Complete and balance the following redox reaction in basic solution

Question

Complete and balance the following redox reaction in basic solution

Cr2O7^2- (aq) + Hg (l) ----> Hg^2+ (aq) + Cr^3+ (aq)

Answer 1

Linda,

Here is a fairly simple step by step process for balancing redox reactions in basic solution.

Separate the oxidation and reduction half reactions and proceed.

(1). Hg ==> Hg²⁺ ... oxidation half reaction

Hg ==> Hg²⁺ + 2e- ... balanced oxidation half reaction

(2). Cr₂O₇^2- ==> Cr³⁺ ... reduction half reaction

Cr₂O₇^2- ==> 2Cr³⁺ ... balanced for Cr

Cr₂O₇^2- ==> 2Cr³⁺ +7H₂O ... balanced for O

Cr₂O₇^2- + 14H₂O ==> 2Cr³⁺ +7H₂O +14OH^- ... balanced for H using basic solution (OH^-)

Cr₂O₇^2- + 14H₂O + 6e- ==> 2Cr³⁺ +7H₂O +14OH^- ... balanced for charge and final balanced equation

(3). Multiply the oxidation half reaction by 3 in order to equalize the electrons being transferred

3Hg ==> 3Hg²⁺ + 6e-

(4). Add the 2 balanced equations together.

3Hg + Cr₂O₇^2- + 14H₂O + 6e- ==> 2Cr³⁺ +7H₂O +14OH^- + 3Hg²⁺ + 6e-

(5). Combine and/or cancel like items to obtain the final balanced redox equation:

3Hg + Cr₂O₇^2- + 7H₂O ===> 2Cr³⁺ + 14OH^- + 3Hg²⁺ ... balanced redox equation