Зв'язок між потенціалом ОВР та константою рівноваги

In the previous lesson, we related electrical work and free energy change. We learned that redox reactions, like all chemical reactions, can be spontaneous or nonspontaneous. So, we linked a voltaic cell's voltage to the Gibbs free energy. But isn't free energy change also correlated to equilibrium? That's correct! We can solve for the equilibrium constant with the free energy change. That means we can also connect it to voltage!

Deriving the Equilibrium Constant from Voltage

Let's use our previous lessons to derive an expression that connects voltage and equilibrium.

Previously we connected free energy and voltage.

$\Delta G^{o}=-nFE^{o}$

Way back, we connected free energy and the equilibrium constant.

$\Delta G^{o}=-RT\ln{(K)}$

What happens if we set the two equations equal to each other?

$-RT\ln{(K)}=-nFE^{o}$

$\ln{(K)}=\frac{nFE^{o}}{RT}$

$K=e^{\frac{nFE^{o}}{RT}}$

• $R$ is the universal gas constant. I recommend using $R=8.3145\frac{J}{mol \times K}$.

• $T$ is the temperature. At standard conditions, it's $298.15K$.

• $n$ is the transferred electrons.

• $F$ is Faraday's constant. It's $96,485\frac{J}{V\times mols}$.

• $E^{o}$ is the cell's voltage from the standard electrode potentials.

Examples of Calculating K from Voltage

1. What is the equilibrium constant of a voltaic cell made from the transition metals, silver and zinc? We are at standard conditions.

Start by writing the two half-reactions.

$Ag^{+}+ e^{-}\rightarrow Ag_{(s)} \, E^{0}=0.80$

$Zn^{2+}+ 2e^{-} \rightarrow Zn_{(s)}\, E^{0}=-0.76$

Immediately, we notice that the moles of electrons are different. So, we need to multiply the top equation by two to match the mole ratios. But remember, voltage is an intensive property. Do you remember why? Voltage has units of $\frac{J}{C}$. That's energy over charge. But, when we double the electrons, we double both. So, it stays the same.

$2Ag^{+}+ 2e^{-}\rightarrow 2Ag_{(s)} \, E^{0}=0.80$

$Zn^{2+}+ 2e^{-} \rightarrow Zn_{(s)}\, E^{0}=-0.76$

Now, we need to find $E_{cell}^{o}$.

$E_{cell}^{o}=E_{cathode}^{o}-E_{anode}^{o}=0.80V-(-0.76V)=1.56V$

Our reaction has a very positive voltage. So, it's a spontaneous reaction. We expect the equilibrium constant to be considerable.

Now, we can calculate the equilibrium constant.

$K=e^{\frac{nFE^{o}}{RT}}=e^{\frac{2\times96,485\times1.56}{8.3145\times 298.15}}=5.47\times 10^{52}$

Our equilibrium constant is huge! Do you remember how to interpret the equilibrium constant's magnitude? Anything greater than one favors the products. Here, the product concentration is so large we would say the reaction went to completion.

1. What's the equilibrium constant between copper and silver?

Again, we should start be writing the standard reduction half-reactions. Notice, that the silver reaction needs to be doubled.

$2Ag^{+}+ 2e^{-}\rightarrow 2Ag_{(s)} \, E^{o}=0.80$

$Cu^{2+}+ 2e^{-} \rightarrow Cu_{(s)}\, E^{o}=0.34$

Now, we need to find $E_{cell}^{o}$.

$E_{cell}^{o}=E_{cathode}^{o}-E_{anode}^{o}=0.80V-0.34V=0.46V$

Now, we can calculate the equilibrium constant.

$K=e^{\frac{nFE^{o}}{RT}}=e^{\frac{2\times96,485\times0.46}{8.3145\times 298.15}}=3.56\times 10^{15}$

1. Calculate the equilibrium constant for the reaction $Cu_{(s)} + 2H^{+} \rightarrow Cu^{2+} + H_{2(g)}$.

First, we need to write the two half-reactions. We see that copper was the reducing agent. The hydrogen ions were the oxidizing agent.

$Cu_{(s)} \rightarrow Cu^{2+}+ 2e^{-}$

$2H^{+} + 2e^{-} \rightarrow H_{2(g)}$

Now, we need to find the standard reduction potentials. The copper reaction is oxidation. But, we still use the standard reduction potential for copper. Why? Because the cell voltage formula accounts for the oxidation of the anode. What about the second reaction? The conversion of hydrogen ions to hydrogen gas is the SHE. So, its voltage is zero.

$Cu_{(s)} \rightarrow Cu^{2+}+ 2e^{-}\, E^{o}=0.34$

$2H^{+} + 2e^{-} \rightarrow H_{2(g)}\, E^{o}=0.00$

We can find the cell's voltage.

$E_{cell}^{o}=E_{cathode}^{o}-E_{anode}^{o}=0.00V-0.34V=-0.34V$

Now, we can solve for the equilibrium constant.

$K=e^{\frac{nFE^{o}}{RT}}=e^{\frac{2\times96,485\times-0.34}{8.3145\times 298.15}}=3.20\times 10^{-12}$

For a nonspontaneous reaction, we expect a minuscule equilibrium constant.

(Video) 19.6 Cell Potential, Delta G, and the Equilibrium Constant | General Chemistry

by Chad's Prep

In this video, Chad explains the connection between voltage, Gibbs free energy, and the equilibrium constant. With one value, you can solve for the others. Chad walks you through an example problem step-by-step. So, you'll learn how to do each of the calculations.