Balance the following redox reaction in an acid solution: MnO4-(aq) + I- (aq) → Mn2+ + I2 (s) a. 5e- + 8H+ + MnO4- → Mn2+ + 2e- + 4H2O b. 10I- + 16H+ + 2MnO4- → 5I2 + 2Mn2+ + 8H2O c. 2MnO4- → 2Mn2+ + 4H2O d. 2I- → I2
Added by Allison Q.
Step 1
Reduction half-reaction: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O Oxidation half-reaction: 2I- → I2 + 2e- Now, we need to balance the electrons in both half-reactions. To do this, we will multiply the reduction half-reaction by 2 and the oxidation half-reaction by Show more…
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