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Cell Potential

Cell Potential. L.O.: Construct redox equations using half-equations or oxidation numbers. Describe how to make an electrochemical cell. An oxidation number is a measure of the number of electrons that an atom uses to bond with atoms of another element.

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Cell Potential

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  1. Cell Potential L.O.: Construct redox equations using half-equations or oxidation numbers. Describe how to make an electrochemical cell.

  2. An oxidation number is a measure of the number of electrons that an atom uses to bond with atoms of another element.

  3. Each element in a compound is given an oxidation number.

  4. The sum of the oxidation numbers must equal the overall charge.

  5. Oxidation states. • F = -1 • Cl = -1 • O = -2 • H = +1 Mg in MgCl2? ? - 2 = 0 ? = +2 Put the + in!

  6. BALANCING REDOX HALF EQUATIONS 1 Work out formulae of the species before and after the change; balance if required 2 Work out oxidation state of the element before and after the change 3 Add electrons to one side of the equation so that the oxidation states balance 4 If the charges on the species (ions and electrons) on either side of the equation do not balance then add sufficient H+ ions to one of the sides to balance the charges 5 If equation still doesn’t balance, add sufficient water molecules to one side Example 2 MnO4¯ being reduced to Mn2+ in acidic solution Step 1 MnO4¯ ———> Mn2+ No need to balance Mn; equal numbers

  7. BALANCING REDOX HALF EQUATIONS 1 Work out formulae of the species before and after the change; balance if required 2 Work out oxidation state of the element before and after the change 3 Add electrons to one side of the equation so that the oxidation states balance 4 If the charges on the species (ions and electrons) on either side of the equation do not balance then add sufficient H+ ions to one of the sides to balance the charges 5 If equation still doesn’t balance, add sufficient water molecules to one side Example 2 MnO4¯ being reduced to Mn2+ in acidic solution Step 1 MnO4¯ ———> Mn2+ Step 2 +7 +2 Overall charge on MnO4¯ is -1; sum of the OS’s of all atoms must add up to -1 Oxygen is in its usual oxidation state of -2; four oxygen atoms add up to -8 To make the overall charge -1, Mn must be in oxidation state +7 ... [+7 + (4x -2) = -1]

  8. BALANCING REDOX HALF EQUATIONS 1 Work out formulae of the species before and after the change; balance if required 2 Work out oxidation state of the element before and after the change 3 Add electrons to one side of the equation so that the oxidation states balance 4 If the charges on the species (ions and electrons) on either side of the equation do not balance then add sufficient H+ ions to one of the sides to balance the charges 5 If equation still doesn’t balance, add sufficient water molecules to one side Example 2 MnO4¯ being reduced to Mn2+ in acidic solution Step 1 MnO4¯ ———> Mn2+ Step 2 +7 +2 Step 3 MnO4¯ + 5e¯ ———> Mn2+ The oxidation states on either side are different; +7 —> +2 (REDUCTION) To balance; add 5 negative charges to the LHS [+7 + (5 x -1) = +2] You must ADD 5 ELECTRONS to the LHS of the equation

  9. BALANCING REDOX HALF EQUATIONS 1 Work out formulae of the species before and after the change; balance if required 2 Work out oxidation state of the element before and after the change 3 Add electrons to one side of the equation so that the oxidation states balance 4 If the charges on the species (ions and electrons) on either side of the equation do not balance then add sufficient H+ ions to one of the sides to balance the charges 5 If equation still doesn’t balance, add sufficient water molecules to one side Example 2 MnO4¯ being reduced to Mn2+ in acidic solution Step 1 MnO4¯ ———> Mn2+ Step 2 +7 +2 Step 3 MnO4¯ + 5e¯ ———> Mn2+ Step 4 MnO4¯ + 5e¯ + 8H+ ———> Mn2+ Total charges on either side are not equal; LHS = 1- and 5- = 6- RHS = 2+ Balance them by adding 8 positive charges to the LHS [ 6- + (8 x 1+) = 2+ ] You must ADD 8 PROTONS (H+ ions) to the LHS of the equation

  10. BALANCING REDOX HALF EQUATIONS 1 Work out formulae of the species before and after the change; balance if required 2 Work out oxidation state of the element before and after the change 3 Add electrons to one side of the equation so that the oxidation states balance 4 If the charges on the species (ions and electrons) on either side of the equation do not balance then add sufficient H+ ions to one of the sides to balance the charges 5 If equation still doesn’t balance, add sufficient water molecules to one side Example 2 MnO4¯ being reduced to Mn2+ in acidic solution Step 1 MnO4¯ ———> Mn2+ Step 2 +7 +2 Step 3 MnO4¯ + 5e¯ ———> Mn2+ Step 4 MnO4¯ + 5e¯ + 8H+ ———> Mn2+ Step 5 MnO4¯ + 5e¯ + 8H+ ———> Mn2+ + 4H2O now balanced Everything balances apart from oxygen and hydrogen O LHS = 4 RHS = 0 H LHS = 8 RHS = 0 You must ADD 4 WATER MOLECULES to the RHS; the equation is now balanced

  11. 5.3 Exercise 1

  12. BALANCING REDOX HALF EQUATIONS Q. Balance the following half equations... Na —> Na+ Fe2+ —> Fe3+ I2 —> I¯ C2O42- —> CO2 H2O2 —> O2 H2O2 —> H2O NO3- —> NO NO3- —> NO2 SO42- —> SO2 REMINDER 1 Work out the formula of the species before and after the change; balance if required 2 Work out the oxidation state of the element before and after the change 3 Add electrons to one side of the equation so that the oxidation states balance 4 If the charges on all the species (ions and electrons) on either side of the equation do not balance then add sufficient H+ ions to one of the sides to balance the charges 5 If the equation still doesn’t balance, add sufficient water molecules to one side

  13. BALANCING REDOX HALF EQUATIONS Q. Balance the following half equations... Na —> Na++ e- Fe2+ —> Fe3+ + e- I2+ 2e- —> 2I¯ C2O42- —> 2CO2 + 2e- H2O2 —> O2 + 2H+ + 2e- H2O2 + 2H+ + 2e- —> 2H2O NO3-+ 4H+ + 3e- —> NO+ 2H2O NO3-+ 2H+ + e- —> NO2+ H2O SO42- + 4H+ + 2e- —> SO2 + 2H2O

  14. COMBINING HALF EQUATIONS A combination of two ionic half equations, one involving oxidation and the other reduction, produces a REDOX equation. The equations are balanced as follows... Step 1 Write out the two half equations Step 2 Multiply the equations so that the number of electrons in each is the same Step 3 Add the two equations and cancel out the electrons on either side Step 4 If necessary, cancel any other species which appear on both sides

  15. COMBINING HALF EQUATIONS A combination of two ionic half equations, one involving oxidation and the other reduction, produces a REDOX equation. The equations are balanced as follows... Step 1 Write out the two half equations Step 2 Multiply the equations so that the number of electrons in each is the same Step 3 Add the two equations and cancel out the electrons on either side Step 4 If necessary, cancel any other species which appear on both sides The reaction between manganate(VII) and iron(II)

  16. COMBINING HALF EQUATIONS A combination of two ionic half equations, one involving oxidation and the other reduction, produces a REDOX equation. The equations are balanced as follows... Step 1 Write out the two half equations Step 2 Multiply the equations so that the number of electrons in each is the same Step 3 Add the two equations and cancel out the electrons on either side Step 4 If necessary, cancel any other species which appear on both sides The reaction between manganate(VII) and iron(II) Step 1 Fe2+ ——> Fe3+ + e¯ Oxidation MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O Reduction

  17. COMBINING HALF EQUATIONS A combination of two ionic half equations, one involving oxidation and the other reduction, produces a REDOX equation. The equations are balanced as follows... Step 1 Write out the two half equations Step 2 Multiply the equations so that the number of electrons in each is the same Step 3 Add the two equations and cancel out the electrons on either side Step 4 If necessary, cancel any other species which appear on both sides The reaction between manganate(VII) and iron(II) Step 1 Fe2+ ——> Fe3+ + e¯ Oxidation MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O Reduction Step 2 5Fe2+ ——> 5Fe3+ + 5e¯ multiplied by 5 MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O multiplied by 1

  18. COMBINING HALF EQUATIONS A combination of two ionic half equations, one involving oxidation and the other reduction, produces a REDOX equation. The equations are balanced as follows... Step 1 Write out the two half equations Step 2 Multiply the equations so that the number of electrons in each is the same Step 3 Add the two equations and cancel out the electrons on either side Step 4 If necessary, cancel any other species which appear on both sides The reaction between manganate(VII) and iron(II) Step 1 Fe2+ ——> Fe3+ + e¯ Oxidation MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O Reduction Step 2 5Fe2+ ——> 5Fe3+ + 5e¯ multiplied by 5 MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O multiplied by 1 Step 3 MnO4¯ + 5e¯ + 8H+ + 5Fe2+ ——> Mn2+ + 4H2O + 5Fe3+ + 5e¯

  19. COMBINING HALF EQUATIONS A combination of two ionic half equations, one involving oxidation and the other reduction, produces a REDOX equation. The equations are balanced as follows... Step 1 Write out the two half equations Step 2 Multiply the equations so that the number of electrons in each is the same Step 3 Add the two equations and cancel out the electrons on either side Step 4 If necessary, cancel any other species which appear on both sides The reaction between manganate(VII) and iron(II) Step 1 Fe2+ ——> Fe3+ + e¯ Oxidation MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O Reduction Step 2 5Fe2+ ——> 5Fe3+ + 5e¯ multiplied by 5 MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O multiplied by 1 Step 3 MnO4¯ + 5e¯ + 8H+ + 5Fe2+ ——> Mn2+ + 4H2O + 5Fe3+ + 5e¯ Step 4 MnO4¯ + 8H+ + 5Fe2+ ——> Mn2+ + 4H2O + 5Fe3+

  20. COMBINING HALF EQUATIONS A combination of two ionic half equations, one involving oxidation and the other reduction, produces a REDOX equation. The equations are balanced as follows... Step 1 Write out the two half equations Step 2 Multiply the equations so that the number of electrons in each is the same Step 3 Add the two equations and cancel out the electrons on either side Step 4 If necessary, cancel any other species which appear on both sides The reaction between manganate(VII) and iron(II) Step 1 Fe2+ ——> Fe3+ + e¯ Oxidation MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O Reduction Step 2 5Fe2+ ——> 5Fe3+ + 5e¯ multiplied by 5 MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O multiplied by 1 Step 3 MnO4¯ + 5e¯ + 8H+ + 5Fe2+ ——> Mn2+ + 4H2O + 5Fe3+ + 5e¯ Step 4 MnO4¯ + 8H+ + 5Fe2+ ——> Mn2+ + 4H2O + 5Fe3+ SUMMARY

  21. COMBINING HALF EQUATIONS A combination of two ionic half equations, one involving oxidation and the other reduction, produces a REDOX equation. The equations are balanced as follows... Step 1 Write out the two half equations Step 2 Multiply the equations so that the number of electrons in each is the same Step 3 Add the two equations and cancel out the electrons on either side Step 4 If necessary, cancel any other species which appear on both sides Q.Construct balanced redox equations for the reactions between... Mg and H+ Cr2O72- and Fe2+ H2O2 and MnO4¯ C2O42- and MnO4¯ S2O32- and I2 Cr2O72- and I¯

  22. BALANCING REDOX EQUATIONS Mg ——> Mg2+ + 2e¯ (x1) H+ + e¯ ——> ½ H2 (x2) Mg + 2H+ ——> Mg2+ + H2 Cr2O72- + 14H+ + 6e¯ ——> 2Cr3+ + 7H2O (x1) Fe2+ ——> Fe3++ e¯ (x6) Cr2O72- + 14H+ + 6Fe2+ ——> 2Cr3+ + 6Fe2+ + 7H2O MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O (x2) H2O2——> O2 + 2H+ + 2e¯ (x5) 2MnO4¯ + 5H2O2 + 6H+ ——> 2Mn2+ + 5O2 + 8H2O MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O (x2) C2O42- ——> 2CO2 + 2e¯(x5) 2MnO4¯ + 5C2O42- + 16H+ ——> 2Mn2+ + 10CO2 + 8H2O 2S2O32- ——> S4O62- + 2e¯ (x1) ½ I2 + e¯ ——> I¯(x2) 2S2O32- + I2 ——> S4O62- + 2I¯ Cr2O72- + 14H+ + 6e¯ ——> 2Cr3+ + 7H2O (x1) ½ I2 + e¯ ——> I¯(x6) Cr2O72- + 14H+ + 3I2 ——> 2Cr3+ + 6I ¯ + 7H2O ANSWERS

  23. Electrode Potentials • know the IUPAC convention for writing half-equations for electrode reactions. • Know and be able to use the conventional representation of cells. • Know that standard electrode potential, E , refers to conditions of 298 K, 100 kPa and 1.00 mol dm−3 solution of ions.

  24. Half-cell: an elements in two oxidation states.

  25. Zn2+(aq) + 2 e–  Zn(s)

  26. Zn2+(aq) + 2e¯ ->Zn(s) E° = - 0.76V The electrode potential of the half cell indicates its tendency to lose or gain electrons.

  27. The standard electrode potential of a half-cell, is the e.m.f of a cell compared with a standard hydrogen half-cell, measured at 298 K with a solution concentration of 1 mol dm-3 and a gas pressure of 100KPa

  28. Standard Conditions Concentration 1.0 mol dm-3 (ions involved in ½ equation) Temperature 298 K Pressure 100 kPa(if gases involved in ½ equation) Current Zero (use high resistance voltmeter)

  29. S tandard H ydrogen E lectrode

  30. - electrode anode oxidation + electrode cathode reduction electron flow At this electrode the metal loses electrons and so is oxidised to metal ions. These electrons make the electrode negative. At this electrode the metal ions gain electrons and so is reduced to metal atoms. As electrons are used up, this makes the electrode positive. Zn Cu Zn  Zn2+ + 2 e- oxidation Cu2+ + 2 e- Cu reduction

  31. Emf = E = E(positive terminal) - E(negativeterminal)

  32. Pt(s) | H2(g) | H+(aq) || Cu2+(aq) | Cu(s)

  33. GOLDEN RULE The more +ve electrode gains electrons (+ charge attracts electrons)

  34. Electrodes with negative emf are better at releasing electrons (better reducing agents).

  35. A2.CHEM5.3.003 5.3 EXERCISE 2 - electrochemical cells

  36. ELECTRODE POTENTIALS – Q1 Emf = Eright - Eleft - 2.71 = Eright - 0 Eright = - 2.71 V

  37. ELECTRODE POTENTIALS – Q2 Emf = Eright - Eleft Emf = - 0.44 - 0.22 Emf= - 0.66 V

  38. ELECTRODE POTENTIALS – Q3 Emf = Eright - Eleft Emf = - 0.13 - (-0.76) Emf= + 0.63 V

  39. ELECTRODE POTENTIALS – Q4 Emf = Eright - Eleft +1.02 = +1.36 - Eleft Eleft = + 1.36 - 1.02 = +0.34 V

  40. ELECTRODE POTENTIALS – Q5 Emf = Eright - Eleft a) Emf = + 0.15 - (-0.25) = +0.40 V b) Emf = + 0.80 - 0.54 = +0.26 V c) Emf = + 1.07 - 1.36 = - 0.29 V

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