Balance the following reaction- FeC2O4 - KMnO4 - H2SO4 - Fe2 SO43 - CO2 - MnSO4 - K2 SO4 - H2O

The correct option is A 10 FeC_2O_4 + 6KMnO_4 + 24H_2SO_4 → 5Fe_2 (SO_4)_3 + 20CO_2 +6 MnSO_4 +3 K_2 SO_4 + 24H_2O The unbalanced redox equation is as foll ...

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Standard XII

Chemistry

Balancing Redox Reaction

Balance the f...

Question

Balance the following reaction.
$F e C_{2} O_{4} + K M n O_{4} + H_{2} S O_{4} \to F e_{2} (S O_{4})_{3} + C O_{2} + M n S O_{4} + K_{2} S O_{4} + H_{2} O$

10 F e C_{2} O_{4} + 6 K M n O_{4} + 24 H_{2} S O_{4} \to 5 F e_{2} (S O_{4})_{3} + 20 C O_{2} + 6 M n S O_{4} + 3 K_{2} S O_{4} + 24 H_{2} O

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5 F e C_{2} O_{4} + 4 K M n O_{4} + 12 H_{2} S O_{4} \to 7 F e_{2} (S O_{4})_{3} + 20 C O_{2} + 6 M n S O_{4} + 3 K_{2} S O_{4} + 24 H_{2} O

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10 F e C_{2} O_{4} + 4 K M n O_{4} + 24 H_{2} S O_{4} \to 6 F e_{2} (S O_{4})_{3} + 20 C O_{2} + 6 M n S O_{4} + 3 K_{2} S O_{4} + 24 H_{2} O

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None of these

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Solution

The correct option is A $10 F e C_{2} O_{4} + 6 K M n O_{4} + 24 H_{2} S O_{4} \to 5 F e_{2} (S O_{4})_{3} + 20 C O_{2} + 6 M n S O_{4} + 3 K_{2} S O_{4} + 24 H_{2} O$
The unbalanced redox equation is as follows:
$F e C_{2} O_{4} + K M n O_{4} + H_{2} S O_{4} \to F e_{2} (S O_{4})_{3} + C O_{2} + M n S O_{4} + K_{2} S O_{4} + H_{2} O$
Balance all atoms other than H and O.
$2 F e C_{2} O_{4} + 2 K M n O_{4} + 6 H_{2} S O_{4} \to F e_{2} (S O_{4})_{3} + 2 C O_{2} + 2 M n S O_{4} + K_{2} S O_{4} + H_{2} O$
The oxidation number of Fe changes from +2 to +3. The change in the oxidation number of one Fe atom is 1.
Total change in the oxidation number of 2 Fe atoms is 2.
The oxidation number of C changes from +3 to +4. The change in the oxidation number of one C atom is 1.
Total change in the oxidation number of 4 C atoms is 4.
The oxidation number of Mn changes from 7 to 2. The change in the oxidation number of Mn is 5.
Total change in the oxidation number of 2 Mn atoms is 10.
The increase in the oxidation number is balanced with decrease in the oxidation number by multiplying
$K M n O_{4}$ and $M n S O_{4}$ with 3 and multiply $F e C_{2} O_{4}$ $F e_{2} (S O_{4})_{3}$ by 5 and $C O_{2}$ by 10. Sulphuric acid and potassium sulphate are also balanced.
$10 F e C_{2} O_{4} + 6 K M n O_{4} + 24 H_{2} S O_{4} \to 5 F e_{2} (S O_{4})_{3} + 20 C O_{2} + 6 M n S O_{4} + 3 K_{2} S O_{4} + H_{2} O$
O atoms are balanced by adding 23 water molecules on RHS.
$10 F e C_{2} O_{4} + 6 K M n O_{4} + 24 H_{2} S O_{4} \to 5 F e_{2} (S O_{4})_{3} + 20 C O_{2} + 6 M n S O_{4} + 3 K_{2} S O_{4} + 24 H_{2} O$
Hydrogen atoms are balanced. Thus, this is the balanced chemical equation.

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Similar questions

Q. In the redox reaction,

10 F e C_{2} O_{4} + x K M n O_{4} + 24 H_{2} S O_{4} \to 5 F e_{2} (S O_{4})_{3} + 20 C O_{2} + y M n S O_{4} + 3 K_{2} S O_{4} + 24 H_{2} O

the value of

x

and

y

are, respectively:

Q. Potassium tetraoxalate crystals

(K H C_{2} O_{4} . H_{2} C_{2} O_{4} .2 H_{2} O)

react with

K M n O_{4}

(oxidising agent) as follows,

4 K M n O_{4} + 5 K H C_{2} O_{4} . H_{2} C_{2} O_{4} .2 H_{2} O + 13 H_{2} S O_{4} \to 4 M n S O_{4} + 9 K H S O_{4} + 20 C O_{2} + 18 H_{2} O

The equivalent weight of potassium tetraoxalate crystals is:

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

The

n -

factor for

F e C_{2} O_{4}

in the above reaction is:

Q. Balance the equation :

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

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}

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Balance the following reaction.FeC2O4+KMnO4+H2SO4→Fe2(SO4)3+CO2+MnSO4+K2SO4+H2O

Balance the following reaction.
$F e C_{2} O_{4} + K M n O_{4} + H_{2} S O_{4} \to F e_{2} (S O_{4})_{3} + C O_{2} + M n S O_{4} + K_{2} S O_{4} + H_{2} O$