In basic solution, Cr(OH)3 is oxidized to CrO4^2- by ClO-, balance this redox eq

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Explanation / Answer

I recommend balancing the reaction for an acidic solution first, then converting it to a basic solution. 1. Break up the reaction into two half-reactions (one for Cr and one for Cl) Cr(OH)3 -----> CrO42- ClO31- -----> Cl- 2. Balance the main atoms, Cr and Cl. They are already balanced in this problem. 3. In acid solution, add H+/H2O as needed. First, balance the O atoms by adding water as needed. Cr(OH)3 + H2O -----> CrO42- ClO31- -----> Cl- + 3H2O 4. Next, balance the H atoms by adding H+ as needed. Cr(OH)3 + H2O -----> CrO42- + 5H+ ClO31- + 6H+ -----> Cl- + 3H2O 5. Balance the charges by adding e- to the most positive side of each half-reaction. Cr(OH)3 + H2O -----> CrO42- + 5H+ + 3e- ClO31- + 6H+ + 6e- -----> Cl- + 3H2O 6. Make the electrons gained and lost equal. Take the top equation times two. 2Cr(OH)3 + 2H2O -----> 2CrO42- + 10H+ + 6e- ClO31- + 6H+ + 6e- -----> Cl- + 3H2O 7. Add the two half-reactions together, simplifying the H+ and H2O where possible. 2Cr(OH)3 + 2H2O -----> 2CrO42- + 10H+ + 6e- (oxidation) ClO31- + 6H+ + 6e- -----> Cl- + 3H2O (reduction) 2Cr(OH)3 + ClO31- -----> 2CrO42- + Cl- + H2O + 4H+ (overall reaction) Notice that both sides have a net charge of -1 and all the atoms are balanced. 8. To convert to a basic solution, add OH- to both sides, equal to the number of H+. 2Cr(OH)3 + ClO31- + 4OH- ------> 2CrO42- + Cl- + H2O + 4H+ + 4OH- 9. Combine the H+ and OH- to make H2O, then simplify the number of H2O if possible. 2Cr(OH)3 + ClO31- + 4OH- ------> 2CrO42- + Cl- + 5H2O It is now balanced for a basic solution. Both sides have a net charge of -5 and all the atoms are balanced.

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