From the skeleton equations below, create a list of balanced...

From the skeleton equations below, create a list of balanced half-reactions in which the strongest oxidizing agent is on top and the weakest is on the bottom: $$ \begin{array}{c}{\mathrm{U}^{3+}(a q)+\mathrm{Cr}^{3+}(a q) \longrightarrow \mathrm{Cr}^{2+}(a q)+\mathrm{U}^{4+}(a q)} \\ {\mathrm{Fe}(s)+\mathrm{Sn}^{2+}(a q) \longrightarrow \operatorname{Sn}(s)+\mathrm{Fe}^{2+}(a q)} \\ {\mathrm{Fe}(s)+\mathrm{U}^{4+}(a q) \longrightarrow \text { no reaction }}\end{array} $$ $$ \begin{array}{c}{\mathrm{Cr}^{3+}(a q)+\mathrm{Fe}(s) \longrightarrow \mathrm{Cr}^{2+}(a q)+\mathrm{Fe}^{2+}(a q)} \\ {\mathrm{Cr}^{2+}(a q)+\mathrm{Sn}^{2+}(a q) \longrightarrow \mathrm{Sn}(s)+\mathrm{Cr}^{3+}(a q)}\end{array} $$

From the skeleton equations below, create a list of balanced half-reactions in which the strongest oxidizing agent is on top and the weakest is on the bottom:
$$
\begin{array}{c}{\mathrm{U}^{3+}(a q)+\mathrm{Cr}^{3+}(a q) \longrightarrow \mathrm{Cr}^{2+}(a q)+\mathrm{U}^{4+}(a q)} \\ {\mathrm{Fe}(s)+\mathrm{Sn}^{2+}(a q) \longrightarrow \operatorname{Sn}(s)+\mathrm{Fe}^{2+}(a q)} \\ {\mathrm{Fe}(s)+\mathrm{U}^{4+}(a q) \longrightarrow \text { no reaction }}\end{array}
$$
$$
\begin{array}{c}{\mathrm{Cr}^{3+}(a q)+\mathrm{Fe}(s) \longrightarrow \mathrm{Cr}^{2+}(a q)+\mathrm{Fe}^{2+}(a q)} \\ {\mathrm{Cr}^{2+}(a q)+\mathrm{Sn}^{2+}(a q) \longrightarrow \mathrm{Sn}(s)+\mathrm{Cr}^{3+}(a q)}\end{array}
$$

Key Concepts

Redox Reactions

Redox reactions involve the transfer of electrons between chemical species. In these reactions, one species is oxidized (loses electrons) while another is reduced (gains electrons). Understanding redox processes is fundamental in various aspects of chemistry, including energy production, corrosion, and biochemical processes.

Half-Reactions

Half-reactions are a method for breaking down redox reactions into two parts: one representing oxidation and the other representing reduction. Each half-reaction is balanced separately for mass and charge, including the electrons lost or gained. This approach simplifies the process of balancing complex redox equations and is critical for electrochemical analysis.

Oxidizing and Reducing Agents

An oxidizing agent is a substance that gains electrons (is reduced) during a redox reaction, while a reducing agent loses electrons (is oxidized). Identifying these agents is essential for predicting the direction of electron flow in a reaction and for understanding the relative strengths of different substances as oxidants or reductants.

Standard Electrode Potentials

Standard electrode potentials provide a quantitative measure of the tendency of a chemical species to be reduced. They are used to rank substances based on their oxidizing and reducing strengths. A higher (more positive) electrode potential indicates a stronger oxidizing agent, which is a key concept when arranging half-reactions in order of oxidizing ability.

Transcript

00:01 Question 137 is a ksp calculation of the insoluble salt calcium oxalate.

00:09 Calcium oxalate, when it reacts with water, the monohydrate, will dissociate into calcium ion and oxalate ion according to its ksp, which will be the product of the calcium ion multiplied by the oxalate anion, which is equal to the 2 .3 times 10 to the negative 9, which is what is provided in this problem.

00:31 In order to get our oxalate, it's going to come from oxalic acid.

00:39 Oxalic acid can react with water and produce h .c2 -04 minus 1 plus the hydronium ion, according to k -a -1.

00:54 Then this can react with water to produce just the oxalate anion according to k -a -2.

01:03 And we can look these values up in the back of the book.

01:06 Ka1 is 5 .6 times 10 to the negative 2, and ka2 is 5 .4 times 10 to the negative 5.

01:14 So in order to understand all of the processes that are occurring, we need to sum up all of the chemical reactions and get an overall chemical reaction.

01:23 If we were to reverse this so that we have the oxalate solid, the product side, because for this problem, we are considering the formation of the solid calcium oxalate as a kidney stone.

01:37 So we want this as a product.

01:40 If we reverse this one and then sum up everything, we'll see that the oxalate anions will cancel and the hydrogen oxalate anions will cancel, and we'll end up with this overall reaction.

01:54 It will have its own equilibrium constant, which will be the product of all of these equilibrium constants, recognizing that because we reverse this one, we now need to take the reciprocal of this value here.

02:06 So this k value will be equal to these two values multiplied by the reciprocal of this value, or 1 ,315.

02:17 In order to determine whether or not this precipitate will form, we need to calculate a q value.

02:28 The q value will be equal to the same ratio that defines the k value.

02:32 It's going to be the product concentrations, which is h3o plus squared.

02:39 This is a solid, so it's not part of the q expression or the equilibrium constant expression.

02:44 Divided by this concentration, whoops, that's water, so we exclude that, divided by this concentration and this concentration.

02:55 The hydronium ion concentration we can determine from the ph.

03:00 It wants to know first whether or not the precipitate calcium oxalate will form at a ph of 5 .5.

03:08 So the hydronium concentration is simply equal to 10 to the negative 5 .5, which we will square...

Please give Ace some feedback