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Oxidation Number Method

Balance the f...

Question

Balance the following equation by oxidation number method.
${A u C l_{4}}^{-} + S n^{2 +} + H^{+} \to S n^{4 +} + A u C l + H C l$

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Solution

Rules for solving or balancing a chemical equation:-
1. Write an unbalanced equation.
2. Assign oxidation number for each atom.
3. Separate out reactions into two half reactions.
4. Balance all other atoms except $H$ and $O$ in each half reaction.
5. Balance the hydrogen by $H^{+}$ or $O H^{-}$ .
6. Balance the oxygen atoms by $H_{2} O$ .
7. Make electron gain equivalent to electron loss in the half reactions by multiplying them with respective numbers.
8. Add and simplify the half reactions together.
9. Check for elements and charges are balanced.

${A u C l_{4}}^{-} + S n^{+ 2} + 3 H^{+} ⟶ S n^{+ 4} + A u C l + 3 H C l$

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Similar questions

Q. Balance the following equation by oxidation number method.

{M n O_{4}}^{2 -} + H^{+} \to M n O_{2} + {M n O_{4}}^{-} + H_{2} O

Q. Balance the following equations by oxidation number method.

{C r_{2} O_{7}}^{2 -} + H N O_{2} + H^{+} \to C r^{3 +} + {N O_{3}}^{-} + H_{2} O

Q. What is the oxidizing agent in the following equation?

H A s O_{2} (a q) + S n^{2 +} (a q) + H^{+} (a q) \to A s (s) + S n^{4 +} (a q) + H_{2} O (l)

Q. The correct code for stability of oxidation states of Sn and Pb is :

[1] P b^{2 +} > P b^{4}, S n^{2 +} > S n^{4 +}

[2] P b^{2 +} < P b^{4}, S n^{2 +} < S n^{4 +}

[3] P b^{2 +} > P b^{4}, S n^{2 +} < S n^{4 +}

[4] P b^{2 +} < P b^{4}, S n^{2 +} > S n^{4 +}

[5] S n^{2 +} < P b^{2 +}, S n^{4 +} > P b^{4 +}

[6] S n^{2 +} < P b^{2 +}, S n^{4 +} < P b^{4 +}

Q. Balance the following redox reaction:

C r^{+} + S n^{4 +} \to C r^{3 +} + S n^{2 +}

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