Balance the following reaction by ion electron method-KMnO 4- H 2 SO 4- HCl - K 2 SO 4- MnSO 4- Cl 2- H 2 OA- 2 KMnO 4-10 HCl -3 H 2 SO 4- 5 Cl 2-2 MnSO 4-8 H 2 OB- 2 KMnO 4-10 HCl - H 2 SO 4- 5 Cl 2-2 MnSO 4-8 H 2 OC- 2 KMnO 4- HCl -3 H 2 SO 4- 5 Cl 2-2 MnSO 4-8 H 2 OD- 2 KMnO 4-10 HCl -2 H 2 SO 4- 5 Cl 2-2 MnSO 4-8 H 2 O

The correct option is A 2KMnO_4 + 10HCl + 3H_2SO_4 → 5Cl_2 + 2MnSO_4 + 8H_2OOxidaiton half: 2Cl^ → Cl_2 + 2e^ —–(1) Reduction half: Setp I : MnO^ _4 → Mn^2+ ...

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Standard XII

Chemistry

Ion Electron Method

Balance the f...

Question

Balance the following reaction by ion electron method:
$K M n O_{4} + H_{2} S O_{4} + H C l \to K_{2} S O_{4} + M n S O_{4} + C l_{2} + H_{2} O$

2 K M n O_{4} + 10 H C l + 2 H_{2} S O_{4} \to 5 C l_{2} + 2 M n S O_{4} + 8 H_{2} O

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2 K M n O_{4} + 10 H C l + 3 H_{2} S O_{4} \to 5 C l_{2} + 2 M n S O_{4} + 8 H_{2} O

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2 K M n O_{4} + 10 H C l + H_{2} S O_{4} \to 5 C l_{2} + 2 M n S O_{4} + 8 H_{2} O

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2 K M n O_{4} + H C l + 3 H_{2} S O_{4} \to 5 C l_{2} + 2 M n S O_{4} + 8 H_{2} O

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Solution

The correct option is B $2 K M n O_{4} + 10 H C l + 3 H_{2} S O_{4} \to 5 C l_{2} + 2 M n S O_{4} + 8 H_{2} O$
Oxidaiton half: $2 C l^{-} \to C l_{2} + 2 e^{-}$ -----(1)
Reduction half:
Setp - I : $M n O_{4}^{-} \to M n^{2 +}$
Step - II : $M n O_{4}^{-} \to M n^{2 +} + 4 H_{2} O$
Step - III : $M n O_{4}^{-} + 8 H^{+} \to M n^{2 +} + 4 H_{2} O$
Step - IV : $M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$ ------(2)
Adding equation (1) $\times 5$ with equation (2) $\times$ 2
$[2 C l^{-} \to C l_{2} + 2 e^{-}] \times 5$
$[M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O] \times 2$
_________________________________________
$2 M n O_{4}^{-} + 10 C l^{-} + 16 H^{+} \to 5 C l_{2} + 2 M n^{2 +} + 8 H_{2} O$

So the balanced equation is
$2 K M n O_{4} + 10 H C l + 3 H_{2} S O_{4} \to 5 C l_{2} + 2 M n S O_{4} + 8 H_{2} O$

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Similar questions

Q. Identify the oxidant and the reductant in the following reaction respectively:

2 K M n O_{4} + 10 K C l + 8 H_{2} S O_{4} \to 2 M n S O_{4} + 6 K_{2} S O_{4} + 8 H_{2} O + 5 C l_{2}

Q. Identify the correctly balanced redox reaction using ion-electron method.

F e S O_{4} + K M n O_{4} + H_{2} S O_{4} \to F e_{2} (S O_{4})_{3} + M n S O_{4} + H_{2} O + K_{2} S O_{4}

2 K M n O_{4} + 5 H_{2} C_{2} O_{4} + 3 H_{2} S O_{4} \to K_{2} S O_{4} + 2 M n S O_{4} + 10 C O_{2} + 8 H_{2} O

How many moles of

M n S O_{4}

are produced when

1.5

mol of

K M n O_{4}, 3.5 m o l

H_{2} C_{2} O_{4}

, and

2.5

mol of

H_{2} S O_{4}

are mixed :

Q. The reaction, is an example of reaction of:

2 M n S O_{4} + 2 K M n O_{4} + 8 H_{2} S O_{4} \to 2 M n S O_{4} + 5 F e_{2} (S O_{4})_{3} + K_{2} S O_{4} + 8 H_{2} O

Q. In the reaction

2 K M n O_{4} + 3 H_{2} S O_{4} ⟶ K_{2} S O_{4} + 2 M n S O_{4} + 3 H_{2} O + 5 (O)

Equivalent weight of

K M n O_{4}

is (when molecular weight of

K M n O_{4} = 158

)

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Ion Electron Method

Standard XII Chemistry

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Balance the following reaction by ion electron method: KMnO4+H2SO4+HCl→K2SO4+MnSO4+Cl2+H2O

Balance the following reaction by ion electron method:
$K M n O_{4} + H_{2} S O_{4} + H C l \to K_{2} S O_{4} + M n S O_{4} + C l_{2} + H_{2} O$