Electrochemistry BasicsTable of contents
As the name suggests, electrochemistry is the study of changes that cause electrons to move. This movement of electrons is called electricity. In electrochemistry, electricity can be generated by movements of electrons from one element to another in a reaction known as a redox reaction or oxidation-reduction reaction.
IntroductionA redox reaction is a reaction which involves a change in oxidation state of one or more elements. When a substance loses its electron, its oxidation state increases, thus it is oxidized. When a substance gains an electron, its oxidation state decreases thus being reduced. For example, for the redox reaction: \[ H_2 + F_2 → 2 HF \] Rewrite it as Oxidation: \( H_2 \rightarrow 2H^+ + 2e^- \) Reduction: \(F_2 + 2e^- \rightarrow 2F^- \) Overall Reaction : \( H_2 + F_2 \rightarrow 2H^+ + 2F^- \) Oxidation reaction lose electrons thus they are being oxidized, while Reduction is gaining electrons thus they are being reduced. In this case, H2 is being oxidized, while F2 is being reduced. OIL RIG Oxidation Is Losing electrons Reduction Is Gaining electrons Practice problemGiven the redox reaction : \( Fe^{3+} + V^{2+}→ Fe^{2+} + V^{3+}\), which is being oxidized and reduced? Which is the reducing agent and oxidizing agent? Answer keyFe3+ is being reduced in to Fe2+, and V2+ is being oxidized into V3+. This is because the oxidized species loses electrons, and the reduced species gains electrons. Iron gains an electron (Fe3+→ Fe2+), and vanadium lose an electron (V2+→ V3+). Thus, Fe3+ is the oxidizing agent because it takes electrons and V2+ is the reducing agent because it loses electrons. Voltaic Cells-Galvanic CellsIn 1793, Alessandro Volta discovered that electricity could be produced by placing different metals on the opposite sides of a wet paper or cloth. He made his first battery by placing Ag and Zn on the opposite sides of a moistened cloth with salt or weak acid solution. Therefore, these batteries got the name Voltaic Cells. Voltaic (Galvanic) Cells are electrochemical cells which contain a spontaneous reaction, and always have a positive voltage. The electrical energy released during the reaction can be used to do work. A voltaic cell consists of two compartments called half-cells . The half-cell where oxidation occurs is called anode. The other half-cell where reduction occurs is called cathode. The Electrons in voltaic cells flow from the negative electrode to the positive electrode, which flows from anode to cathode.(See, figure below) (Note: electrodes=where the oxidation/reduction reactions occur). Red Cat and An Ox Reduction Cathode and Anode Oxidation is a helful way to remember which reaction occurs on which half cell. In order for oxidation-reduction reaction to occur, the two substances in each respective half-cell are connected by closed circuit so that electrons can flow from the reducing agent to the oxidizing agent. It also needs a salt bridge so that reaction can keep proceeding. Galvanic Cell Needs electrical current from outside for it to work:
Notice how there is a voltmeter. Galvanic cell take energy sources from outside to make itself work. The figure above shows that Zn(s) is continuously oxidized, producing Zn2+ into the solution. Zn(s)--> Zn2+(aq)+2e-. Conversely, in the cathode, Cu2+ is reduced and continuously deposits on to the Cu(s) metal bar Cu2+(aq)+2e--->Cu(s). As a result, the solution containing Zn(s) becomes more positively charged while the solution containing Cu(s) becomes more negatively charged. In order for the voltaic cell to work, the solutions in the two half-cells must remain electrically neutral. Therefore, a salt bridge containing KNO3 is added to keep the solutions neutral by adding NO3-, an anion into the anode solution and K+, a cation into the cathode solution. As oxidation and reduction proceed, ions from the salt bridge migrate to neutralize charge in the cell compartments. The diagram of this electrical cell : Zn(s) │Zn2+(aq) ║ Cu2+(aq) │ Cu(s) -use ║ to separate anode(-) and cathode(+) and so represents the salt bridge in between. -use │to separate different states of matter on the same side -use , to separate same states of matter on the same side. Example: │Fe2+(aq),Fe3+(aq)║Ag+(aq) │ Ag(s) - The anode (where the oxidation occurs) is placed on the left side of the ║ - The cathode (where reduction occurs) is placed on the right side of the ║ Practice problem: Write out the Electrical cell diagram for this reaction: Cu(s) + 2Ag+(aq) →Cu2+(aq) + 2Ag(s) Answer key: Cu(s) │Cu2+(aq) ║ Ag+(aq) │ Ag(s)
Practice problem: Write cell reactions for this cell diagram. Al(s) │Al3+(aq) ║ Sn2+(aq) │ Sn(s) Answer key: Oxidation: {Al(s) → Al3+(aq) +3e-} x 2 Reduction: {Sn2+(aq) +2e- → Sn(s)} x 3 Net: 2Al(s) + 3Sn2+(aq) → 2Al3+(aq) + 3Sn(s)
Cell PotentialThe oxidation of Zn(s) into Zn2+ and the reduction of Cu2+ to Cu(s) occur spontaneously. In other words, the redox reaction between Zn and Cu2+ is spontaneous. This is caused by the difference in potential energy between the two substances. The difference in potential energy between the anode and cathode dictates the direction of electrons movement. Electrons move from area of higher potential energy to area of lower potential energy. In this case, the anode has a higher potential energy so electrons move from anode to cathode. The potential difference between the two electrodes is measured in units of volts. One volt (V) is the potential difference necessary to generate a charge of 1 coulomb (C) from 1 Joule (J) of energy.
For a voltaic cell, this potential difference is called the cell potential, and is denoted Ecell. For a spontaneous reaction, Ecell is positive and ΔG (Gibbs free energy that can be used to determine if a reaction occurs spontaneously) is negative. Thus, when ΔG is negative the reaction is spontaneous. By merging electrochemistry with thermodynamics we get this formula: ΔG = −n× F × Ecell(or EMF). EMF= Electromotive Force. Cell potential is different for each voltaic cell, its value depends upon the concentrations of specific reactants and products as well as temperature of the reaction. For standard cell potential, temperature of the reaction is assumed to be at 25o Celsius, the concentration of the reactants and products is 1M and reaction occurs at 1 atm pressure. The standard cell potential is denoted Eocell. The Practice Problem: For the following redox reaction, calculate Eocell 2Al(s) + 3Sn2+(aq) → 2Al3+(aq) + 3Sn(s) Answer: Oxidation:{Al(s) → Al3+(aq) +3e-} x 2 -Eo = +1.676V Reduction:{Sn2+(aq) +2e- → Sn(s)} x 3 Eo = -0.137V Net:2Al(s) + 3Sn2+(aq) → 2Al3+(aq) + 3Sn(s) Eocell = -0.137V - (-1.676V) Eocell = +1.539 V.
Since the values of the standard potential are given in the form of standard reduction potential for each half-reaction, in order to calculate the standard cell potential Eocell, we need to find out which substance is being oxidized, then subtract the standard reduction potential of the oxidation reaction from the standard reduction potential of the reducing reaction. For example: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) From the equation above, we know that Zn is being oxidized, and Cu is being reduced. For the redox reaction above, the potential for the two half reaction is given in the reduction form: Zn(s) → Zn2+(aq) + e- Cu2+(aq) + e- → Cu(s) By looking at it's cell potential we can know which one is anode and cathode. Cathode has a more positive potential energy, thus Cu is the cathode Zn is the anode. In order to calculate Eocell, subtract Zn of the oxdized species from the Cu of the reduced species, which is Eocell =(cathode)-E(Anode) (Note: Keep Cell potential in reduction form) Oxidation: Zn(s) → Zn2+(aq) + e- - Eo= +0.763V Reduction: Cu2+(aq) + e- → Cu(s) Eo= +0.340V Or, you can rewrite the anode half reaction into it's oxidation form, then add to the reduction half reaction of cathode to find Eocell: Oxidation: Zn2+(aq) + e- → Zn(s) Eo= -0.763V Reduction: Cu2+(aq) + e- → Cu(s) Eo= +0.340V Net: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) Since standard reduction potential have intensive properties, so changing the stoichiometric coefficient in a half reaction does not affect the value of the standard potential. For example, Oxidation:{Al(s) → Al3+(aq) +3e-} x 2 is still Eo= -1.676 Reduction:{Sn2+(aq) +2e- → Sn(s)} x 3 is still Eo= -0.137 When mulitplying the stoichiometric coefficient by 2, the standard potential does not change: Practice ProblemGiven the redox reaction Fe3+ + V2+ → Fe2+ + V3+ calculate Answer Key: Balancing Redox Reactions
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