Question

In each of the following elements:-Assign oxidation numbers.-Identify the oxidizing and reducing agents.-Identify the change in oxidation number.2Cr^3+ + H2O + 6ClO3^- —> Cr2O7^2- + 6ClO2 + 2H^+

Answer 1

Given equation:

`2Cr^(3+)+H_2O+6ClO^-_3\rightarrow Cr_2O^(2-)_7+6ClO_2+2H^+`

Assign oxidation numbers

The oxidation number of H is + 1

The oxidation number of Cr in Cr³⁺ is +3

The oxidation number of O in ClO₃⁻ is +2

The oxidation number of Cl in ClO₃⁻ is -7

The oxidation number of Cr in Cr₂O₇²⁻ is +6

The oxidation number of Cl in ClO₂ is +4

The oxidation number of O in ClO₂ is -2

Oxidizing and reducing agents.

An oxidizing agent is a substance that causes oxidation by accepting electrons; therefore, it gets reduced. A reducing agent is a substance that causes reduction by losing electrons; therefore it gets oxidized.

ClO₃⁻ is the oxidizing agent and Cr³⁺ is the reducing agent.

Change in oxidation number.

The oxidation number of Cr changes from +3 in Cr³⁺ to +6 in Cr₂O₇²⁻

The oxidation number O changes from +2 in ClO₃⁻ to -2 in ClO₂

The oxidation number of Cl changes from -7 in ClO₃⁻ to +4 in ClO₂