Question

Complete and balance the following redox reaction in basic solution: Cr₂O₇^2-(aq) + Hg(l) ---> Hg²⁺(aq) + Cr³⁺(aq)

Answer 1

Step 1: The half-reactions are

Cr₂O₇^2-(aq) ---> Cr³⁺(aq)

Hg(l) ---> Hg²⁺(aq)

Step 2: Balance elements other than O and H.

Cr₂O₇^2-(aq) ---> 2Cr³⁺(aq)

Hg(l) ---> Hg²⁺(aq)

Step 3: Add H2O to balance oxygen.

Cr₂O₇^2-(aq) ---> 2Cr³⁺(aq) + 7H₂O

Hg(l) ---> Hg²⁺(aq)

Step 4: Balance hydrogen with protons.

Cr₂O₇^2-(aq) + 14H⁺ ---> 2Cr³⁺(aq) + 7H₂O

Hg(l) ---> Hg²⁺(aq)

Step 5: Balance the charge with electrons

Cr₂O₇^2-(aq) + 14H⁺ + 6e^- ---> 2Cr³⁺(aq) + 7H₂O

Hg(l) ---> Hg²⁺(aq) + 2e^-

Step 6: Scale the reactions with equal amount of electrons.

Cr₂O₇^2-(aq) + 14H⁺ + 6e^- ---> 2Cr³⁺(aq) + 7H₂O

3Hg(l) ---> 3Hg²⁺(aq) + 6e^-

Step 7: Add the reactions and cancel the electrons.

Cr₂O₇^2-(aq) + 14H⁺ + 3Hg(l) ---> 2Cr³⁺(aq) + 3Hg²⁺(aq) + 7H₂O

Step 8: Add OH^- to balance H⁺.

Cr₂O₇^2-(aq) + 14H⁺ + 3Hg(l) + 14OH^- ---> 2Cr³⁺(aq) + 3Hg²⁺(aq) + 7H₂O + 14OH^-

Step 9: Combine OH- and H+ ions to form water

Cr₂O₇^2-(aq) + 14H₂O + 3Hg(l) ---> 2Cr³⁺(aq) + 3Hg²⁺(aq) + 7H₂O + 14OH^-

Or,

Cr₂O₇^2-(aq) + 7H₂O + 3Hg(l) ---> 2Cr³⁺(aq) + 3Hg²⁺(aq) + 14OH^-(Balanced redox reactions)