Question

Balance the following equation by oxidation number mothod:
Cr2O7^-2 +H2SO3 gives Cr^+3 +HSO4^-​

Answer 1

This is an oxidation-reduction (redox) reaction:

3 SIV - 6 e- → 3 SVI

(oxidation)

2 CrVI + 6 e- → 2 CrIII

(reduction)

H2SO3 is a reducing agent, Cr2O72- is an oxidizing agent.

Reactants:

Cr2O72-

H+

H2SO3 – Sulfurous acid source: wikipedia, accessed: 2019-09-28source: wikidata, accessed: 2019-09-02

Other names: S(O)(OH)2 source: wikidata, accessed: 2019-09-02, Sulphurous acid source: wikidata, accessed: 2019-09-02

Products:

HSO4-

Cr3+

H2O – Water, oxidane source: wikipedia, accessed: 2019-09-27

Other names: Water (H2O) source: wikipedia, accessed: 2019-09-27, Hydrogen hydroxide (HH or HOH) source: wikipedia, accessed: 2019-09-27, Hydrogen oxide source: wikipedia, accessed: 2019-09-27

Appearance: White crystalline solid, almost colorless liquid with a hint of blue, colorless gas source: wikipedia, accessed: 2019-09-27

Search by reactants (Cr2O72-, H+, H2SO3)

1 H:+ + Cr2O7:2- + H2SO3 → H2O + Cr:3+ + HSO4:-

Search by products (HSO4-, Cr3+, H2O)

1 H:+ + Cr2O7:2- + SO3:2- → H2O + Cr:3+ + HSO4:-

2 SO2 + H:+ + Cr2O7:2- → H2O + Cr:3+ + HSO4:-

3 H:+ + Cr2O7:2- + HSO3:- → H2O + Cr:3+ + HSO4:-

4 Cr2O7:2- + H3O:+ + HSO3:- → H2O + Cr:3+ + HSO4:-

5 H:+ + Cr2O7:2- + H2SO3 → H2O + Cr:3+ + HSO4:-

Search by reactants (Cr2O72-, H+, H2SO3) and by products (HSO4-, Cr3+, H2O)

1 H:+ + Cr2O7:2- + H2SO3 → H2O + Cr:3+ + HSO4:-

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Answer 2

Explanation:

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