O_{2} + S b O_{2}^{-} \to C l O_{2}^{-} + S b (O H)_{6}^{-}$
The balanced form of the above reaction in basic medium using oxidation number method is: