Cl 2 g - OH - aq - OCl - aq - Cl - aq Balance the above reaction using oxidation number method-A- 3 Cl - 2 g -5 OH - aq -ightarrow OCl - aq - Cl - aq -2 H-2 O l B- Cl - 2 g -2 OH - aq - rightarrow OCl - aq - Cl - aq - H- 2 O l C- 2 Cl - 2 g - OH - aq - 5 OCl - aq - Cl - aq -3 H-2 O l D- 2 Cl - 2 g -2 OH - aq - rightarrow OCl - aq -6 Cl - aq - H2 O l

The correct option is B Cl_2 (g) + 2OH^ (aq) → OCl^ (aq) + Cl^ (aq) + H _2O(l)0Cl_2 (g) + OH^ (aq) → O+1Cl^ (aq) + 1Cl^ (aq) n_f=(|O.S._Product O.S._Re ...

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Byju's Answer

Standard XII

Chemistry

Oxidation Number Method

Cl 2 g + OH -...

Question

$C l_{2} (g) + O H^{-} (a q) \to O C l^{-} (a q) + C l^{-} (a q)$
Balance the above reaction using oxidation number method.

3 C l_{2} (g) + 5 O H^{-} (a q) \to O C l^{-} (a q) + C l^{-} (a q) + 2 H_{2} O (l)

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C l_{2} (g) + 2 O H^{-} (a q) \to O C l^{-} (a q) + C l^{-} (a q) + H_{2} O (l)

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2 C l_{2} (g) + 2 O H^{-} (a q) \to O C l^{-} (a q) + 6 C l^{-} (a q) + H_{2} O (l)

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2 C l_{2} (g) + O H^{-} (a q) \to 5 O C l^{-} (a q) + C l^{-} (a q) + 3 H_{2} O (l)

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Solution

The correct option is B $C l_{2} (g) + 2 O H^{-} (a q) \to O C l^{-} (a q) + C l^{-} (a q) + H_{2} O (l)$
$0 C l_{2} (g) + O H^{-} (a q) \to O + 1 C l^{-} (a q) + - 1 C l^{-} (a q)$

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

$(n_{f})_{O C l^{-}} = | + 1 - 0 | \times 1 = 1$
$(n_{f})_{C l^{-}} = | - 1 - 0 | \times 1 = 1$

Cross multiplying with $n_{f}$ values.

$C l_{2} (g) \to 1 O C l^{-} (a q) + 1 C l^{-} (a q)$

Elements except $O$ and $H$ are already balanced.

Now, balancing oxygen by adding $H_{2} O$

$C l_{2} (g) + H_{2} O (l) \to O C l^{-} (a q) + C l^{-} (a q)$

Balancing hydrogen by adding $H^{+}$
$C l_{2} (g) + H_{2} O (l) \to O C l^{-} (a q) + C l^{-} (a q) + 2 H^{+} (a q)$

For balancing in basic medium adding $2 O H^{-}$ to both sides
$C l_{2} (g) + H_{2} O (l) + 2 O H^{-} (a q) \to O C l^{-} (a q) + C l^{-} (a q) + 2 H^{+} (a q) + 2 O H^{-} (a q)$

$C l_{2} (g) + H_{2} O (l) + 2 O H^{-} (a q) \to O C l^{-} (a q) + C l^{-} (a q) + 2 H_{2} O (l)$

Canceling $H_{2} O$
$C l_{2} (g) + 2 O H^{-} (a q) \to O C l^{-} (a q) + C l^{-} (a q) + H_{2} O (l)$

Balancing charge:

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

So the balanced equation is
$C l_{2} (g) + 2 O H^{-} (a q) \to O C l^{-} (a q) + C l^{-} (a q) + H_{2} O (l)$

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Similar questions

C l_{2} (g) + O H^{-} (a q) \to O C l^{-} (a q) + C l^{-} (a q)

Balance the above reaction using oxidation number method.

Balance the following redox reactions:

I_{2} (s) + O {C l}^{-} (a q) ⟶ {I O}_{3}^{-} (a q) + {C l}^{-} (a q)

Q. Find the heat change in the reaction

N H_{3} (g) + H C l (g) \to N H_{4} C l (s)

from the following data

N H_{3} (g) + a q \to N H_{3} (a q),

Δ H = - 8.4 K . C a l .

H C l (g) + a q \to H C l (a q),

Δ H = - 17.3 K . C a l .

N H_{3} (a q) + H C l (a q) \to N H_{4} C l (a q), Δ H = - 12.5 K . C a l s .

N H_{4} C l (s) + a q \to N H_{4} C l (a q), Δ H = + 3.9 K . C a l .

Q. Consider a galvanic cell using solid

C u

and

F e

metals with their corresponding solutions.
What is the

E_{c e l l}^{\circ}

Standard Potential (V)	Reduction Half-Reaction
$2.87$	$F_{2} (g) + 2 e^{-} \to 2 F^{-} (a q)$
$1.51$	$M n O_{4}^{-} (a q) + 8 H^{+} (a q) + 5 e^{-} \to M n^{2 +} (a q) + 4 H_{2} O (l)$
$1.36$	$C l_{2} (a q) + 3 e^{-} \to 2 C l^{-} (a q)$
$1.33$	$C r_{2} O_{7}^{2 -} (a q) + 14 H^{+} (a q) + 6 e^{-} \to 2 C r^{3 +} (a q) + 7 H_{2} O (l)$
$1.23$	$O_{2} (g) + 4 H^{+} (a q) + 4 e^{-} \to 2 H_{2} O (l)$
$1.06$	$B r_{2} (l) + 2 e^{-} \to 2 B r^{-} (a q)$
$0.96$	$N O_{3}^{-} (a q) + 4 H^{+} (a q) + 3 e^{-} \to N O (g) + H_{2} O (l)$
$0.80$	$A g^{+} (a q) + e^{-} \to A g (s)$ $
$0.77$	$F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q)$
$0.68$	$O_{2} (g) + 2 H^{+} (a q) + 2 e^{-} \to H_{2} O_{2} (a q)$
$0.59$	$M n O_{4}^{-} (a q) + 2 H_{2} O (l) + 3 e^{-} \to M n O_{2} (s) + 4 O H^{-} (a q)$
$0.54$	$I_{2} (s) + 2 e^{-} \to 2 I^{-} (a q)$
$0.40$	$O_{2} (g) + 2 H_{2} O (l) + 4 e^{-} \to 4 O H^{-} (a q)$
$0.34$	$C u^{2 +} (a q) + 2 e^{-} \to C u (s)$
$0$	$2 H^{+} (a q) + 2 e^{-} \to H_{2} (g)$
$- 0.28$	$N i^{2 +} (a q) + 2 e^{-} \to N i (s)$
$- 0.44$	$F e^{2 +} (a q) + 2 e^{-} \to F e (s)$
$- 0.76$	$Z n^{2 +} (a q) + 2 e^{-} \to Z n (s)$
$- 0.83$	$2 H_{2} O (l) + 2 e^{-} \to H_{2} (g) + 2 O H^{-} (a q)$
$1.66$	$A l^{3 +} (a q) + 3 e^{-} \to A l (s)$
$- 2.71$	$N a^{+} (a q) + e^{-} \to N a (s)$
$- 3.05$	$L i^{+} (a q) + e^{-} \to L i (s)$

Δ G^{o}

of the cell reaction

A g C l (s) + 1 / 2 H_{2} (g) \to A g (s) + H^{+} (a q .) + {C l}^{-} (a q)

- 21.52 k J

Calculate the EMF for the cell reaction:

2 A g C l (c) + H_{2} (g) \to A g (s) + 2 H^{+} (a q .) + 2 {C l}^{-} (a q .)

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Oxidation Number Method

Standard XII Chemistry

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Cl2(g)+OH−(aq)→OCl−(aq)+Cl−(aq) Balance the above reaction using oxidation number method.

$C l_{2} (g) + O H^{-} (a q) \to O C l^{-} (a q) + C l^{-} (a q)$
Balance the above reaction using oxidation number method.