Examples of balancing oxidation-reduction reaction equations

On the basis of writing and balancing the electrode reaction equation, the balancing of the redox reaction equation can be further completed
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(1) Denote the oxidation-reduction reaction as two electric pairs respectively
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(2) Balance the electrode reactions of the two electric pairs respectively
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(3) Adjust the stoichiometric number so that the number of electrons gained and lost in the reaction of the two electrodes is equal
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(4) Combine the two electrode reactions, eliminate electrons, and complete the balance of the redox reaction equation
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Finally, check whether the balanced equation is correct by checking the number of atoms and the number of charges on both sides of the reaction equation
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[Example 8-1] Use the ion-electron method to balance the redox equation

Mn ²⁺ + NABIO _.
3 → of MnO _.
4 ^- + of Bi ³⁺

Solution From the existence of Mn ²⁺ and Bi ³⁺ , it can be judged that this reaction is carried out under acidic conditions
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(1) Write this reaction as two redox couples

Of MnO _.
4 ^- / Mn ²⁺NABIO _.
3 / of Bi ³⁺

(2) Balance the electrode reactions of these two pairs separately

Of MnO _.
4 ^- + 8H ⁺ + 5E ^- = Mn ²⁺ + 4H ₂ O ①

NABIO _.
3 + 6H ⁺ + 2E ^- = of Bi ³⁺ + of Na ⁺ + 3H ₂ O ②

(3) Adjust the stoichiometric numbers in the two formulas so that the e-quantities in the two formulas are equal

① × 2 obtained 2MnO _.
4 ^- + 16H ⁺ + 10e ^- = 2Mn ²⁺ + 8H ₂ O ③

② × 5 obtained 5NaBiO _.
3 + 30H ⁺ + 10e ^- = 5Bi ³⁺ + 5Na ⁺ + 15H ₂ O ④

(4) Subtract ③ from ④ to combine the reactions of the two electrodes to obtain a balanced equation

5NaBiO _.
3 + 2Mn ²⁺ + 14H ⁺ = 5Bi ^{.
3 +} + 2MnO _.
4 ^- + 5Na ⁺ + 7H ₂ O

It can be seen from the balance of this oxidation-reduction reaction that this is an ionic reaction with incomplete reactants and products.
While the ion-electron method is used to balance the ionic reaction, the incomplete reactants and products are also eliminated.
Find and complete it
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Therefore, this method is very suitable for balancing this type of reaction
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[Example 8-2] Balance the following reaction formula

+ The CN of CuS ^- → [a Cu (the CN) _.
4 ] ^3- + S ^2- + the NCO ^-

Solution From the existence of CN ^- and S ^2- in the reaction formula , it can be judged that the reaction is carried out in alkaline medium
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(1) Write this reaction as two redox couples

Of CuS / [a Cu (the CN) _.
4 ] ^3-the CN ^- / the NCO ^-

(2) Balance the electrode reactions of these two pairs separately

+ 4CN of CuS ^- + E ^- = [a Cu (the CN) _.
4 ] ^3- + S ^2-①

The CN ^- + 2OH ^- = the NCO ^- + H ₂ O + 2E ^-②

(3) Adjust the stoichiometric number in the electrode reaction equation to make the e-measurement number in the two equations equal

+ 8CN 2CuS ^- + 2E ^- = 2 [a Cu (the CN) _.
4 ] ^3- + 2S ^2-            ③

The CN ^- + 2OH ^- = the NCO ^- + H ₂ O + 2E ^-            ②

(4) ③+②, eliminate the electrons, and get the balanced equation

+ 9 cN 2CuS ^- + 2OH ^- = 2 [a Cu (the CN) _.
4 ] ^3- + 2S ^2- + the NCO ^- + H ₂ O

In this reaction formula trim, it is unnecessary to consider the CN ^- and the NCO C, the oxidation number N, and even do not have to distinguish between the CN ^- and NCO ^- which one is oxidized, which is reduced, as long as the number of atoms flat, with The electronic balance can be used to balance the charge
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