# Video: Calculating the Equilibrium Constant for the Galvanic Cell of Tin and Copper Using the Standard Electrode Potentials

Using the standard electrode potentials in the table below, calculate to 2 significant figures the equilibrium constant at 25°C for a galvanic cell with the following overall reaction. Sn(s) + 2Cu²⁺(aq) ⇌ Sn²⁺(aq) + 2Cu⁺(aq)

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### Video Transcript

Using the standard electrode potentials in the table below, calculate to two significant figures the equilibrium constant at 25 degrees Celsius for a galvanic cell with the following overall reaction. Sn solid plus 2Cu²⁺ aqueous in equilibrium with Sn²⁺ aqueous and 2Cu⁺ aqueous.

Standard electrode potentials are measured relative to a hydrogen electrode which has a defined voltage of zero volts. For instance, the standard electrode potential for the process by which Sn²⁺ ions are turned into Sn solid is 0.1375 volts less than for the equivalent process for H⁺. We can label standard electrode potentials as 𝐸 standard red for reduction. Because standard electrode potentials are always expressed as reduction half equations.

The equivalent standard potentials for the oxidation processes have voltages that are exactly opposite to those for the reduction potential. The process we are looking to examine is the reaction of solid tin with copper two ions producing tin two ions and copper one ions. So, in the forward reaction, solid tin is being oxidized to tin two and copper two is being reduced to copper one. In a galvanic cell, the electrode where oxidation occurs is called the anode and the electrode where reduction occurs is called the cathode.

In cell notation, we put the anodic reaction on the left and the cathodic reaction on the right. So, we can start building up the cell notation for this galvanic cell. We start with the reactant that is going to be oxidized, which in this case is the solid tin. Next, we have the oxidized product, which is tin two plus. We put a line between the tin solid and the tin two plus aqueous to indicate the solid-liquid interface. And next, we put in a double line to indicate the salt bridge.

On the cathodic reaction’s side, we start with the reactant that is going to be reduced. That’s our copper two aqueous ions. Next, we have our reduced product, which is copper plus ions, which are in the same phase as our copper two plus. So, we just have a comma between them. We’re not quite done because we need an electrode to deliver electrons. I’m going to assume that we’re going to use a nonreactive platinum electrode. However, there are other options.

Now we’ve properly analyzed and described this chemical system, let’s go back to the question. Our job is to use the information in the question to figure out the equilibrium constant for this process. To figure out the equilibrium constant, we’re going to need to work out the cell potential. Once we have the cell potential, we can figure out the equilibrium constant using the Nernst equation.

This is the form of the Nernst equation we’ll need. This relates the cell potential with the equilibrium constant via the gas constant, the temperature in Kelvin, the number of electrons in the balanced equation, and the Faraday constant. So, let’s start off by figuring out the cell potential. The standard cell potential is equal to the standard oxidation potential for the reaction at the anode plus the standard of reduction potential at the cathode.

You might’ve seen this written this way, where we take away the standard of reduction potential for the anode reaction from the standard of reduction potential of the cathode reaction. These statements are essentially equivalent. For this case, the oxidation potential for the anodic reaction is positive 0.1375 volts and the standard reduction potential for the cathodic reaction is positive 0.153 volts. So, the cell potential for the reaction of tin with copper two to produce tin two and copper one is 0.2905 volts. Since it’s positive, the reaction is spontaneous.

We can store away that value for now and move on to part two. The first step is to rearrange the Nernst equation in terms of the equilibrium constant. To start with, let’s just swap around the terms so that we have the equilibrium constant on the left and the cell potential on the right. We can then multiply both sides by 𝑛𝐹 and then divide both sides by 𝑅𝑇. This leaves us with the natural logarithm of the equilibrium constant on the left. The natural logarithm is a logarithm to the base 𝑒.

If log to the base 𝑒 of 𝑥 is equal to 𝑦, then 𝑒 to the power of 𝑦 is equal to 𝑥. So, by rearranging the Nernst equation, we’ve shown that the equilibrium constant is equal to 𝑒 to the power of the number electrons multiplied by the Faraday constant multiplied by the cell potential over the gas constant multiplied by the temperature in Kelvin. To make sure we don’t make a mistake, it’s helpful to work out each of the terms separately.

The equation given to us is already balanced. We have one tin atom turning into one tin ion, and that takes two electrons. So, the number of electrons in the balanced chemical equation is two, and therefore 𝑛 equals two. 𝐹 is Faraday’s constant, which has a value of 96485.3 columbs per mole. This number is equivalent to the charge for each mole of electrons. The gas constant is equal to the energy per mole of gas particles per unit temperature. Here, I’ve chosen the value of the gas constant with energy in units of joules and temperature in units of Kelvin.

The question states the temperature in degrees Celsius. But Celsius is not an absolute temperature scale, so we need to convert Celsius to Kelvin in order to proceed. So, our temperature is equal to 25 degrees Celsius plus 273.15, which is 298.15 Kelvin. You can see that our temperature units now match up and so do our units of energy and our units of volts. Volts can otherwise be expressed as joules per columb. Now, all we need to do is plug these values into the equilibrium constant expression.

This means our equilibrium constant is equal to 𝑒 to the power of two times 96485.3 columbs per mole multiplied by 0.2905 volts divided by 8.31446 joules per mole per Kelvin multiplied by 298.15 Kelvin. If we cancel our Kelvin and our per mole units, we’re left with columbs, joules, and volts. As I’ve already mentioned, volts can be expressed as joules per columb. So, we can cancel those units and evaluate the expression. This gives us 𝑒 to the power of 22.6135, which evaluates to 6.62114 times 10 to the nine.

Such a large value for the equilibrium constant means that the reaction is highly product-favored. The question asked us to give our answer to two significant figures. So, the final value for our equilibrium constant is 6.6 times 10 to the power of nine. Now, what you could do is take this very large value for the equilibrium constant and assume that mixing tin with copper two will produce tin two and copper one in very high yield.

In actual fact, the reduction potential for the addition of two electrons to copper two is even greater than for the addition of one. This would yield a value for the cell potential greater than our 0.2905 volts, meaning that when we mix tin and copper two our expected products are tin two and copper solid. However, this question takes the reaction to produce copper one in isolation, and it is perfectly valid to calculate the equilibrium constant for that process alone. So, our answer is 6.6 times 10 to the power of nine.