Cr2O7^2- + I^- ---> Cr^3+ + IO3^- Solution Break up the reaction into two half-r
ID: 731231 • Letter: C
Question
Cr2O7^2- + I^- ---> Cr^3+ + IO3^-Explanation / Answer
Break up the reaction into two half-reactions -- one for Cr and one for I. Cr2O72- -----> 2Cr3+ I- ------> IO3- After balancing the Cr and I, balance the O by adding H2O. Cr2O72- -----> 2Cr3+ + 7H2O I- 3H2O ------> IO3- Next, balance the H by adding H+ Cr2O72- + 14H+ -----> 2Cr3+ + 7H2O I- + 3H2O ------> IO3- + 6H+ Balance the charges by adding e- to the more positive side of each half-reaction. Cr2O72- + 14H+ + 6e- -----> 2Cr3+ + 7H2O (reduction) I- + 3H2O ------> IO3- + 6H+ + 6e- (oxidation) Since each half-reaction has 6e- involved, just add the half-reactions together. Simplify the water and H+. Cr2O72- + 14H+ + 6e- -----> 2Cr3+ + 7H2O I- + 3H2O ------> IO3- + 6H+ + 6e- _____________________________________ I- + Cr2O72- + 8H+ -----> 2Cr3+ + 4H2O