Balance the following redox reactions, and calculate the...

Balance the following redox reactions, and calculate the standard-state potential and the equilibrium constant for each. Assume that the $\left[\mathrm{H}_3 \mathrm{O}^{+}\right]$is $1 \mathrm{M}$ for acidic solutions, and that the $\left[\mathrm{OH}^{-}\right]$is $1 \mathrm{M}$ for basic solutions. a. $$ \begin{aligned} & \mathrm{MnO}_4-(a q)+\mathrm{H}_2 \mathrm{SO}_3(a q) \rightleftharpoons \mathrm{Mn}^{2+}(a q)+ \\ & \mathrm{SO}_4{ }^{2-}(a q) \quad \text { (acidic solution) } \end{aligned} $$ b. $\mathrm{IO}_3^{-}-(a q)+\mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{I}_2(s) \quad$ (acidic solution) c. $\mathrm{ClO}^{-}(a q)+\mathrm{I}^{-} \rightleftharpoons \mathrm{IO}_3^{-}(a q)+\mathrm{Cl}^{-}(a q)$ (basic solution)

Balance the following redox reactions, and calculate the standard-state potential and the equilibrium constant for each. Assume that the $\left[\mathrm{H}_3 \mathrm{O}^{+}\right]$is $1 \mathrm{M}$ for acidic solutions, and that the $\left[\mathrm{OH}^{-}\right]$is $1 \mathrm{M}$ for basic solutions.
a.
$$
\begin{aligned}
& \mathrm{MnO}_4-(a q)+\mathrm{H}_2 \mathrm{SO}_3(a q) \rightleftharpoons \mathrm{Mn}^{2+}(a q)+ \\
& \mathrm{SO}_4{ }^{2-}(a q) \quad \text { (acidic solution) }
\end{aligned}
$$
b. $\mathrm{IO}_3^{-}-(a q)+\mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{I}_2(s) \quad$ (acidic solution)
c. $\mathrm{ClO}^{-}(a q)+\mathrm{I}^{-} \rightleftharpoons \mathrm{IO}_3^{-}(a q)+\mathrm{Cl}^{-}(a q)$ (basic solution)

Key Concepts

Redox Reactions

Redox (reduction-oxidation) reactions involve the transfer of electrons between chemical species. In these reactions, one species loses electrons (oxidation) while another gains electrons (reduction). Understanding the flow of electrons is fundamental for writing balanced chemical equations and analyzing energy changes in the system.

Half-Reaction Method

The half-reaction method is a systematic approach for balancing redox reactions by separating the oxidation and reduction processes. Each half-reaction is balanced individually for mass and charge, including the addition of water, hydrogen ions, or hydroxide ions as needed, depending on whether the reaction occurs in acidic or basic solution.

Standard Reduction Potentials

Standard reduction potentials represent the tendency of a chemical species to gain electrons under standard conditions. They are measured in volts and are used to predict the direction of electron flow in redox reactions. These potentials allow for the calculation of cell voltages and are critical in assessing the feasibility of redox processes.

Equilibrium Constant Calculation

The equilibrium constant of a redox reaction quantifies the ratio of product concentrations to reactant concentrations at equilibrium. It can be determined from the standard-state potential through the Nernst equation, linking the thermodynamics of the reaction with its observed equilibrium composition.

Acidic and Basic Media Considerations

The conditions of the solution, whether acidic or basic, affect the balancing of redox reactions especially through the role of H3O+ in acid and OH- in base. These ions are used to balance oxygen and hydrogen atoms in the half-reaction method, ensuring that the overall reaction is balanced in terms of both mass and charge.

Transcript

00:01 So this problem is asking us to take a good look at and how to balance redox reactions.

00:06 For acid and base, the process is very similar.

00:10 Base contains one more step than the acidic.

00:12 So i'm going to start in reverse.

00:13 I'm going to go e through a in this video.

00:16 So the first step is to identify what is being oxidized, what is being reduced.

00:21 So i have to check the oxidation states.

00:23 The oxidation state of this manganese is plus two, right? if it's just an ion by itself, the charred.

00:30 Matches the oxidation state.

00:35 For the bio3 minus, well, we know that auction is always minus 2, so minus 2 times 3 is negative 6.

00:43 So what plus negative 6 equals negative 1? well, positive 5.

00:47 So this bismuth is positive 5.

00:50 On the other side, bismith becomes 3 plus.

00:55 In manganese, we have to figure that one out.

00:58 Minus 1, this is always an oxygen always minus 2, so minus 8, well plus negative 8 equals negative 1, well, it's positive 7.

01:07 So now we have to identify oxidation reduction.

01:10 The oxidation is a loss of electrons, reduction is the gain.

01:15 And in this case, you can see that manganese had to lose electrons, which makes it oxidation.

01:30 Whereas bismith had to gain electrons.

01:34 Which makes it the reduction.

01:37 So now we have to break them into the two half reactions.

01:39 So first we're going to work with our manganese for our oxidation half reaction.

01:48 So mn plus two becomes mn of four minus.

01:54 First step is to balance the reaction for the elements.

02:03 They're not oxygen or hydrogen, so we're good there.

02:05 The next thing we do is balance oxygens using water and then we balance hydrants using h -pluses.

02:16 The last step is to balance for charge using electrons.

02:22 So in this case we have plus two on the left and on the right we have plus seven.

02:30 So we have eight plus one, so it's plus seven.

02:33 So now we have to balance by using electrons.

02:36 Since oxidation's loss, electrons should be a product.

02:41 Now for your reduction half reaction, you have bio3 minus, becoming bi3 plus.

02:51 Well, just like last time, we balanced anything other than hydrogen and oxygen, so it's okay.

02:57 Next thing we add is three waters to balance out the oxygen.

03:02 And finally, we have to add hydrogens to balance out.

03:11 The hydrogen.

03:12 The h plus is the balance of the water is the other side.

03:15 And then finally we balance for charge.

03:17 So on the right we have plus three.

03:19 On the left we have plus five.

03:20 So we have to add in two electrons.

03:25 And then we have to combine them.

03:28 But we can only combine them if the number of electrons are the same.

03:32 So we have two...

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