Standard Electrode Potentials for Redox Reactions Tutorial

Some Standard Reduction Potentials

Oxidants

Reductants

E^{0} (volts)

Weakest Oxidant

K^{+}+e

K(s)

-2.94

Strongest Reductant

↓

Ba^{2+}+2e

Ba(s)

-2.91

↑

↓

Ca^{2+}+2e

Ca(s)

-2.87

↑

↓

Na^{+}+e

Na(s)

-2.71

↑

↓

Mg^{2+}+2e

Mg(s)

-2.36

↑

↓

Al^{3+}+3e

Al(s)

-1.68

↑

↓

Mn^{2+}+2e

Mn(s)

-1.18

↑

↓

H_{2}O+e

½H_{2}(g)+OH^{-}

-0.83

↑

↓

Zn^{2+}+2e

Zn(s)

-0.76

↑

↓

Fe^{2+}+2e

Fe(s)

-0.44

↑

↓

Ni^{2+}+2e

Ni(s)

-0.24

↑

↓

Sn^{2+}+2e

Sn(s)

-0.14

↑

↓

Pb^{2+}+2e

Pb(s)

-0.13

↑

↓

H^{+}+e

½H_{2}(g)

0.00

↑

↓

Cu^{2+}+2e

Cu(s)

0.34

↑

↓

Cu^{+}+e

Cu(s)

0.52

↑

↓

½I_{2}(s)+e

I^{-}

0.54

↑

↓

½I_{2}(aq)+e

I^{-}

0.62

↑

↓

Fe^{3+}+e

Fe^{2+}

0.77

↑

↓

Ag^{+}+e

Ag(s)

0.80

↑

↓

½Br_{2}(l)+e

Br^{-}

1.08

↑

↓

½Br_{2}(aq)+e

Br^{-}

1.10

↑

↓

½Cl_{2}(g)+e

Cl^{-}

1.36

↑

↓

½Cl_{2}(aq)+e

Cl^{-}

1.40

↑

Strongest Oxidant

½F_{2}(g)+e

F^{-}

2.89

Weakest Reductant

Key Concepts

The standard electrode potential for a redox reaction, E^{o}_{(redox)}, is the sum of the standard electrode potential for the oxidation reaction, E^{o}_{(oxidation)}, and the standard electrode potential for the reduction reaction, E^{o}_{(reduction)}:

Use tabulated values to find the standard electrode potential for each half-equation:
Y^{+}+e^{-} → Y E^{o}_{(reduction of Y+)} X → X^{+}+e^{-} E^{o}_{(oxidation of X+)}

Add the two half-reaction equations together, and add together their standard electrode potentials:
Y^{+}+e^{-} → Y E^{o}_{(reduction of Y+)} X → X^{+}+e^{-} E^{o}_{(oxidation of X+)} Y^{+}+X → Y+X^{+}

E^{o}_{(redox)} = E^{o}_{(reduction of Y+)} + E^{o}_{(oxidation of X+)}

Note: changing the stoichiometric coefficients for reactions and products when balancing a redox reaction equation does not change the value of E^{o} for either the oxidation or the reduction reaction.

1. Calculate the standard electrode potential, E^{o}, for the following redox reaction in which all species are present in their standard states:

Zn^{2+} + Pb → Zn + Pb^{2+}

Write the two balanced half-reaction equations:

reduction of Zn^{2+}: Zn^{2+} + 2e^{-} → Zn

oxidation of Pb: Pb → Pb^{2+} + 2e^{-}

Use tabulated values to find the standard electrode potentail for each half-equation:

Zn^{2+} + 2e^{-} → Zn E^{o} = -0.76 V

Pb → Pb^{2+} + 2e^{-} E^{o} = +0.13 V
(Note: E^{o} for the oxidation of Pb to Pb^{2+} is equal to -E^{o} for the reduction of Pb^{2+} to Pb)

Add the two half-equations together, and add together their standard electrode potentials:

Zn^{2+} + 2e^{-}

→

Zn

E^{o} = -0.76 V

Pb

→

Pb^{2+} + 2e^{-}

E^{o} = +0.13 V

Zn^{2+} + Pb

→

Zn + Pb^{2+}

E^{o} = (-0.76) + (+0.13) = -0.63 V

2. Calculate the standard electrode potential, E^{o}, for the following redox reaction in which all species are present in their standard states:

2Ag^{+} + Cu → 2Ag + Cu^{2+}

Write the two balanced half-reaction equations:

reduction of 2Ag^{+}: 2Ag^{+} + 2e^{-} → 2Ag

oxidation of Cu: Cu → Cu^{2+} + 2e^{-}

Use tabulated values to find the standard electrode potentail for each half-equation:

Ag^{+} + e^{-} → Ag E^{o} = +0.80 V

So 2Ag^{+} + 2e^{-} → 2Ag E^{o} = +0.80 V (Note: changing the stoichiometric coefficients of reactants and products does not change the value of E^{o} for the reaction.)

Cu → Cu^{2+} + 2e^{-} E^{o} = -0.34 V
(Note: E^{o} for the oxidation of Cu to Cu^{2+} is equal to -E^{o} for the reduction of Cu^{2+} to Cu)

Add the two half-equations together, and add together their standard electrode potentials: