Question #17435

1 Answer
Jan 13, 2016

Answer:

#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.