Electrochemistry is the study of reduction and oxidation (redox) reactions. Redox reactions occur when electrons are transferred between reactants and products.
Oxidation numbers are used to compare atomic charges on the reactants and products side of a reaction to see which elements or species are oxidized and which are reduced. Though most elements have oxidation numbers that change, the following rules can usually be followed:
Reduction is a gain of electrons from the reactants to products side. This means that the charge of the ion decreases. Reduction occurs at the cathode of an electrochemical cell, and a reduction half reaction shows electrons added on the reactants side. Remember the acronyms OIL RIG (oxidation is loss, reduction is gain) or LEO the lion says GER (Loss of electrons is oxidation, gain of electrons is reduction)
Oxidation is a loss of electrons from the reactants to products side. This means that the charge of the ion increases. Oxidation occurs at the anode of an electrochemical cell, and an oxidation half reaction shows electrons added on the products side.
Any redox reaction can be separated into its reduction and oxidation half reaction. To begin this process, identify the oxidation numbers of all the ions in the reaction to determine which ions are oxidized and which are reduced. Remember that species that are oxidized have an increase in oxidation number, while species that are reduced have a reduction in oxidation number.
Next, separate the species that are reduced and oxidized into two separate reactions. In the reduction reaction, add electrons to the reactants side to show that electrons are gained. The number of electrons added should be consistent with the change in oxidation number. Do the same for the species that are oxidized, but add electrons to the products side to show that electrons are lost. Again, the number of electrons added should be consistent with the change in oxidation number.
Once both reactions are set up, multiply either of the reactions by any coefficient to make the total number of electrons exchanged equal. To check your work, make sure that when the reactions are added the electrons cancel from both sides and what remains is the original equation.
Sometimes, a net ionic redox reaction will show a reactant (usually oxygen) that does not appear on the products side. This is not a mistake, although it might appear to be. This means that the reaction takes place in acidic conditions and water is a product, and the water molecules are left out of the equation. To balance a reaction in acidic conditions, start as you would with any other redox problem: determine oxidation numbers and separate the reduction and oxidation parts of the reaction. Now, since oxygen is on the reactants side and not the products side, add water molecules to the products side of the equation where oxygen is a reactant (usually the reduction equation). Add as many water molecules are needed to balance out the oxygens. Then, to balance the hydrogen atoms that come with the water molecules, add as many H+ ions as are needed to the reactants side. The other reaction stays as it is. To check your work, add the two equations together and make sure all the charges and elements are balanced.
Balancing in basic conditions is similar to balancing in acidic conditions and is most often required when oxygen is a product but not a reactant. Start with the same process as balancing in acidic conditions. Usually, to balance the equation water molecules will be added on the reactants side and H+ ions will be added to the products side of the oxidation equation. After balancing as you would in acidic conditions, add as many OH- ions as H+ ions are present to both sides of the equation. This will cause all of the H+ to turn into water which will cancel out with some of the water molecules from the other side of the equation. You are left with OH- on one side and H2O on the other. To check your work, add the two equations together and make sure all the charges and elements are balanced.
As shown in the image above, a Galvanic cell consists of six major components: two electrodes, two solutions, a salt bridge, and a wire connection between the two electrodes. A voltmeter can also be added to measure the voltage between the electrodes. The solutions in the cell contain the ion of the metal in the electrode. The reaction in a Galvanic cell results in electrons leaving the anode (remember anode/away) and moving through the wire towards the cathode. As electrons leave the anode, metal ions come off the electrode and enter the solution, making it more concentrated. When the electrons reach the cathode, ions come out of solution and become uncharged solid on the electrode. This means that the anode is corroded (loses mass) while the cathode is plated (gains mass). Meanwhile, all the electrons on the cathode solution attract cations in the salt bridge, while the lack of electrons near the anode attracts anions. Which electrode is the cathode or anode can be determined by seeing which metal is reduced and which is oxidized. The shorthand form of an electrochemical cell is anode/solution//solution/cathode. For the cell above, the shorthand is Zn/Zn2+//Cu2+/Cu.
Note that under standard conditions, the half reaction with the smaller reduction potential will be reduced and the other half reaction will become the oxidation half reaction (more on reduction potentials below).
Concentration cells are another form of electrochemical cell that use two of the same electrodes and different concentrations of solutions. The cathode of the concentration cell has the higher cation concentration in solution so electrons flow towards it from the anode.
All redox reactions have a standard reduction potential, notated by E, which can be looked up in a table of values. These values correspond to standard conditions of 25 degrees celsius and 1 molar solutions in the electrochemical cell and are used to determine the electromotive force that a cell produces. To find overall Ecell, use either Ered(cathode)-Ered(anode) or Ered(cathode)+Eox(anode). The signs can become very confusing in these questions because the reduction potential needs to be negated when it is used as the potential for the oxidation reaction. In other words, if using two values straight out of a table, subtract the value of the reduction potential being used for the oxidation reaction from the reduction reaction potential. Or, if you have already flipped the sign of the reduction potential to use as the oxidation potential, you can add it to the reduction potential of the reduction reaction.
A spontaneous redox reaction will have a positive overall Ecell. Under standard conditions, Ecell can be related to ΔG (Gibbs Free Energy) through the equation ΔG0=-nFE0 (the zeros being pronounced G naught and E naught to show standard conditions) where n is the number of moles of electrons transferred and F is Faraday’s constant: 96495 coulombs/mole. Since a spontaneous redox reaction will have a positive overall Ecell, this corresponds to the negative ΔG of a spontaneous reaction
Under nonstandard conditions, the Nernst Equation must be used to find ΔG. Given ΔG = ΔG0 + RTlnQ and substituting the above equation for ΔG0, we find that Ecell = E0 - (RT/nF)lnQ where E0 is standard cell potential, R is the gas constant (8.314 J/Mol K), T is temperature is Kelvin, n is the number of moles of electrons transferred, F is Faraday’s constant (96495 coulombs/mole) and Q is the equilibrium comparison Q.
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