Redox Reactions AND Electrode potentials

Redox reactions What happens when you add Mg to CuSO 4 ? Why? Which element is oxidised and which is reduced? Mg Cu 2+ Mg 2+ Cu 2e _ E L O R E G: Electrons Lost Oxidation Reduction Electron Gain L E O G E R: Lose Electrons Oxidation Gain Electron Reduction O I L R I G: Oxidation Is Loss of electrons Reduction Is Gain on electrons

Redox Reactions CuSO 4 and Fe Mg and CuSO 4 KI and conc H 2 SO 4 KBr and KMnO 4 Cl 2 and NaOH WHAT IS OXIDISED AND REDUCED

Oxidation Numbers RULES Uncombined element = 0 (eg H in H 2, O in O 3 ) Ion = charge (eg Na in Na + = 1, Cl in Cl -1 = -1)  ON in compound = 0 ( eg in NaCl ON Na + ON Cl = 0)  ON in ion = 0 (eg in SO 4 -2 ON S + 4xON O =-2) In a compound F = -1 Group 1 metal = 1 Group 2 metal = +2 H = +1 (unless bonded to metal) O = -2 (unless bonded to F or in peroxide)

Oxidation Numbers Increase in Oxidation = Number of electrons lost Decrease in Reduction = Number of electrons gained Eg KBr + KMnO 4 5KBr + KMnO 4 + 4H 2 SO 4  2.5Br 2 + MnSO 4 + 3K 2 SO 4 + 4H 2 O K does not change: Starts and finishes as +1 H does not change: Starts and finishes as +1 O does not change: Starts and finishes as -2 S does not change: Starts and finishes as +6 THESE ELEMENTS ARE NOT OXIDISED OR REDUCED Br changes: Starts -1 and finishes as 0, hence each Br LOSES one electron Mn changes: Starts +7 and finishes as +2, hence each Br Gains 5 electrons NOTE 5 Br react with 1 Mn (so same number of electrons are lost and gained)

5KBr + KMnO 4 + 4H 2 SO 4  2.5Br 2 + MnSO 4 + 3K 2 SO 4 + 4H 2 O Usually written as 5Br - + MnO H +  2.5Br 2 + Mn H 2 O The spectator ions K + and SO 4 -2 ignored NOTE 10 Br react with 2 Mn (so same number of electrons are lost and gained) H and O balanced by protons and water (hydroxide ions and water if alkaline conditions) If we know the oxidation changes we can work out the balanced equation

Fe 2+ + MnO 4 -  Fe 3+ + Mn 2+ + H 2 O FeSO 4 + KMnO 4 = Fe 2+ and MnO 4 - without spectator ions +7  +2 GAIN 5electrons +2  +3 LOSS 1 electron 55 Need 5 Fe 2+ to react Hence 5 Fe 3+ formed

Fe 2+ + MnO 4 -  5 Fe 3+ + Mn 2+ + H 2 O 8H on RHS so must have 8H + as reactants 5 4O on LHS Hence 4 H 2 O formed (to balance) + 8H + 4

Cu+ Cr 2 O 7 -2  Cu Cr 3+ + H 2 O Cu + K 2 Cr 2 O 7 = Cu 2+ and Cr 2 O 7 -2 without spectator ions +6  +3 GAIN 3 electrons BUT 2 Cr so total gain is 6 electrons 0  +2 LOSS 2 electron 33 Need 3 Cu to react Hence 3 Cu 2+ formed

Cu+ Cr 2 O 7 -2  3 Cu 2+ + Cr 3+ + H 2 O 14H on RHS so must have 14H + as reactants 3 7O on LHS Hence 7H 2 O formed (to balance) + 14H + 7

Half Equations Split the reduction from the oxidation Show electron Loss and Electron Gain

Half Equations CuSO 4 and Fe Cu e -  Cu Fe  Fe e - Mg and CuSO 4 Cu e -  Cu Mg  Mg e - I - and H 2 SO 4 2I -  I 2 + 2e - 8H + + 8e - + H 2 SO 4  H 2 S + 4H 2 O

Making electrons do some work Mg Cu 2+ Mg 2+ Cu 2e - Salt bridge Rebalances charges NO 3 - K+K+ K+K+ Production of electrons here: Negative electrode: THE ANODE Electrons destroyed here: Positive electrode: THE CATHODE Mg = Mg e - Cu = Cu e -

Making electrons do some work Mg Cu 2+ 2e - Salt bridge Rebalances charges NO 3 - K+K+ K+K+ Production of electrons would be here: Hence THE ANODE for positive voltage Electrons would be destroyed here: Hence THE CATHODE for positive voltage V Voltmeter prevents electron flow Measures potential Mg = Mg e - Cu = Cu e -

(standard) (workings)

Measuring the potential how-electrochemical-cells-work/ how-electrochemical-cells-work/