According to Table 7.2, what are the equilibrium constants...

According to Table $7.2,$ what are the equilibrium constants for the following reactions at $25^{\circ} \mathrm{C} ?$ $(a) \quad H^{+}(a o)+L i(s)=L i^{+}(a o)+\frac{1}{2} H_{2}(g)$ $(\boldsymbol{b}) \quad 2 \mathbf{H}^{+}(\mathbf{a o})+\mathbf{P b}(\mathbf{s})=\mathbf{P b}^{2+}(\mathbf{a o})+\mathbf{H}_{2}(\mathbf{g})$ $(c) \quad 3 \mathrm{H}^{+}(\mathrm{ao})+\mathrm{Au}(\mathrm{s})=\mathrm{Au}^{3+}(\mathrm{ao})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{g})$

According to Table $7.2,$ what are the equilibrium constants for the following reactions at $25^{\circ} \mathrm{C} ?$
$(a) \quad H^{+}(a o)+L i(s)=L i^{+}(a o)+\frac{1}{2} H_{2}(g)$
$(\boldsymbol{b}) \quad 2 \mathbf{H}^{+}(\mathbf{a o})+\mathbf{P b}(\mathbf{s})=\mathbf{P b}^{2+}(\mathbf{a o})+\mathbf{H}_{2}(\mathbf{g})$
$(c) \quad 3 \mathrm{H}^{+}(\mathrm{ao})+\mathrm{Au}(\mathrm{s})=\mathrm{Au}^{3+}(\mathrm{ao})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{g})$

Key Concepts

Equilibrium Constant (K)

The equilibrium constant quantifies the ratio of product to reactant activities at equilibrium for a chemical reaction. In electrochemistry, it reflects the extent to which a redox reaction will proceed under standard conditions, with a high K value indicating a reaction that strongly favors the products.

Relation between Free Energy, Electrode Potentials, and Equilibrium Constant

A key connection in electrochemistry is the relationship ?G° = -nFE° = -RT ln K. This relation links the standard electrode potential (E°) of a reaction to its equilibrium constant by connecting the thermodynamic free energy change to the logarithm of K. It demonstrates how measurements of electrical potential can be used to infer the favorability of an electron-transfer reaction.

Electrochemical Reaction Stoichiometry

The stoichiometry of an electrochemical reaction, particularly the number of electrons transferred (n), plays a crucial role in applying the relationship between ?G°, E°, and K. Correctly accounting for n ensures that the calculated free energy change, and thereby the equilibrium constant, accurately represents the redox process involved.

Transcript

00:01 So in this problem we are given with a cell reaction and we are asked to find out the equilibrium constant for this cell reaction.

00:11 So to do that what we have to do first is we have to set up a relation between the equilibrium constant and the standard cell potential because the information given here if you want to find out the equilibrium constant from the information we have to set up the relation between the standard cell potential and the equilibrium constant and the relation is as follows the standard cell potential is equal to 2 .303rt over nf times log of k where k is the equilibrium constant and you notice the standard cell potential and since we are asked to find out the value in 25 degrees c as we know that at 25 degrees c we have a value for this part which is 0 .0592 so we can rearrange this to write down the or find out the equilibrium constant value and this will give us in e not over 0 .0 592 volt so now now we have to know the standard cell potential value and we have to know the value of n.

01:41 If we know these two values you can find out the equivalent constant.

01:44 So to find out the value of standard cell potential, let's write down the cell half cell equations and from there we'll be able to find out the entire cell potential.

01:55 So our anode half fraction here would be f .e2 plus neckos form it will become fe3 plus by donating one electron and the standard cell potential for this half reaction is equal to minus 0 .77 volt and then for the cathode hydration we have oxygen reacting with four hydrogen and it will require four electron to to give us two water molecule so as we see that we need four electron to reduce the oxygen and so that means we'll have to multiply the first equation by four so that we'll get four electron here so if we just write this as four so it will be four and then four here and four here and the standard cell potential for this reaction is minus 0 .77 volt so if we now just sum up these two values so this four is it this so it is 4 api so if we now add them up we'll get 4 f2 plus plus oxygen plus 4 plus and we will cut off these electrons and on the right -hand side we'll get 4 a f3 plus syn a quos plus 2h2 and then the overall cell potential will be the summation of this two so the overall cell potential is minus 0 point oh okay so so the standard cell potential for, yeah, the standard sell potential for this reaction is 1 .23 volt...

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