Question

Write balanced half-reactions for the following redox reaction: 512(3)+2 Mn²+ (aq)+160H (aq) → 101 (aq)+2 MnO2 (aq)+8 H20(1) reduction: oxidation:

Answer 1

Following is the - complete Answer -&- Explanation: for the given: Question, in.....typed format....

$\Rightarrow$Answer:

Following are the balanced half reactions, for oxidation, and reduction, respectively:

1. Oxidation half reaction: 2 Mn2+ (aq) + 16 OH - (aq) $\rightleftharpoons$ 2 MnO4 - (aq) + 10 e - + 8 H2O
2. Reduction half raction:   5 I2 (s) + 10 e-    $\rightleftharpoons$10 I - (aq)

$\Rightarrow$Explanation:

Following is the complete Explanation, for the above Answer:

• Given:

​​​​​​​Folowing is the given Redox-reaction:

$\Rightarrow$5 I2 (s) + 2 Mn2+ (aq) + 16 OH - (aq)   $\rightleftharpoons$ 10 I - (aq) + 2 MnO4 - (aq) + 8 H2O (l) ------ Eq. ( 1 )

• Step - 1:

​​​​​​​We know the following half reactions:

1. Reduction:   I2 (s) + 2 e-    $\rightleftharpoons$   2 I - (aq) -------------------------------------------------- Eq. ( 2 )
2. Oxidation: Mn2+ (aq) + 8 OH - (aq) $\rightleftharpoons$ MnO4 - (aq) + 5 e - + 4 H2O -------- Eq. ( 3 )

• ​​​​​​​Step - 2:

​​​​​​​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-    $\rightleftharpoons$10 I - (aq)	Reduction (half reaction)
Eq.(3) x 2 ---- Eq. (5)	2 Mn2+ (aq) + 16 OH - (aq) $\rightleftharpoons$ 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) $\rightleftharpoons$ 10 I - (aq) + 2 MnO4 - (aq) + 8 H2O	overall redox reaction

$\Rightarrow$ 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...

$\Rightarrow$ We can also see, as marked by deep brown color : 10 e-  , are getting cancelled out, on LHS nad RHS, of the reaction.

• Step - 3:

​​​​​​​The desired, half reactions, for oxidation, and reduction, will be the following:

1. Oxidation half reaction: 2 Mn2+ (aq) + 16 OH - (aq) $\rightleftharpoons$ 2 MnO4 - (aq) + 10 e - + 8 H2O
2. Reduction half raction:   5 I2 (s) + 10 e-    $\rightleftharpoons$10 I - (aq)

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