For the following half-reaction, ℰ^∘=-2.07 V : AlF6^3-(a q)+3 e^- ⟶Al(s)+6 F^-(a q) Using data f

step 1 So we are asked to find K for this equation here. And we're given an equation that it's a reduct

For the following half-reaction, ℰ^∘=-2.07 V : AlF6^3-(a...

For the following half-reaction, $\mathscr{E}^{\circ}=-2.07 \mathrm{V} :$ $$\mathrm{AlF}_{6}^{3-}(a q)+3 e^{-} \longrightarrow \mathrm{Al}(s)+6 \mathrm{F}^{-}(a q)$$ Using data from Table $18.1,$ calculate the equilibrium constant at $25^{\circ} \mathrm{C}$ for the reaction $$\mathrm{Al}^{3+}(a q)+6 \mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{AlF}_{6}^{3-}(a q) \quad K=?$$

For the following half-reaction, $\mathscr{E}^{\circ}=-2.07 \mathrm{V} :$
$$\mathrm{AlF}_{6}^{3-}(a q)+3 e^{-} \longrightarrow \mathrm{Al}(s)+6 \mathrm{F}^{-}(a q)$$
Using data from Table $18.1,$ calculate the equilibrium constant at $25^{\circ} \mathrm{C}$ for the reaction
$$\mathrm{Al}^{3+}(a q)+6 \mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{AlF}_{6}^{3-}(a q) \quad K=?$$

Key Concepts

Standard Electrode Potential

This is a measure of the intrinsic tendency of a chemical species to be reduced under standard conditions (typically 1 M concentration, 1 atm pressure, and a specified temperature such as 25°C). It provides a reference for comparing different redox reactions and is central to predicting the direction of electron flow in electrochemical cells.

Gibbs Free Energy Change

The Gibbs free energy change (?G°) quantifies the spontaneity of a reaction under standard conditions and is directly related to the standard electrode potential by the equation ?G° = –nFE°. This connection bridges electrochemistry and thermodynamics by linking the electrical work obtainable from a redox reaction with its chemical energy change.

Equilibrium Constant

The equilibrium constant (K) defines the ratio of the concentrations of products to reactants at equilibrium in a chemical reaction. It is fundamentally connected to the standard Gibbs free energy change via the relationship ?G° = –RT ln K, allowing one to determine the extent to which a reaction will proceed under standard conditions.

Nernst Equation

The Nernst equation relates the potential of an electrochemical cell at non?standard conditions to the standard electrode potential, temperature, and the activities (or concentrations) of the reacting species. It is an essential tool for understanding how deviation from standard conditions affects the cell potential and thus the position of equilibrium in redox reactions.

Transcript

00:01 So we are asked to find k for this equation here.

00:06 And we're given an equation that it's a reduction equation, and we're given the standard reduction potential.

00:14 So how are we going to find k for one equation when we're given data about another? well, if you have a redox situation, you can always add the reduction and oxidation, potentials together, and that will give you e0 for the overall reaction.

00:39 So if there's a reaction that we could add to this one, add things, add it together, cancel out things around both sides, and then add the reduction in oxidation potentials, we could get e0 for the reaction, and from there, we can get k.

00:56 And that's where we use delta g, which just kind of lives in the middle between those things, allows us to calculate back and forth.

01:04 So if we look at table 18 .1, we need to do something to get aluminum 3 plus in the equation because it's missing from there's no aluminum 3 plus in this equation.

01:19 So we need to find something with aluminum 3 plus.

01:21 Well, in table 18 .1, we have the equation and potential for the reduction of aluminum 3 plus.

01:31 So now, if we add these two reactions together as is, things are not on the right side.

01:38 So what we could do is flip the first equation around and turn it into oxidation.

01:44 When we do that, our e0 flips, so that would be, it flips the signs, that would be 2 .07.

01:52 So if we add these together, things cancel out.

01:56 The electrons cancel the aluminum cancels.

02:00 And we're left with the equation we need to find k4, which is great.

02:05 So what we just do is add the oxidation.

02:08 So this has become e oxidation and this is e reduction.

02:13 So when we add them together, we get 0 .41 volts.

02:19 And so now we can use e0 to get delta g...

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