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MCAT Physical : Half Reactions and Balancing Equations

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Example Question #1 : Oxidation Reduction

Consider the redox reaction:

$MnO_4^-+I^-\rightarrow Mn^{2+}+I_2(s)$ $\displaystyle MnO_4^-+I^-\rightarrow Mn^{2+}+I_2(s)$

What is the balanced equation for the above redox reaction in acidic solution?

Possible Answers:

$2I^-(aq)+MnO_4^-(aq)+8H^+(aq)\rightarrow Mn^{2+}(aq)+I_2(s)+4H_2O(l)$ $\displaystyle 2I^-(aq)+MnO_4^-(aq)+8H^+(aq)\rightarrow Mn^{2+}(aq)+I_2(s)+4H_2O(l)$

$10I^-(aq)+2MnO_4^-(aq)\rightarrow 2Mn^{2+}(aq)+5I_2(s)+8H_2O(l)+16OH^-(aq)$ $\displaystyle 10I^-(aq)+2MnO_4^-(aq)\rightarrow 2Mn^{2+}(aq)+5I_2(s)+8H_2O(l)+16OH^-(aq)$

$10I^-(aq)+2MnO_4^-(aq)+16H^+(aq)\rightarrow 2Mn^{2+}(aq)+5I_2(s)+8H_2O(l)$ $\displaystyle 10I^-(aq)+2MnO_4^-(aq)+16H^+(aq)\rightarrow 2Mn^{2+}(aq)+5I_2(s)+8H_2O(l)$

$2I^-(aq)+MnO_4^-(aq)+8H^+(aq)+5e^-\rightarrow Mn^{2+}(aq)+I_2(s)+4H_2O(l)+2e^-$ $\displaystyle 2I^-(aq)+MnO_4^-(aq)+8H^+(aq)+5e^-\rightarrow Mn^{2+}(aq)+I_2(s)+4H_2O(l)+2e^-$

$10I^-(aq)+2MnO_4^-(aq)+8H_2O(l)\rightarrow 2Mn^{2+}(aq)+5I_2(s)+16OH^-(aq)$ $\displaystyle 10I^-(aq)+2MnO_4^-(aq)+8H_2O(l)\rightarrow 2Mn^{2+}(aq)+5I_2(s)+16OH^-(aq)$

Correct answer:

$10I^-(aq)+2MnO_4^-(aq)+16H^+(aq)\rightarrow 2Mn^{2+}(aq)+5I_2(s)+8H_2O(l)$ $\displaystyle 10I^-(aq)+2MnO_4^-(aq)+16H^+(aq)\rightarrow 2Mn^{2+}(aq)+5I_2(s)+8H_2O(l)$

Explanation:

$MnO_4^-+I^-\rightarrow Mn^{2+}+I_2(s)$ $\displaystyle MnO_4^-+I^-\rightarrow Mn^{2+}+I_2(s)$

1. Determine the oxidation number of each element.

Reactants: $Mn=+7,\ O=-2,\ I=-1$ $\displaystyle Mn=+7,\ O=-2,\ I=-1$

Products: $Mn=+2,\ I=0$ $\displaystyle Mn=+2,\ I=0$

2. Write and balance the half reactions.

Oxidation (loss of electrons): $2I^-(aq)\rightarrow I_2(s)+2e^-$ $\displaystyle 2I^-(aq)\rightarrow I_2(s)+2e^-$

Reduction (gain of electrons): $MnO^-_4+5e^-\rightarrow Mn^{2+}$ $\displaystyle MnO^-_4+5e^-\rightarrow Mn^{2+}$

3. Balance oxygen by adding water (from the solution) into the half reactions.

Oxidation: $2I^-(aq)\rightarrow I_2(s)+2e^-$ $\displaystyle 2I^-(aq)\rightarrow I_2(s)+2e^-$

Reduction: $MnO^-_4+5e^-\rightarrow Mn^{2+}+{4H_2O}$ $\displaystyle MnO^-_4+5e^-\rightarrow Mn^{2+}+{4H_2O}$

4. Balance hydrogen by adding $\small H^ +$ $\displaystyle \small H^ +$ ions from the acid into the half reactions.

Oxidation: $2I^-(aq)\rightarrow I_2(s)+2e^-$ $\displaystyle 2I^-(aq)\rightarrow I_2(s)+2e^-$

Reduction: $MnO^-_4+5e^-+ 8H^+\rightarrow Mn^{2+}+4H_2O$ $\displaystyle MnO^-_4+5e^-+ 8H^+\rightarrow Mn^{2+}+4H_2O$

5. Multiply each half reaction times an integer such that the electrons cancel when the equations are added. The oxidation reaction will be multiplied by five, and the reduction reaction will be multiplied by two.

Oxidation times 5: $10I^-(aq)\rightarrow 5I_2(s)+10e^-$ $\displaystyle 10I^-(aq)\rightarrow 5I_2(s)+10e^-$

Reduction times 2: $2MnO^-_4+10e^-+ 16H^+\rightarrow 2Mn^{2+}+8H_2O$ $\displaystyle 2MnO^-_4+10e^-+ 16H^+\rightarrow 2Mn^{2+}+8H_2O$

6. Add the half reactions together. The electrons will cancel from both the reactants and products.

$10I^-(aq)+2MnO_4^-(aq)+16H^+(aq)\rightarrow 2Mn^{2+}(aq)+5I_2(s)+8H_2O(l)$ $\displaystyle 10I^-(aq)+2MnO_4^-(aq)+16H^+(aq)\rightarrow 2Mn^{2+}(aq)+5I_2(s)+8H_2O(l)$

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MCAT Physical : Half Reactions and Balancing Equations

Study concepts, example questions & explanations for MCAT Physical

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Example Question #1 : Oxidation Reduction

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