Balance each of thefollowing redox reactions occurring in acidic solution a) SO

Question

Balance each of thefollowing redox reactions occurring in acidic solution a) SO32- (s)+MNO4- ---> SO42- +Mn2+ b) S2O32-+ Cl2---> SO42-+Cl- Balance each of thefollowing redox reactions occurring in acidic solution a) SO32- (s)+MNO4- ---> SO42- +Mn2+ b) S2O32-+ Cl2---> SO42-+Cl- b) S2O32-+ Cl2---> SO42-+Cl-

Explanation / Answer

a) SO32- + H2O =>    SO42- + 2e- +2H+          (S goes from 4+ to 6+, hence 2e-. Add H2O to left to balance O, addH+ on right to balance H)     MnO4-   + 5e- + 8H+ => Mn2+    + 4H2O (Mn goes from 7+ to 2+, need 5e-) multiply the first by 5 and the second by 2 and add (eliminateredundant species) 5SO32- + 5H2O + 2MnO4- + 16H+ =>    5SO42- + 10H+ +2Mn2+ + 8H2O 5SO32- + 2MnO4- + 6H+ =>    5SO42- +2Mn2+ + 3H2O b) S2O32-+ Cl2 --->SO42-+Cl- Cl2 + 2e- =>2Cl-                            (1) S2O32- =>2SO42- (S goes from 2+ to6+) S2O32- =>2SO42-+ 8e- S2O32- + 5H2O =>2SO42-+ 8e- +10H+           (2) Multiply equation 1 by 4 and add to (2) to eliminate electrons: S2O32- + 5H2O + 4Cl2 =>2SO42-+ 10H+ + 8Cl-

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