Chemistry, asked by Anonymous, 9 months ago

Q. Explain steps of redox reaction with example.

Answers

Answered by Anonymous

The reduction is the gain of electrons whereas oxidation is the loss of electrons.

The combination of reduction and oxidation reaction together refers to redox reaction/redox process.

As discussed, it is very important to understand “balancing redox reactions”

.There are generally two methods for balancing redox reactions.

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Answered by Anonymous

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$\bold{\huge{\underline{\underline\red{ANSWER: }}}}$

Redox Reaction - The reaction in which oxidation and reduction takes place at same time.

Let's take the following reaction to understand steps of balancing of redox reaction.

$FeSO_4 + KMnO_4 + H_2SO_4 \longrightarrow Fe_2(SO_4)_3 + MnSO_4 + H_2O + K_2SO_4$

Step - 1 $\implies$ Identify the elements going oxidation and reduction.

In this equation -

${Fe}^{2 + } \longrightarrow {Fe}^{3 + }$

As the charge is increasing, Fe is going oxidation.

${MnO_4}^{ - } \longrightarrow {Mn}^{ 2+}$

As oxygen is removed, it is going reduction.

Step - 2 $\implies$ Balance the atom other than oxygen and hydrogen in the elements going oxidized and reduced.

${Fe}^{2 + } \longrightarrow {Fe}^{3 + }$

${MnO_4}^{ - } \longrightarrow {Mn}^{ 2+}$

Fe and Mn are balanced on both sides.

Step - 3 $\implies$ Now balance O and H atoms by $H_2O$ and ${H}^{ + }$ respectively by following way :- For one excess oxygen atom , add one $H_2O$ on the other side and two ${H}^{ + }$ on the same side.

${Fe}^{2 + } \longrightarrow {Fe}^{3 + } ( no \: oxygen \: atom )$ (i)

${8H}^{ + } + {MnO_4}^{ - } \longrightarrow {Mn}^{ 2+} +4H_2O$ (ii)

Step - 4 $\implies$ Equation (i) & (ii) are balanced atomwise. Now balance charge . To balance charge , add electrons to electrically positive charge.

${Fe}^{2 + } \longrightarrow {Fe}^{3 + } + {e}^{ - }$

${5e}^{ - }{8H}^{ + } + {MnO_4}^{ - } \longrightarrow {Mn}^{ 2+} +4H_2O$

Step - 5 $\implies$ The number of electrons gained and lost are equalized by multiplying both reactions by cross multiplication of number of electrons. Here we will multiply (i) with 5 and (ii) with 1

$( {Fe}^{2 + } \longrightarrow {Fe}^{3 + } + {e}^{ - } ) × 5$

= ${5Fe}^{2 + } \longrightarrow {5Fe}^{3 + } + {5e}^{ - }$

${5e}^{ - }{8H}^{ + } + {MnO_4}^{ - } \longrightarrow {Mn}^{ 2+} +4H_2O$

Now we get balanced redox reaction in ionic form.

Step - 6 $\implies$ Now convert ionic reaction into molecular form.

$5 FeSO_4 + KMnO_4 +4 H_2SO_4 \longrightarrow\frac{5}{2} Fe_2(SO_4)_3 + MnSO_4 + 4H_2O +\frac{1}{2} K_2SO_4$

Or

$10 FeSO_4 + 2KMnO_4 + 8H_2SO_4 \longrightarrow 5Fe_2(SO_4)_3 + 2MnSO_4 + 8H_2O + K_2SO_4$

$\bold{\huge{\underline{\underline\red{THANKS}}}}$

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Q. Explain steps of redox reaction with example.​

Answers

Redox Reaction - The reaction in which oxidation and reduction takes place at same time.

Step - 1 Identify the elements going oxidation and reduction.

Step - 2 Balance the atom other than oxygen and hydrogen in the elements going oxidized and reduced.

Step - 3 Now balance O and H atoms by and respectively by following way :- For one excess oxygen atom , add one on the other side and two on the same side.

Step - 4 Equation (i) & (ii) are balanced atomwise. Now balance charge . To balance charge , add electrons to electrically positive charge.

Step - 5 The number of electrons gained and lost are equalized by multiplying both reactions by cross multiplication of number of electrons. Here we will multiply (i) with 5 and (ii) with 1

Step - 6 Now convert ionic reaction into molecular form.