A disproportionation reaction involves a substance that acts...

A disproportionation reaction involves a substance that acts as both an oxidizing and a reducing agent, producing higher and lower oxidation states of the same element in the products. Which of the following disproportionation reactions are spontaneous under standard conditions? Calculate $\Delta G^{\circ}$ and $K$ at $25^{\circ} \mathrm{C}$ for those reactions that are spontaneous under standard conditions. a. $2 \mathrm{Cu}^{+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{Cu}(s)$ b. $3 \mathrm{Fe}^{2+}(a q) \longrightarrow 2 \mathrm{Fe}^{3+}(a q)+\mathrm{Fe}(s)$ c. $\mathrm{HClO}_{2}(a q) \longrightarrow \mathrm{ClO}_{3}^{-}(a q)+\mathrm{HClO}(a q) \quad$ (unbalanced) Use the half-reactions: $\mathrm{ClO}_{3}^{-}+3 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}_{2}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.21 \mathrm{V}$ $\mathrm{HClO}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.65 \mathrm{V}$

A disproportionation reaction involves a substance that acts as both an oxidizing and a reducing agent, producing higher and lower oxidation states of the same element in the products. Which of the following disproportionation reactions are spontaneous under standard conditions? Calculate $\Delta G^{\circ}$ and $K$ at $25^{\circ} \mathrm{C}$ for those reactions that are spontaneous under standard conditions.
a. $2 \mathrm{Cu}^{+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{Cu}(s)$
b. $3 \mathrm{Fe}^{2+}(a q) \longrightarrow 2 \mathrm{Fe}^{3+}(a q)+\mathrm{Fe}(s)$
c. $\mathrm{HClO}_{2}(a q) \longrightarrow \mathrm{ClO}_{3}^{-}(a q)+\mathrm{HClO}(a q) \quad$ (unbalanced)
Use the half-reactions:
$\mathrm{ClO}_{3}^{-}+3 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}_{2}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.21 \mathrm{V}$
$\mathrm{HClO}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.65 \mathrm{V}$

Key Concepts

Standard Conditions

Standard conditions (usually 25°C, 1 atm pressure, and 1 M concentrations for solutions) provide a consistent reference point for comparing reaction spontaneity and electrochemical potentials. They enable the use of tabulated standard reduction potentials and facilitate the calculation of ?G° and K.

Equilibrium Constant and its Relationship to ?G°

The equilibrium constant (K) for a reaction is linked to the Gibbs free energy change via the equation ?G° = -RT lnK. A negative ?G° indicates a large equilibrium constant, meaning the reaction strongly favors product formation at equilibrium.

Gibbs Free Energy and its Relationship to Electrochemical Potential

The standard Gibbs free energy change (?G°) is directly related to the standard cell potential (E°cell) by the equation ?G° = -nFE°cell, where n is the number of electrons transferred and F is the Faraday constant. This relationship helps determine whether a reaction is spontaneous (?G° < 0) under standard conditions.

Half-Reaction Balancing

The half-reaction method involves splitting the overall redox reaction into separate oxidation and reduction processes. Each half-reaction is balanced for mass and charge, and then they are combined to yield the overall balanced reaction. This approach simplifies the analysis and calculation of reaction parameters.

Disproportionation Reaction

A disproportionation reaction is a type of redox reaction in which a single element undergoes both oxidation and reduction, forming two different products with higher and lower oxidation states. This concept is central to understanding how one species can act simultaneously as an oxidant and a reductant.

Standard Reduction Potential

Standard reduction potentials, given under standard conditions, quantify the tendency of a species to gain electrons and be reduced. These potentials are used to determine the overall cell potential when the half-reactions are combined, thereby indicating the driving force behind the reaction.

Transcript

00:01 So the goal with these problems is to calculate delta g if they, if the reaction depicted is spontaneous.

00:09 And so in order to get at if something is spontaneous or not, we need to look for the standard cell potential.

00:18 And so spontaneity would be positive.

00:20 So that's the key to getting started here because we don't actually calculate delta g or k according to the problem if it's not spontaneous.

00:28 So we have to look in table 18 .1 for the half reactions that would give us the equation that's given in the problem.

00:38 So we need to find the half reactions that have copper plus and copper 2 plus.

00:43 And so here they are in the table written in red.

00:47 And so this is a table of reduction reaction.

00:50 So we need to rearrange the equations because something's going to be reduced and something's going to be oxidized.

00:56 Well, if we flip equation one, that would put cu 2 plus on the right, which is exactly what we need.

01:03 So when we flip those equations, we get everything on the correct side.

01:08 So that's good.

01:09 Now, we need to balance everything because that will give us the moles of electrons.

01:14 And so in order to do that, we need to multiply the second equation by two.

01:20 Now, multiplying by two doesn't do anything to the cell potential.

01:24 It just allows you to figure out the moles of electrons.

01:27 But really we should take a moment just to make sure that this is spontaneous.

01:33 So because we flipped equation one, this cell potential becomes negative.

01:38 And so this, if we look in this purple part down here, this would now be 0 .52 minus 0 .34, which is a positive number.

01:49 So this is, in fact, spontaneous.

01:52 So now that we've balanced the equation, we see that the moles of electrons is two.

02:00 We know this is spontaneous, so we're going to move over to calculating delta g.

02:06 So that would be delta g equals negative nf e0.

02:11 When i plug in the values, two moles of electrons, faraday's constant, and then the e0, which is 0 .18 joules per coulum.

02:22 Remember that a volt is a jewel per coulone.

02:25 So the units cancel out very nicely moles.

02:29 Coolums and we're left with joules.

02:31 Delta g often expressed in kilojoules.

02:33 So when you do this math, you get negative 34 .7 kilojoules.

02:37 Now to calculate k, we can use delta g equals negative rtlnk.

02:44 We plug in the numbers and we end up with lnk being, excuse me, being equal to 14.

02:51 Take e, raise it to both sides.

02:53 Cancel out that natural log and we're left with 1 .21 times 10 to the 6.

03:01 Now for part b, we need to do the same thing with table 18 .1, finding the equations that when combined give us the equation in the problem.

03:15 So we have fe2 plus and fe3 plus...

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