Balance the following oxidation–reduction reactions that...

Balance the following oxidation–reduction reactions that occur in acidic solution using the half-reaction method. a. $\mathbf{I}^{-}(a q)+\mathrm{ClO}^{-}(a q) \rightarrow \mathrm{I}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)$ b. $\mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{NO}(g)$ c. $\mathrm{Br}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{Br}_{2}(l)+\mathrm{Mn}^{2+}(a q)$ d. $\mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \rightarrow \mathrm{CH}_{2} \mathrm{O}(a q)+\mathrm{Cr}^{3+}(a q)$

Balance the following oxidation–reduction reactions that occur in acidic solution using the half-reaction method.
a. $\mathbf{I}^{-}(a q)+\mathrm{ClO}^{-}(a q) \rightarrow \mathrm{I}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)$
b. $\mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{NO}(g)$
c. $\mathrm{Br}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{Br}_{2}(l)+\mathrm{Mn}^{2+}(a q)$
d. $\mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \rightarrow \mathrm{CH}_{2} \mathrm{O}(a q)+\mathrm{Cr}^{3+}(a q)$

Key Concepts

Electron Balance

Central to redox reactions is the concept of electron balance, where the total electrons lost in the oxidation half-reaction must equal the total electrons gained in the reduction half-reaction. Balancing electrons is crucial for maintaining charge neutrality in the overall reaction.

Balancing in Acidic Solutions

When redox reactions occur in acidic solutions, auxiliary species like H? and H?O are used to balance hydrogen and oxygen atoms. This method accounts for the acidic environment and ensures that both mass and charge are conserved in the final balanced equation.

Oxidation?Reduction Reactions

These are chemical reactions in which electrons are transferred between species, leading to changes in oxidation states. The process involves one species being oxidized (losing electrons) and another being reduced (gaining electrons), highlighting the fundamental principle of electron conservation in chemical transformations.

Half-Reaction Method

This is a systematic approach for balancing redox reactions by dividing the overall reaction into two separate half-reactions: one for oxidation and one for reduction. Each half-reaction is independently balanced for atoms and charge before combining them, ensuring that electrons lost equal electrons gained.

Transcript

00:01 Let's balance the following oxidation reduction reactions that occur in acidic solution using the half reaction method.

00:07 To use the half reaction method, we have to separate into two half reactions, a reduction, half reaction, and not reactant reaction, and use the major oh method for balancing.

00:18 For a, we're going to split this up into two half reactions.

00:23 Balance the major, and we have minus three on the left, minus one on the right, add two electrons to balance the charge.

00:35 And we'll have clo minus to c l minus the major chlorine is balanced add h2o to balance the oxygen add 2h plus to balance the electrons and add two electrons to balance charge two electrons will cancel and we will get 3i minus add clo minus add 2h plus to i3 minus add cl minus and h2o and there is our balanced half reaction in acidic solution for b one half reaction be as 203 to h3 aso4 arsenic is our major atom we'll put a 2 here to balance that that gives us 2 arsenic on each side i have 8 oxygen on the right hand side 3 oxygen on the left so i need 5h2o for 8 oxygen on each side 10 hydrogen on the left 6 hydrogen on the right so i need 4h plus to balance that and to balance the charge add 4 electrons second half reaction n03 minus to n0 major nitrogen is balanced 3 oxygen on the left 1 oxygen on the right so 2h2o to balance the oxygen 4 hydrogen on the right, so i need four hydrogen on the left, and charge here, plus three, so i need three electrons to balance the charge.

02:22 Now we need to multiply by a common factor of the electrons, and i'm going to multiply the oxidation half reaction.

02:30 Common multiple is 12, so times three and times four.

02:35 Let's rewrite these here.

02:38 Multiplying by three, i get 15 each two o.

02:44 3 as203 to 6h3 aso4, 12h plus 12 electrons, and multiplying by 4 would give me 4 times 3 is 12 electrons, 16h plus 4 -10 -03 minus 4 -0 8h -20.

03:17 Cancel out the electrons 12 on each side i can simplify the hydrogen's this leaves me with 4 here symbolify the waters this leaves me with 7 here and my final balanced equation is 3 as 2 0 add 4 and 03 minus add 4h plus add 7h20 to 6 h3 as 04 and 4 n oh, and there is my balanced redox reaction.

04:01 For c, first half reaction would be br minus to br2.

04:09 Balance the major, balance the charge.

04:14 Second half reaction is mno4 minus to mn2 plus.

04:20 Mnines is balanced on both sides.

04:23 Four oxygens on the left, so 4h20 on the right, eight hydrogens on the left, so 8h plus on the right.

04:31 So left there...

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