Following is the - complete Answer -&- Explanation: for the given: Question, in.....typed format....
Answer:
Following are the balanced half reactions, for oxidation, and reduction, respectively:
Explanation:
Following is the complete Explanation, for the above Answer:
Folowing is the given Redox-reaction:
5 I2 (s) + 2 Mn2+ (aq) + 16 OH - (aq) 10 I - (aq) + 2 MnO4 - (aq) + 8 H2O (l) ------ Eq. ( 1 )
We know the following half reactions:
Therefore, if we multiply Eq. ( 2 ), by five ( 5 ), and multiply Eq. ( 3 ), by two ( 2 ), and then add the resulting oxidation, and reduction -------- half reactions, we will get the following final redox reaction, as desired..
Equation ID | Half reaction(s) & Overall reaction | Type of Half Reaction |
Eq. (2) x 5 ---- Eq. (4) | 5 I2 (s) + 10 e- 10 I - (aq) | Reduction (half reaction) |
Eq.(3) x 2 ---- Eq. (5) | 2 Mn2+ (aq) + 16 OH - (aq) 2 MnO4 - (aq) + 10 e - + 8 H2O | Oxidation ( half reaction ) |
Eq.(4) + Eq.(5) --- Eq.(6) | 5 I2 (s) + 2 Mn2+ (aq) + 16 OH - (aq) 10 I - (aq) + 2 MnO4 - (aq) + 8 H2O | overall redox reaction |
As we can see above, by adding the Reduction, and Oxiadtion, half reactions, we are being able to obtain the overall Redox reaction, as desired...
We can also see, as marked by deep brown color : 10 e- , are getting cancelled out, on LHS nad RHS, of the reaction.
The desired, half reactions, for oxidation, and reduction, will be the following:
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