Balance the following oxidation-reduction reactions that...

Balance the following oxidation-reduction reactions that occur in basic solution. a. $\mathrm{Al}(s)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{MnO}_{2}(s)+\mathrm{Al}(\mathrm{OH})_{4}^{-}(a q)$ b. $\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{Cl}^{-}(a q)+\mathrm{OCl}^{-}(a q)$ c. $\mathrm{NO}_{2}^{-}(a q)+\mathrm{Al}(s) \rightarrow \mathrm{NH}_{3}(g)+\mathrm{AlO}_{2}^{-}(a q)$

Balance the following oxidation-reduction reactions that occur in basic solution.
a. $\mathrm{Al}(s)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{MnO}_{2}(s)+\mathrm{Al}(\mathrm{OH})_{4}^{-}(a q)$
b. $\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{Cl}^{-}(a q)+\mathrm{OCl}^{-}(a q)$
c. $\mathrm{NO}_{2}^{-}(a q)+\mathrm{Al}(s) \rightarrow \mathrm{NH}_{3}(g)+\mathrm{AlO}_{2}^{-}(a q)$

Key Concepts

Oxidation-Reduction Reactions

Oxidation-reduction (redox) reactions are processes in which electrons are transferred between chemical species. In these reactions, one species loses electrons (is oxidized) while another gains electrons (is reduced). Understanding redox processes is fundamental in predicting the flow of electrons and the changes in oxidation states of the involved elements.

Half-Reaction Method

The half-reaction method is a systematic approach to balance redox reactions by separating them into two parts: one for oxidation and one for reduction. Each half-reaction is balanced individually for mass and charge, and then the electrons lost and gained are equalized when the half-reactions are combined. This method is especially useful for complex reactions in various chemical environments.

Oxidation States

Oxidation states are numbers assigned to atoms in compounds that indicate the degree of oxidation. They provide insight into how electrons are distributed in chemical species, which is critical for identifying which species undergo oxidation and which undergo reduction. Accurate assignment of oxidation states is a key step in analyzing and balancing redox reactions.

Balancing in Basic Solution

Balancing redox reactions in a basic solution requires special considerations, as it involves using water (H?O) and hydroxide ions (OH?) to balance oxygen and hydrogen atoms. This contrasts with acidic environments, where protons (H?) are used. Mastery of balancing redox equations in basic conditions is essential for accurately representing reaction stoichiometry in alkaline systems.

Transcript

00:01 This problem has three chemical reactions, redox reactions that we need to balance in basic solution.

00:05 We'll balance them first in acidic solution by writing the half reactions.

00:09 First one is aluminum going to aloh4 minus, and then we have mno4 minus going to mno2.

00:19 For the first half reaction, we have one aluminum on each side, so we're good, but we have four oxygens on the right -hand side, so we need four waters on the left -hand side.

00:27 Now to make the hydrogens balanced with eight on the left -hand side and four on the right, we need four more hydrogen ions on the right -hand side.

00:35 Then, in order to have the right -hand side have a zero total charge, we need three electrons.

00:41 The manganese are balanced in the second half -reaction.

00:44 We have four oxygens and two oxygens, so we need two more by adding water to the right -hand side, two of them.

00:50 We then introduced four hydrogen, so we'll add four hydrogen ions to the left -hand side, and then to make the left -hand side equal to zero, just like the right -hand side.

00:57 Hand side, we'll add three electrons.

01:00 When we sum these up, the three electrons will cancel straight up.

01:04 We will end up with a net amount of two waters on the left -hand side.

01:09 We have two waters here on the right, four on the left, so we've got a net of two.

01:16 And then the hydrogen ions will cancel completely, so there will not be any more hydrogen ions present.

01:22 Now if there were hydrogen ions present, then we would have to go one step further and add the same number of hydroxides to both sides as there were hydrogen ions to balance it in basic solution, which will do with some of these other problems.

01:35 So for the second one we've got cl2 going to cl minus and then cl2 going to ocl minus.

01:42 We'll balance the chlorines first in the first half reaction and then balance the charges by adding two electrons to the left hand side so that the charge is negative two on both sides.

01:53 Then we'll balance the chlorine in the second half reaction by adding a two.

01:57 We'll bounce the oxygens with water.

01:59 We have two oxygens on the right hand side, so we'll add two waters to the left -hand side...

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