Question

For a particular redox reaction Cr is oxidized to CrO42– and Ag is reduced to Ag. Complete and balance the equation for this reaction in basic solution. Phases are optional.

Answer 1

Cr in Cr has oxidation state of 0

Cr in CrO4-2 has oxidation state of +6

So, Cr in Cr is oxidised to CrO4-2

Ag in Ag+ has oxidation state of +1

Ag in Ag has oxidation state of 0

So, Ag in Ag+ is reduced to Ag

Reduction half cell:

Ag+ + 1e- --> Ag

Oxidation half cell:

Cr --> CrO4-2 + 6e-

Balance number of electrons to be same in both half reactions

Reduction half cell:

6 Ag+ + 6e- --> 6 Ag

Oxidation half cell:

Cr --> CrO4-2 + 6e-

Lets combine both the reactions.

6 Ag+ + Cr --> 6 Ag + CrO4-2

Balance Oxygen by adding water

6 Ag+ + Cr + 4 H2O --> 6 Ag + CrO4-2

Balance Hydrogen by adding H+

6 Ag+ + Cr + 4 H2O --> 6 Ag + CrO4-2 + 8 H+

Add equal number of OH- on both sides as the number of H+

6 Ag+ + Cr + 4 H2O + 8 OH- --> 6 Ag + CrO4-2 + 8 H+ + 8 OH-

Combine H+ and OH- to form water

6 Ag+ + Cr + 4 H2O + 8 OH- --> 6 Ag + CrO4-2 + 8 H2O

Remove common H2O from both sides

Balanced Eqn is

6 Ag+ + Cr + 8 OH- --> 6 Ag + CrO4-2 + 4 H2O

This is balanced chemical equation in basic medium

Answer:

6 Ag⁺ + Cr + 8 OH^- --> 6 Ag + CrO₄^2- + 4 H₂O

Feel free to comment below if you have any doubts or if this answer do not work. I will edit it if you let me know