Question

Question #17435

Answer 1

`3UO^(2+)+Cr_2O_7^(2-)+8H^(+)->3UO_2^(2+)+2Cr^(3+)+4H_2O`

Explanation:

This is a redox reaction:

`UO^(2+)(aq)+Cr_2O_7^(2-)(aq)->UO_2^(2+)(aq)+Cr^(3+)(aq)`

The oxidizing agent is the `Cr_2O_7^(2-)` and the reducing agent is the `UO^(2+)`.

`U` is getting oxidized since its oxidation number is increasing from `+4` to `+6`.

`Cr` is getting reduced since its oxidation number is decreasing from `+6` to `+3`.

To balance this reaction we will split it into two half equations and we will assume that is happening in acidic medium:

Oxidation: `(UO^(2+)+H_2O->UO_2^(2+)+2H^++2e^-)color(red)(xx3)`

Reduction: `Cr_2O_7^(2-)+14H^(+)+6e^(-)->2Cr^(3+)+7H_2O`

RedOx: `color(red)(3)UO^(2+)+Cr_2O_7^(2-)+color(blue)8H^(+)->color(red)(3)UO_2^(2+)+color(green)(2)Cr^(3+)+color(blue)(4)H_2O`

Here is a video that explains how to balance RedOx reactions in Acidic Medium:
Balancing Redox Reactions | Acidic Medium.

Basic Medium:
Balancing Redox Reactions | Basic Medium.