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

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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.

Each element in a compound is given an oxidation number.

SpeciesOxidation number examples Uncombined element 0C, Na, O 2, H 2, P 4, Cl 2 Combined oxygen -2H 2 O, CaO Combined Hydrogen +1NH 3, H 2 S Simple ionCharge on ionNa +, Mg 2+, Cl -1 Combined fluorine NaF, CaF 2

SpeciesOxidation number examples Combined group 1 +1 NaCl Combined group 2 +2MgCl 2

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

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

No need to balance Mn; equal numbers 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 MnO 4 ¯ being reduced to Mn 2+ in acidic solution Step 1 MnO 4 ¯ ———> Mn 2+

BALANCING REDOX HALF EQUATIONS Overall charge on MnO 4 ¯ 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] 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 MnO 4 ¯ being reduced to Mn 2+ in acidic solution Step 1 MnO 4 ¯ ———> Mn 2+ Step 2 +7 +2

BALANCING REDOX HALF EQUATIONS 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 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 MnO 4 ¯ being reduced to Mn 2+ in acidic solution Step 1 MnO 4 ¯ ———> Mn 2+ Step 2 +7 +2 Step 3 MnO 4 ¯ + 5e¯ ———> Mn 2+

BALANCING REDOX HALF EQUATIONS 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 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 MnO 4 ¯ being reduced to Mn 2+ in acidic solution Step 1 MnO 4 ¯ ———> Mn 2+ Step 2 +7 +2 Step 3 MnO 4 ¯ + 5e¯ ———> Mn 2+ Step 4MnO 4 ¯ + 5e¯ + 8H + ———> Mn 2+

Example 2 MnO 4 ¯ being reduced to Mn 2+ in acidic solution Step 1 MnO 4 ¯ ———> Mn 2+ Step 2 +7 +2 Step 3 MnO 4 ¯ + 5e¯ ———> Mn 2+ Step 4MnO 4 ¯ + 5e¯ + 8H + ———> Mn 2+ Step 5MnO 4 ¯ + 5e¯ + 8H + ———> Mn 2+ + 4H 2 O now balanced BALANCING REDOX HALF EQUATIONS Everything balances apart from oxygen and hydrogenO LHS = 4RHS = 0 H LHS = 8RHS = 0 You must ADD 4 WATER MOLECULES to the RHS; the equation is now balanced 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

5.3 Exercise 1

BALANCING REDOX HALF EQUATIONS 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 Q. Balance the following half equations... Na —> Na + Fe 2+ —> Fe 3+ I 2 —> I¯ C 2 O 4 2- —> CO 2 H 2 O 2 —> O 2 H 2 O 2 —> H 2 O NO 3 - —> NO NO 3 - —> NO 2 SO 4 2- —> SO 2

BALANCING REDOX HALF EQUATIONS Q. Balance the following half equations... Na —> Na + + e - Fe 2+ —> Fe 3+ + e - I 2 + 2e - —>2I¯ C 2 O 4 2- —> 2CO 2 + 2e - H 2 O 2 —> O 2 + 2H + + 2e - H 2 O 2 + 2H + + 2e - —> 2H 2 O NO 3 - + 4H + + 3e - —> NO + 2H 2 O NO 3 - + 2H + + e - —> NO 2 + H 2 O SO 4 2- + 4H + + 2e - —> SO 2 + 2H 2 O

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

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)

COMBINING HALF EQUATIONS The reaction between manganate(VII) and iron(II) Step 1Fe 2+ ——> Fe 3+ + e¯Oxidation MnO 4 ¯ + 5e¯ + 8H + ——> Mn 2+ + 4H 2 OReduction 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

COMBINING HALF EQUATIONS The reaction between manganate(VII) and iron(II) Step 1Fe 2+ ——> Fe 3+ + e¯Oxidation MnO 4 ¯ + 5e¯ + 8H + ——> Mn 2+ + 4H 2 OReduction Step 2 5Fe 2+ ——> 5Fe 3+ + 5e¯multiplied by 5 MnO 4 ¯ + 5e¯ + 8H + ——> Mn 2+ + 4H 2 Omultiplied by 1 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

COMBINING HALF EQUATIONS The reaction between manganate(VII) and iron(II) Step 1Fe 2+ ——> Fe 3+ + e¯Oxidation MnO 4 ¯ + 5e¯ + 8H + ——> Mn 2+ + 4H 2 OReduction Step 2 5Fe 2+ ——> 5Fe 3+ + 5e¯multiplied by 5 MnO 4 ¯ + 5e¯ + 8H + ——> Mn 2+ + 4H 2 Omultiplied by 1 Step 3MnO 4 ¯ + 5e¯ + 8H + + 5Fe 2+ ——> Mn 2+ + 4H 2 O + 5Fe 3+ + 5e¯ 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

COMBINING HALF EQUATIONS The reaction between manganate(VII) and iron(II) Step 1Fe 2+ ——> Fe 3+ + e¯Oxidation MnO 4 ¯ + 5e¯ + 8H + ——> Mn 2+ + 4H 2 OReduction Step 2 5Fe 2+ ——> 5Fe 3+ + 5e¯multiplied by 5 MnO 4 ¯ + 5e¯ + 8H + ——> Mn 2+ + 4H 2 Omultiplied by 1 Step 3MnO 4 ¯ + 5e¯ + 8H + + 5Fe 2+ ——> Mn 2+ + 4H 2 O + 5Fe 3+ + 5e¯ Step 4 MnO 4 ¯ + 8H + + 5Fe 2+ ——> Mn 2+ + 4H 2 O + 5Fe 3+ 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 1Fe 2+ ——> Fe 3+ + e¯Oxidation MnO 4 ¯ + 5e¯ + 8H + ——> Mn 2+ + 4H 2 OReduction Step 2 5Fe 2+ ——> 5Fe 3+ + 5e¯multiplied by 5 MnO 4 ¯ + 5e¯ + 8H + ——> Mn 2+ + 4H 2 Omultiplied by 1 Step 3MnO 4 ¯ + 5e¯ + 8H + + 5Fe 2+ ——> Mn 2+ + 4H 2 O + 5Fe 3+ + 5e¯ Step 4 MnO 4 ¯ + 8H + + 5Fe 2+ ——> Mn 2+ + 4H 2 O + 5Fe 3+ COMBINING HALF EQUATIONS SUMMARY 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

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 COMBINING HALF EQUATIONS Q. Construct balanced redox equations for the reactions between... Mg and H + Cr 2 O 7 2- and Fe 2+ H 2 O 2 and MnO 4 ¯ C 2 O 4 2- and MnO 4 ¯ S 2 O 3 2- and I 2 Cr 2 O 7 2- and I¯

Mg ——> Mg 2+ + 2e ¯ (x1) H + + e ¯ ——> ½ H 2 (x2) Mg + 2H + ——> Mg 2+ + H 2 Cr 2 O 7 2- + 14H + + 6e ¯ ——> 2Cr 3+ + 7H 2 O (x1) Fe 2+ ——> Fe 3+ + e ¯ (x6) Cr 2 O 7 2- + 14H + + 6Fe 2+ ——> 2Cr 3+ + 6Fe 2+ + 7H 2 O MnO 4 ¯ + 5e¯ + 8H + ——> Mn 2+ + 4H 2 O (x2) H 2 O 2 ——> O 2 + 2H + + 2e ¯ (x5) 2MnO 4 ¯ + 5H 2 O 2 + 6H + ——> 2Mn 2+ + 5O 2 + 8H 2 O MnO 4 ¯ + 5e¯ + 8H + ——> Mn 2+ + 4H 2 O (x2) C 2 O 4 2- ——> 2CO 2 + 2e ¯ (x5) 2MnO 4 ¯ + 5C 2 O 4 2- + 16H + ——> 2Mn 2+ + 10CO 2 + 8H 2 O 2S 2 O 3 2- ——> S 4 O 6 2- + 2e¯ (x1) ½ I 2 + e¯ ——> I ¯ (x2) 2S 2 O 3 2- + I 2 ——> S 4 O 6 2- + 2I ¯ Cr 2 O 7 2- + 14H + + 6e ¯ ——> 2Cr 3+ + 7H 2 O (x1) ½ I 2 + e¯ ——> I ¯ (x6) Cr 2 O 7 2- + 14H + + 3I 2 ——> 2Cr 3+ + 6I ¯ + 7H 2 O BALANCING REDOX EQUATIONS ANSWERS

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.

Half-cell: an elements in two oxidation states.

Zn 2+ (aq) + 2 e –  Zn(s)

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

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

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)

S tandard H ydrogen E lectrode

Zn Zn  Zn 2+ + 2 e - oxidation Cu 2+ + 2 e -  Cu reduction - 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. Cu

Emf = E = E (positive terminal) - E (negative terminal )

Pt(s) | H 2 (g) | H + (aq) || Cu 2+ (aq) | Cu(s)

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

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

A2.CHEM5.3.003 5.3 EXERCISE 2 - electrochemical cells

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

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

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

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

Emf = E  right - E  left ELECTRODE POTENTIALS – Q5 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

Emf = E  right - E  left ELECTRODE POTENTIALS – Q6 a) E  right = +2.00 - 2.38 = - 0.38 V Ti 3+ (aq) + e -  Ti 2+ (aq) b) E  left = -2.38 - 0.54 = - 2.92 V K + (aq) + e -  K(aq) c) E  right = - 3.19 + 0.27 = - 2.92 V Ti 3+ (aq) + e -  Ti 2+ (aq)

ELECTRODE POTENTIALS – Q7 Emf = -0.76 - (-0.91) = +0.15 V a) Cr(s) | Cr 2+ (aq) || Zn 2+ (aq) | Zn(s) Emf = +0.77 - 0.34 = +0.43 V b) Cu(s) |Cu 2+ (aq)|| Fe 3+ (aq),Fe 2+ (aq)| Pt(s) Emf = +1.51 – 1.36 = +0.15 V c) Pt(s) | Cl - (aq)| Cl 2 (g) || MnO 4 - (aq),H + (aq),Mn 2+ (aq)| Pt(s)

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