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.

Each element in a compound is given an oxidation number.

Species Oxidation number

examples

Uncombined element

0 C, Na, O2, H2, P4, Cl2

Combined oxygen

-2 H2O, CaO

Combined Hydrogen

+1 NH3, H2S

Simple ion Charge on ion Na+, Mg2+, Cl-1

Combined fluorine

-1 NaF, CaF2

Species Oxidation number

examples

Combined group 1

+1 NaCl

Combined group 2

+2 MgCl2

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

Oxidation states.

•F = -1

•Cl = -1

•O = -2

•H = +1

Mg in MgCl2?

? - 2 = 0

? = +2

Put the + in!

No need to balance Mn; equal numbers

BALANCING REDOX HALF EQUATIONSBALANCING 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+


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 -8To make the overall charge -1, Mn must be in oxidation state +7 ... [+7 + (4x -2) = -1]




Step 2 +7 +2


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




Step 2 +7 +2Step 3 MnO4¯ + 5e¯ ———> 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




Step 2 +7 +2Step 3 MnO4¯ + 5e¯ ———> Mn2+

Step 4 MnO4¯ + 5e¯ + 8H+ ———> Mn2+



Step 2 +7 +2Step 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 = 0H LHS = 8 RHS = 0

You must ADD 4 WATER MOLECULES to the RHS; the equation is now balanced


5.3 Exercise 1


REMINDER1 Work out the formula of the species before and after the change; balance if required2 Work out the oxidation state of the element before and after the change3 Add electrons to one side of the equation so that the oxidation states balance4 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 charges5 If the equation still doesn’t balance, add sufficient water molecules to one side

Q. Balance the following half equations...

Na —> Na+

Fe2+ —> Fe3+

I2 —> I¯

C2O42- —> CO2

H2O2 —> O2

H2O2 —> H2O

NO3- —> NO

NO3- —> NO2

SO42- —> SO2


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

COMBINING HALF EQUATIONSCOMBINING 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 equationsStep 2 Multiply the equations so that the number of electrons in each is the sameStep 3 Add the two equations and cancel out the electrons on either sideStep 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¯ OxidationMnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O Reduction






Step 2 5Fe2+ ——> 5Fe3+ + 5e¯ multiplied by 5MnO4¯ + 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+







Step 4 MnO4¯ + 8H+ + 5Fe2+ ——> Mn2+ + 4H2O + 5Fe3+


SUMMARY






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¯

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- ——> S4O6

2- + 2e¯ (x1)

½ I2 + e¯ ——> I¯ (x2)

2S2O32- + I2 ——> S4O6

2- + 2I¯

Cr2O72- + 14H+ + 6e¯ ——> 2Cr3+ + 7H2O (x1)

½ I2 + e¯ ——> I¯ (x6)

Cr2O72- + 14H+ + 3I2 ——> 2Cr3+ + 6I ¯ + 7H2O

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.

Zn2+(aq) + 2 e– Zn(s)

Zn2+(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 Zn2+ + 2 e-

oxidation

Cu2+ + 2 e- Cureduction

- electrode

anodeoxidation

+ electrodecathode

reductionelectron 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)

H2 at 100 kPa

o

o

o

o

o

o

o

o

o

o

o

o

salt bridge

1.0 M H+(aq)

Pt

temperature= 298 K

1.0 M Cu2+(aq)

V

Cu

high resistancevoltmeter

H2 at 100 kPa

o

o

o

o

o

o

o

o

o

o

o

o

salt bridge

1.0 M H+(aq)

Pt

temperature= 298 K

1.0 M Cu2+(aq)

V

Cu

high resistancevoltmeter

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

Standard electrode potentials E/V

F2(g) + 2 e- 2 F-(aq) + 2.87

MnO42-(aq) + 4 H+(aq) + 2 e- MnO2(s) + 2 H2O(l) + 1.55

MnO4-(aq) + 8 H+(aq) + 5 e- Mn2+(aq) + 4 H2O(l) + 1.51

Cl2(g) + 2 e- 2 Cl-(aq) + 1.36

Cr2O72-(aq) + 14 H+(aq) + 6 e- 2 Cr3+(aq) + 7 H2O(l) + 1.33

Br2(g) + 2 e- 2 Br-(aq) + 1.09

Ag+(aq) + e- Ag(s) + 0.80

Fe3+(aq) + e- Fe2+(aq) + 0.77

MnO4-(aq) + e- MnO4

2-(aq) + 0.56

I2(g) + 2 e- 2 I-(aq) + 0.54

Cu2+(aq) + 2 e- Cu(s) + 0.34

Hg2Cl2(aq) + 2 e- 2 Hg(l) + 2 CI-(aq) + 0.27

AgCl(s) + e- Ag(s) + Cl-(aq) + 0.22

2 H+(aq) + 2 e- H2(g) 0.00

Pb2+(aq) + 2 e- Pb(s) - 0.13

Sn2+(aq) + 2 e- Sn(s) - 0.14

V3+(aq) + e- V2+(aq) - 0.26

Ni2+(aq) + 2 e- Ni(s) - 0.25

Fe2+(aq) + 2 e- Fe(s) - 0.44

Zn2+(aq) + 2 e- Zn(s) - 0.76

Al3+(aq) + 3 e- Al(s) - 1.66

Mg2+(aq) + 2 e- Mg(s) - 2.36

Na+(aq) + e- Na(s) - 2.71

Ca2+(aq) + 2 e- Ca(s) - 2.87

K+(aq) + e- K(s) - 2.93

Increasingreducing

power

Increasingoxidising

power

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

http://A2.CHEM5.3.003.doc/

Emf = Eright - Eleft

ELECTRODE POTENTIALS – Q1

- 2.71 = Eright - 0

Eright = - 2.71 V



Emf = - 0.44 - 0.22

Emf = - 0.66 V



Emf = - 0.13 - (-0.76)

Emf = + 0.63 V



+1.02 = +1.36 - Eleft

Eleft = + 1.36 - 1.02 = +0.34 V



a) Emf = + 0.15 - (-0.25) = +0.40 Vb) Emf = + 0.80 - 0.54 = +0.26 Vc) Emf = + 1.07 - 1.36 = - 0.29 V



a) Eright = +2.00 - 2.38 = - 0.38 V

Ti3+(aq) + e- Ti2+(aq)

b) Eleft = -2.38 - 0.54 = - 2.92 V

K+(aq) + e- K(aq)c) Eright = - 3.19 + 0.27 = - 2.92 V Ti3+(aq) + e- Ti2+(aq)


Emf = -0.76 - (-0.91) = +0.15 V

a) Cr(s) | Cr2+(aq) || Zn2+(aq) | Zn(s)

Emf = +0.77 - 0.34 = +0.43 V

b) Cu(s) |Cu2+(aq)|| Fe3+(aq),Fe2+(aq)| Pt(s)

Emf = +1.51 – 1.36 = +0.15 V

c) Pt(s) | Cl-(aq)| Cl2(g) || MnO4-(aq),H+(aq),Mn2+(aq)| Pt(s)

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Cell Potential