5.3 REDOX Reactions Half-reactions from Full Redox Equations If a piece of magnesium is placed in an aqueous solution of copper (II) sulfate, the magnesium displaces the copper in a single displacement reaction. This reaction is represented by the following equation: Mg (s) + CuSO 4(aq) Cu (s) + MgSO 4(aq) Rewriting the equation to show the dissociated ions gives you: The sulfate ion, SO4 2-, is a spectator ion not involved in the chemical reaction. Cancelling out the sulfate ion gives you the net ionic equation for the reaction: Looking at the net ionic equation you can see the magnesium atom loses electrons to form a magnesium ion, while the copper ion gains electrons to form a copper atom. In this reaction the magnesium atom is oxidized because it lost electrons, while the copper (II) ion gained electrons and was reduced. Since oxidation and reduction both occur, the reaction is known as an oxidationreduction, or redox reaction. A half-reaction refers to either a reduction reaction or an oxidation reaction. It can be found from the net ionic equation by considering just one species and using the appropriate number of electrons to balance the charges on both sides of the reaction arrow. Half-reactions are most useful since in many devices each component or compartment is the site for just one half-reactions. Reduction Potential
The reduction potential is an indication of the spontaneous direction a redox reaction will take. Its symbol is E o. The higher the E o, the more the reaction will tend to orient itself spontaneously towards the right, favouring a reduction reaction. The lower the E o (the more negative), the more the reaction will tend to orient itself spontaneously towards the left, favouring an oxidation reaction, and viceversa. So, in writing a full redox reaction, you must invert the half-reaction which has the lowest reduction potential. Electric potential is measured in volts (V) and the electrical potential difference is often referred to as the voltage difference or just voltage. Standard Cell Potential The standard cell potential is the maximum electrical potential difference the cell can produce under standard ambient temperature and pressure, or SATP, (1 atm, 298 K, 1 M solutions) and is given by: E o = E o E o (cell) (cathode) (anode) This order of subtraction is useful to predict the spontaneity of the reaction; if the sign of E o is positive, the reaction will proceed spontaneously. Note that these overall cell potentials can be determined by adding the reduction potential of the reduced substance and the oxidation potential of the oxidized substance. Example A zinc electrode is placed in a solution of ZnSO 4 (1 M at 25 o C) and a copper electrode is placed in a solution of CuSO 4 (1 M at 25 o C). Solution The ZnSO 4 is the source of Zn 2+ ions. The CuSO 4 is the source of Cu 2+ ions. Zn 2+ (aq) + 2é ----> Zn (s) E o = -0.76 V (#1) Cu 2+ (aq) + 2é ----> Cu (s) E o = 0.34 V(#2) And, since the reactants are always on the left side of the equation, equation #1 must be re-written: Zn (s) ----> Zn 2+ (aq) + 2é E o = 0.76 V(#3)
This positive value of E o represents the oxidation potential (the opposite of the reduction potential) and follows from the fact that we have reversed reaction #1. Also, the number of electrons given during reaction #1, must be equal to the number of electrons received in reaction #3 which is already the case here. In order to obtain the overall redox reaction of the standard cell, reaction #2 and #3 must be added: Zn (s) ----> Zn 2+ (aq) + 2é E o = 0.76 V (#3) Cu 2+ (aq) + 2é ----> Cu (s) E o = 0.34 V (#2) ========================================= Zn (s) + Cu 2+ (aq) ----> Zn 2+ (aq) + Cu (s) E o = 1.10 V (#4) So, the overall reaction (#4) tends to spontaneously go from left to right and the value of E o is 1.10 V. We could have calculated this same overall reduction potential of the cell by: E o = E o E o = 0.34 V (-0.76 V) = 1.10 V (cell) (cathode) (anode) This method would yield the correct result and show that the reaction is indeed spontaneous but fail to detail the electron transfer entailed in each half-reaction. Special Note The reduced substance is always the substance which has the highest reduction potential (E o ) value. Any element whose electronegativity is higher than that of hydrogen will have a negative voltage, and vice-versa. The higher the reduction potential (E o ), the more the element has a tendency to gain electrons and form negative ions. The opposite occurs when talking about oxidation potential. The lower the E o, the stronger of a reducing agent (reductant) the substance is; it is a good electron donor.
Many redox reactions in cells are set up to produce a maximum potential difference (voltage) from the two metals. However, if the two substances, usually metals, are not chosen carefully the reduction potential of the cell will be less than that of either substance. Example A Cu electrode in placed in a solution of Cu 2+ (aq) 1 M at 25 o C and a Ag (s) electrode is placed in a solution of Ag + (aq) 1 M at 25 o C. Solution Before the circuit is closed: Cu 2+ (aq) + 2é -----> Cu (s) E o = 0.34 V (#1) Ag + (aq) + 1é -----> Ag (s) E o = 0.80 V (#2) After the circuit has been closed, you must invert equation #1, so E o becomes negative: Cu (s) -----> Cu 2+ (aq) + 2é E o = -0.34 V (#3) Ag + (aq) + 1é -----> Ag (s) E o = 0.80 V (#2) But there are 2 é in equation #3 and only 1 é in equation #2, therefore we must multiply equation #2 by 2, but the value of E o remains the same. Cu (s) -----> Cu 2+ (aq) + 2é E o = -0.34 V (#3) 2(Ag + (aq) + 1é -----> Ag (s) ) E o = 0.80 V (#2) ============================================================= Cu (s) + 2Ag + (aq) -----> Cu 2+ (aq) + 2Ag (s) E o = 0.46 V (#4) So, in this cell, the value of E o is 0.46 V, and the reaction will spontaneously tend towards the right Electrochemical Series (Displacement Series) (a.k.a Activity Series) A list of metals arranged in order of their electrode potentials. A metal will displace, from their salts, metals lower down in the series. You might wonder what the significance of the positive and negative reduction potentials stems from; it is a consequence of the measurement technique. It is impossible to measure the reduction potential of only half of a redox reaction since the electrons must be both donated and accepted for current to flow. Consequently, the hydrogen half-cell has been arbitrarily chosen as the reference for all other substances. Thus, copper's reduction potential was measured with a hydrogen half-cell and this provides the 0 level it is the substance's reduction potential relative to hydrogen.
Worksheet 5.3: Redox Reactions 1. What is the E o for the reaction Cl 2 + 2I - 2Cl - + I 2? 2. Find E o for the reaction 4Fe 2+ + O 2 + 4H + 4Fe 3+ + 2H 2 O 3. Find E o for the reaction 2Fe 3+ + Fe 3Fe 2+ 4. The net reaction of the mercury cell is Zn + HgO + H 2 O Zn(OH) 2 + Hg What substance is oxidized at the anode of this cell? 5. The overall reaction of the silver oxide battery, which is used in watches, may be written as Ag 2 O + Zn 2Ag + ZnO. The cathode of the battery is: 6. What is the resulting voltage produced for a cell which is made with the following reactions? Mg 2+ + 2é Mg (1) Ag + + é Ag (2) 7. If a piece of copper metal is dipped into a solution containing Cr 3+ ions, what will happen? Explain using E o values. 8. A silver ring is dropped in a solution of Pb(NO 3 ) 2 for a certain period. Will you find it intact? Explain. 9. Consider the following metals: Pb / Cu / Ni / Fe / Al / Zn / Mg State which metals will react in each of the following solutions. a) HCl b) CuSO 4 c) Fe 2 (NO 3 ) 3 d) Ca(NO 3 ) 2 10. In the following reaction: 2FeCl 2 + SnCl 4 2FeCl 3 + SnCl 2 a) Which particle is oxidised? b) Which particle is reduced? 11. What is the maximum potential difference that can be obtained from a cell which operates according to the following global reaction? Cr + Au 3+ Cr 3+ + Au