Standard Electrode Potentials for Reduction and Oxidation Reactions Chemistry Tutorial

Key Concepts

• A standard electrode potential¹ is a measure of the energy per electron in a given reaction.
• E^o is the symbol used for standard electrode potentials.²
• E^o is measured in volts (V).
• Values of standard electrode potentials are tabulated for reactions in which the reactants and products are in their standard states:

  ⚛ Standard state of an element is its state at a temperature of 25^oC and a pressure of 100 kPa.

  ⚛ Standard state of substances in aqueous solution refers to a concentration³ of 1 mol L^-1

• Tables of E^o values are usually given for the reduction reaction:

  For a metallic element, M_(s), the reduction reaction is the one in which ions of the metal in aqueous solution, M⁺_(aq), gain electrons to produce the solid metal:

  M⁺_(aq) + e^- → M_(s)

  The value of E^o_(reduction) for this reduction reaction is tabulated.

• You will find E^o values referred to in many ways, such as :

  ⚛ standard reduction potentials (if the value is for the reduction reactions)
  ⚛ standard electrode potentials
  ⚛ standard potentials
  ⚛ standard redox potentials

• E^o for the oxidation reaction has the opposite sign to the E^o value for the reduction reaction:

  For a metal, M_(s), which loses an electron to produce metal ions in aqueous solution, M⁺_(aq), the oxidation reaction is:

  M_(s) → M⁺_(aq) + e^-

  The standard electrode potential for this oxidation reaction might be given the symbol E^o_(oxidation)

  and the value of E^o_(oxidation) is the same as the value of E^o_(reduction) for the reduction reaction (as tabulated) BUT with the opposite sign:

  E^o_(oxidation) = -E^o_(reduction)

• E^o values do not change if we change the stoichiometric coefficients of the reactants and products:

  For metal ions M⁺_(aq) being reduced to metal M:
  M⁺_(aq) + e^- → M     E^o_(reduction)
  2M⁺_(aq) + 2e^- → 2M     E^o_(reduction)
  ½M⁺_(aq) + ½e^- → ½M     E^o_(reduction)

  E^o_(reduction) for all these reactions is the same because changing the stoichiometric coefficient does not change the conditions of the reaction and therefore the energy per electron for the reaction does not change.

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Table of Standard Reduction Potentials

You will need access to a table of standard electrode potentials.
This will probably be provided on a data sheet, or, in a data booklet.
We will use the table of standard reduction potentials given below:

The standard electrode potentials are measured under standard conditions:

• temperature = 25^o C (≈ 298 K)
• pressure = 100 kPa
• concentration of species in aqueous solution = 1 mol L^-1

This is a list of standard reduction potentials because all the reactions are given as reduction equations, that is, a species gains electrons.

M⁺_(aq) + e^- → M_(s)
This is a reduction reaction.
Ions of element M, M⁺_(aq), gain an electron (e^-) to produce the element in its standard state, M_(s)

There is one reduction equation for which E^o = 0.00 V

H⁺_(aq) + e^- ⇋ ½H_2(g)     E^o = 0.00 V

This is because all the other electrode potentials listed have been measured relative to this reaction, known as the standard hydrogen electrode.

Notice that metals that are the least reactive (least active metals) like silver, Ag_(s), are found towards the bottom of the list.

We know that these metals are not very active (reactive) because they can be found in nature as the native metal (that is, as the elements and not combined with other elements in a compound).
This means that these metals do NOT oxidise easily, that is, these metals do NOT readily form ions.
Any metal ions that do form will readily gain electrons to re-form the solid metal, that is, the reduction of metal ions to solid metal is the favoured reaction:

Ag⁺ + e^- → Ag_(s)
Reduction of Ag⁺ to Ag_(s) is a favoured reaction

Let's find the E^o value for this reaction in the Table of Standard Reduction Potentials above:

Ag⁺ + e^- → Ag_(s)     E^o = +0.80 V
E^o is positive

The value of E^o is positive for the reduction of metal ions if the metal is of low activity (if the metal is not reactive).

Notice that very reactive metals (active metals), those from Group 1 (potassium, K_(s), and sodium, Na_(s)) and Group 2 (barium, Ba_(s), calcium, Ca_(s), magnesium, Mg_(s)), are right at the top of the list of standard reduction potentials.

Very active (reactive) metals readily lose electons to form ions, that is they are easy to oxidise, for example, sodium, Na_(s), readily reacts by losing an electron to form sodium ions, Na⁺.
You won't find lumps of sodium metal lying around in nature because metallic sodium is just too reactive, but you will find sodium ions, Na⁺ in many compounds in nature and even floating around in water as hydrated sodium ions, Na⁺_(aq).
This tells us that the oxidation of active (reactive) metals like sodium for example, is a highly favoured reaction:

Na_(s) → Na⁺ + e^-
Oxidation of Na_(s) to Na⁺ is a highly favoured reaction.

If the oxidation of metallic sodium to sodium ions is the favoured reaction, then that must mean that the opposite reaction does NOT eadily occur.
That is, the reduction of sodium ions, Na⁺, to sodium metal, Na_(s), is NOT the favoured reaction:

Na⁺ + e^- → Na_(s)
Reduction of Na⁺ to Na_(s) is NOT a favoured reaction.

Let's use our Table of Standard Reduction Potentials to find the value of E^o for the reduction of sodium metal:

Na⁺ + e^- → Na_(s)     E^o = -2.71 V
E^o for the reduction of an active (reactive) metal is negative.

We can generalise and say that as the activity (reactivity) of a metal decreases, the value of the standard reduction potential increases, and the forward reaction (the reduction reaction) is MORE favoured:

relative activity	metal	reduction of metal ions	standard reduction potential Eo / V	trend in Eo	reduction is
most active	potasium	K+ + e- → K(s)	-2.94	lowest	NOT favoured
↑	calcium	Ca+ + e- → Ca(s)	-2.87	↓	↓
↑	sodium	Na+ + e- → Na(s)	-2.71	↓	↓
↑	magnesuim	Mg2+ + 2e- → Mg(s)	-2.36	↓	↓
↑	aluminium	Al3+ + 3e- → Al(s)	-1.68	↓	↓
↑	zinc	Zn2+ + 2e- → Zn(s)	-0.76	↓	↓
↑	iron	Fe2+ + 2e- → Fe(s)	-0.44	↓	↓
↑	nickel	Ni2+ + 2e- → Ni(s)	-0.24	↓	↓
↑	tin	Sn2+ + 2e- → Sn(s)	-0.14	↓	↓
↑	lead	Pb2+ + 2e- → Pb(s)	-0.13	↓	↓
↑	copper	Cu2+ + 2e- → Cu(s)	+0.34	↓	↓
least active	silver	Ag+ + e- → Ag(s)	+0.80	highest Eo	Favoured

A reaction that is favoured has positive value for E^o.
A reaction that is not favoured has negative value for E^o.

IF the reduction of metal ions to solid metal is favoured then E^o_(reduction) is positive:

reduction favoured: M²⁺ + 2e^- → M_(s)   E^o positive

then the same reaction in reverse, the oxidation of solid metal to metal ions, is NOT favoured and E^o(oxidation) is negative:

oxidation NOT favoured: M_(s) → M²⁺ + 2e^-   E^o negative

This means, that in order to find the standard electrode potential for a given oxidation reaction, we only need to look up the standard reduction potential and reverse the sign, that is, change + to - OR change - to +

reduction of Ag⁺ to Ag_(s) : Ag⁺ + e^- → Ag_(s)
    E^o_{(reduction of Ag⁺)} = +0.80 V (from tables of E^o values above)

oxidation of Ag_(s) to Ag⁺ : Ag_(s) → Ag⁺ + e^-
    E^o_{(oxidation of Ag)} = -E^o_{(reduction of Ag⁺)} = -(+0.80 V) = -0.80 V

A metal that is easily oxidised is said to be a strong reductant (or a stong reducing agent) because it can cause another substance to be reduced easily.

A metal which is easily oxidised, such as sodium metal, is highly active (very reactive) and is positioned close to the top of the Table of Standard Reduction Potentials.

A metal which is not easily oxidised, such as silver, is not very active (not very reactive) and is positioned close to the bottom of the Table of Standard Reduction Potentials.

A metal that is easily oxidised is said to be a strong reductant (reducing agent) and is found towards the top of the Table of Standard Reduction Potentials.

A metal that is easily NOT oxidised is said to be a weak reductant (reducing agent) and is found towards the bottom of the Table of Standard Reduction Potentials.

The Table of Standard Reduction Potentials lists the metals in order from strongest reductant (reducing agent) to weakest reductant (reducing agent)

relative activity	Strength of Reductant	metal	reduction of metal ions	standard reduction potential Eo / V	trend in Eo	reduction is
most active	strongest reductant	potasium	K+ + e- → K(s)	-2.94	lowest	NOT favoured
↑	↑	calcium	Ca+ + e- → Ca(s)	-2.87	↓	↓
↑	↑	sodium	Na+ + e- → Na(s)	-2.71	↓	↓
↑	↑	magnesuim	Mg2+ + 2e- → Mg(s)	-2.36	↓	↓
↑	↑	aluminium	Al3+ + 3e- → Al(s)	-1.68	↓	↓
↑	↑	zinc	Zn2+ + 2e- → Zn(s)	-0.76	↓	↓
↑	↑	iron	Fe2+ + 2e- → Fe(s)	-0.44	↓	↓
↑	↑	nickel	Ni2+ + 2e- → Ni(s)	-0.24	↓	↓
↑	↑	tin	Sn2+ + 2e- → Sn(s)	-0.14	↓	↓
↑	↑	lead	Pb2+ + 2e- → Pb(s)	-0.13	↓	↓
↑	↑	copper	Cu2+ + 2e- → Cu(s)	+0.34	↓	↓
least active	weakest reductant	silver	Ag+ + e- → Ag(s)	+0.80	highest Eo	Favoured

Remember that the Table of Standard Reduction Potentials lists E^o values for substances in their standard states, for aqueous solutions this refers to a concentration of 1 mol L^-1.

For the reduction of silver ions in aqueous solution to silver metal, the reduction reaction equation given in the Table of Standard Reduction Potentials is:

Ag⁺_(aq) + e^- → Ag_(s)     E^o = +0.80 V

The concentration of silver ions in solution is 1 mol L^-1 (Ag⁺_(aq) in its standard state!)

If we chose to balance the reduction reaction equation in a different way, as shown in the example below:

3Ag⁺_(aq) + 3e^- → 3Ag_(s)

the concentration of silver ions in aqueous solution is still the same, [Ag⁺_(aq)] = 1 mol L^-1 because the silver ions must still be in their standard state!

Since the concentration of silver ions will always be 1 mol L^-1, irrespective of how we balance the reduction reaction equation, then the concentration of electrons required will always be the same, so the electrical energy required for the reaction will always be the same, so, the value of E^o for the reduction reaction (standard state) will not change.

Changing the stoichiometric cofficients of reactants and products in a reduction reaction equation in which all the species are present in their standard states DOES NOT change the value, nor the sign, of E^o

For Ag⁺_(aq) and Ag_(s) in their standard states:

Ag⁺_(aq) + e^- → Ag_(s)     E^o = +0.80 V

2Ag⁺_(aq) + 2e^- → 2Ag_(s)     E^o = +0.80 V

3Ag⁺_(aq) + 3e^- → 3Ag_(s)     E^o = +0.80 V

½Ag⁺_(aq) + ½e^- → ½Ag_(s)     E^o = +0.80 V

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Worked Examples: Determining E^o For Reduction and Oxidation Reactions

Question 1. What is the standard electrode potential for the reduction of Cu²⁺_(aq) ions in a standard solution to copper metal?

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

   Find the standard electrode potential
   E^o = ? V

2. What data (information) have you been given in the question?

   Extract the data from the question:

   Cu²⁺_(aq) (copper(II) ions in their standard state have a concentration of 1 mol L^-1)

   Cu_(s) (elemental copper in its standard state is a solid)

   Cu²⁺_(aq) is being reduced to Cu_(s)

3. What is the relationship between what you know and what you need to find out?

   Definition: A reduction reaction is one in which a species gains electrons

   Cu²⁺ with a charge of +2 will need to gain 2 negative charges, that is, gain 2 electrons (e^-), in order to balance the charge on the ion and produce a neutral atom of copper.

   Write the equation for the reduction of Cu²⁺_(aq) to Cu_(s)

   Cu²⁺_(aq) + 2e^- → Cu_(s)

   Values of E^o are tabulated for reduction reactions.

4. Find the value of E^o for this reaction in the Table of Standard Reduction Potentials

   Cu²⁺_(aq) + 2e^- → Cu_(s)     E^o = +0.34 V

5. Is your answer plausible?

   Think about the activity (reactivity) of copper metal.
   Copper metal has been beaten into different shapes since ancient times for use as plates, drinking vessels etc, and in more recent times, it is used as wire to carry electricity.
   In order for copper to be used in this way, it must not be an active (reactive) metal.
   Metals that are not very active have higher values for E^o.
   +0.34 V is a moderately high value.
   So our answer is plausible.

6. State your solution to the problem "find the standard electrode potential for the reduction reaction":

   E^o = +0.34 V

Question 2. What is the standard electrode potential for the oxidation of metallic iron to Fe²⁺_(aq)?

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

   Determine the value of E^o for the oxidation reaction
   E^o = ? V

2. What data (information) have you been given in the question?

   Extract the data from the question:

   Fe_(s)
   (metallic iron, iron in its standard state)

   Fe²⁺_(aq)
   (iron(II) ions must be in their standard state, [Fe²⁺_(aq)] = 1 mol L^-1)

   oxidation of Fe_(s) to Fe²⁺_(aq)

3. What is the relationship between what you know and what you need to find out?

   Definition: oxidation is the loss of electrons from a species.

   Fe_(s) must lose 2 negative charges (2 electrons) in order to produce an ion with a charge of 2+ (Fe²⁺).

   Write the equation for the oxidation of Fe_(s) to Fe²⁺_(aq)

   oxidation: Fe_(s) → Fe²⁺_(aq) + 2e^-     E^o_(oxidation)

   Note that the Table of Standard Reduction Potentials gives the value of E^o for the reduction reaction which is the reverse of the oxidation reaction:

   reduction: Fe²⁺_(aq) + 2e^- → Fe_(s)     E^o_(reduction)

   Look up the value of E^o_(reduction) in the Table of Standard Reduction Potentials:

   reduction: Fe²⁺_(aq) + 2e^- → Fe_(s)     E^o_(reduction) = -0.44 V

   BUT, E^o_(oxidation) has the same numerical value as, but the opposite sign to, E^o_(reduction)

   E^o_(oxidation) = -E^o_(reduction)

4. Determine the value of E^o_(oxidation)

   E^o_(oxidation) = -E^o_(reduction)

   E^o_(oxidation) = -(-0.44) = +0.44 V

5. Is your answer plausible?

   Iron is not found in nature as native, or elemental iron so it must react in nature with other elements to form compounds.
   But you can leave a lump of pure iron lying around for a quite long time before it shows signs of oxidation, so it must be only moderately (active) reactive.
   A value of +0.44 V indicated that the oxidation of metallic ion is favoured, but the value is not very large so the reaction is only moderately favoured, which agrees with our understanding of the activity (reactivity) of metallic iron, so our answer is plausible.

6. State your solution to the problem " determine the value of E^o for the oxidation reaction":

   E^o = +0.44 V

Question 3. What is the standard electrode potential for the reaction given below?
        2Br^-_(aq) → Br_2(l) + 2e^-

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

   Determine E^o for the reaction
   E^o = ? V

2. What data (information) have you been given in the question?

   Extract the data from the question:

   Equation for the reaction:

   2Br^-_(aq) → Br_2(l) + 2e^-
   Species are Br^-_(aq) and Br_2(l)

3. What is the relationship between what you know and what you need to find out?

   Definition: this is an oxidation reaction, Br^-_(aq) is losing electrons to form Br_2(l)

   oxidation: 2Br^-_(aq) → Br_2(l) + 2e^-

   Species must be in their standard states if we are to find the Standard Electrode Potential for the reaction:
   Concentration of Br^-_(aq) is 1 mol L^-1

   Tabulated values of E^o are for the reduction reaction, so write the oxidation equation in reverse:

   reduction: Br_2(l) + 2e^- → 2Br^-_(aq)     E^o_(reduction)

   Note that the tabulated value is for a reaction with different stoichiometric coefficients:

   reduction: ½Br_2(l) + e^- → Br^-_(aq)     E^o_(reduction) = +1.08 V

   But because the species are in their standard states, what stoichiometric coefficients we use to balance the equation does NOT effect the value of E^o_(reduction) for the equation, so

   reduction: ½Br_2(l) + e^- → Br^-_(aq)     E^o_(reduction) = +1.08 V
   reduction: Br_2(l) + 2e^- → 2Br^-_(aq)     E^o_(reduction) = +1.08 V

   So, E^o for the oxidation reaction will have the same numerical value as, but the opposite sign to, the value of E^o for the reduction reaction (under standard conditions).

   E^o_(oxidation) = -E^o_(reduction)

4. Determine the value of E^o_(oxidation)

   E^o_(oxidation) = -E^o_(reduction)

   E^o_(oxidation) = -(+1.08) = -1.08 V

5. Is your answer plausible?

   Bromine is a Group 17 element (a halogen).
   The reactivity of halogens decreases as you go down the Group from highly reactive fluorine to not so reactive iodine.
   Bromine is likely to be only moderately reactive.
   Halogens tend to gain an electron in order to form a negatively charged ion.
   We expect the reduction reaction in which bromine atoms gain an electron to form negatively charged bromide ions to be moderately favoured, that is, E^o for the reduction of bromine atoms should be moderately large and positive.
   Therefore, E^o for the reverse reaction, the oxidation reaction, should be moderately large and negative (that is, the oxidation of bromide ions to bromine atoms is not favoured).
   Since we found E^o for the oxidation to be negative and moderately large (-1.08) we are reasonably confident that our answer is plausible.

6. State your solution to the problem "determine E^o for the reaction":

   E^o_(oxidation) = -1.08 V