Electrochemical cells are devices that produce electrical energy from a chemical reaction. These cells consist of two electrodes, an electrolyte, and an external electrical circuit.
Electrochemical cells can be either primary (pile) or secondary (battery). In a primary cell, the chemical reactions are not reversible, thus no electrical energy is produced. In a secondary cell, the chemical reaction is reversible, producing electrical energy as it switches between oxidation and reduction states.
The simplest electrochemical cell consists of aqueous solution (water based) with two copper wires acting as electrodes. When connected to an external circuit, this cell will produce a small voltage and negligible current. This is due to the poor separation of charges on the wires and in the liquid solution.
This article will discuss how to determine the internal voltahe in an electrochemical cell that produces no external voltage.
Calculate the equilibrium concentration
In order to calculate the equilibrium concentration, you need to know the electrical force, the equilibrium constant, and the concentrations of reactants and products.
In this case, you have the electrical force (electrochemical cell voltage), the equilibrium constant (overall cell voltage), and concentrations of reactants (cell oxidant and cell reductant).
You need to find the concentration of products in order to calculate what they will be at equilibrium. Since you know what Q is at equilibrium, you can use that value!
At equilibrium, Q = 0 so 0/0 = 1 which means that there is no change in concentration or amount of product formed. This is because at equilibrium, there is a balance between formation and breakdown.
There is no difference in concentration or amount of product formed at equilibrium.
Determine the voltage of the reaction
The voltage of the reaction is determined by the electron transfer between the anode and the cathode reactants. The difference in electron charge between anodic and cathodic reactants determines the voltage of the reaction.
More electron rich reactants will have a lower voltage requirement to achieve an oxidation-reduction reaction. In other words, less electronegative atoms will require a higher voltage to be reduced.
This is because there are more electrons being transferred to them from more electronegative atoms. They will have a higher electric potential as a result.
Ions can affect the cell voltage as well. If there are ions present in the solution, they may need to be taken into account when calculating cell voltage. They can either increase or decrease the cell voltage depending on their charge.
Determine the current of the reaction
Once you have the voltage and equilibrium constant, you can determine the current of the reaction. The current is how many electrons pass through a point in a particular direction per second.
Electrochemical cells use Faraday’s law to determine the current of the reaction. Faraday’s law states that if there is a change in potential energy, then there is a corresponding change in current.
In an electrochemical cell, there is a difference in potential energy between the half-cells, which means that there is a current due to this difference.
By using Faraday’s law with Q = 0 and K = 0, you can conclude that Ecell = 0. There is no potential energy change between the reactants and products, so there is no current.
Interpret your results
Now that you have your results, it is time to interpret them. How much energy is required to charge the battery and what is the potential difference of the battery?
Energy required to charge the battery is determined by Q, which was measured in this experiment. The higher the Q value, the longer it will take to charge the battery.
Potential difference of the battery is determined by Ecell, which was calculated in this experiment. The higher Ecell, the stronger the battery!
This experiment also calculated K, or how easily electrons move between atoms. The lower K indicates a harder time moving electrons, making it harder to charge the battery.
These three parameters can all influence one another, making this experiment relevant and important for further research.
What assumptions did you make?
You assumed that the two half-reactions were balanced, which is a very important assumption to make. If one of the half-reactions was not balanced, then the voltage of the cell would be different.
You also assumed that the electron transfer was instantaneous, or very fast. If this were not the case, then there would be additional time where the cell was not producing a voltage and therefore the cell voltage would be lower.
You assumed that the concentrations of all molecules in each solution were equal, which is obviously not true but was an assumption that you had to make in order to solve for Ecell and E∘cell. If one solution had a higher concentration than the other, then there would be a difference in voltage between the solutions due to what is called membrane polarization.
Leave a Reply