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Calculating E^{o} for a Redox Reaction Given the Reduction and Oxidation Equations
If you are asked to find the value for E^{o} for a redox reaction and you have been given the equations for the reduction reaction and the oxidation equation with the reactants and products present in their standard states, you only need to find the E^{o} values for each equation and add them together to determine the E^{o} value for the overall redox equation.
Y^{2+}_{(aq)} + 2e^{} → Y_{(s)}
This is an equation for a reduction reaction, Y^{2+}_{(aq)} is being reduced to Y_{(s)}
X_{(s)} → X^{+}_{(aq)} + e^{}
This is an equation for an oxidation reaction, X_{(s)} is being oxidised to X^{+}_{(aq)}
To calculate the value of the standard electrode potential for the overall redox reaction, E^{o}_{(redox)}:
Step 1: Write the two balanced halfreaction equations
reduction reaction: Y^{2+}_{(aq)} + 2e^{} → Y_{(s)}
oxidation reaction: X_{(s)} → X^{+}_{(aq)} + e^{}
Step 2: Use tabulated values to find the standard electrode potential for each halfequation:
E^{o} values are usually tabulated for reduction reactions (and are therefore referred to as Standard Reduction Potentials)
(a) E^{o} for the reduction reaction can be looked up straight away in the tables and recorderd:
reduction reaction: Y^{2+}_{(aq)} + 2e^{} → Y_{(s)} E^{o}_{(reduction)}
(b) In order to find the E^{o} value for the oxidation reaction, you first need to reverse the reaction and write it as a reduction equation (remember, the tables list E^{o} values for reduction reactions):
oxidation reaction: X_{(s)} → X^{+}_{(aq)} + e^{}
reverse reaction is a reduction reaction which will be tabulated: X^{+}_{(aq)} + e^{} → X_{(s)}
Look up the value of E^{o} for this reversed reaction:
X^{+}_{(aq)} + e^{} → X_{(s)} E^{o}_{(rev)}
The E^{o} value for the oxidation reaction is the same as for the reversed reaction BUT the sign will be the opposite (change + to , or, change  to +)
X_{(s)} → X^{+}_{(aq)} + e^{} E^{o}_{(oxidation)} = E^{o}_{(rev)}
Step 3: Add together the standard electrode potentials for the two halfequations
E^{o}_{(redox)} = E^{o}_{(reduction)} + E^{o}_{(oxidation)}
Calculating E^{o} for a Given Redox Reaction
To calculate the value of E^{o}_{(redox)} for a redox reaction that is given to you:
2X_{(s)} + Y^{2+}_{(aq)} → 2X^{+}_{(aq)} + Y_{(s)}
you will first need to split the redox reaction up into two balanced halfequations, one equation for the reduction reaction and one equation for the oxidation reaction, before you can look up the relevant electrode potentials.
Note: species must be present in their standard states if you want to use a table of standard electrode potentials.
For elements this means their state at 100 kPa, for aqueous solutions this refers to a concentration of 1 mol L^{1})
Step 1: Write the two balanced halfreaction equations
reduction of Y^{2+}_{(aq)} to Y_{(s)}:
Y^{2+}_{(aq)} + 2e^{} → Y_{(s)}
oxidation of 2X_{(s)} to 2X^{+}_{(aq)}:
2X_{(s)} → 2X^{+}_{(aq)} + 2e^{}
Step 2: Use tabulated values to find the standard electrode potential for each halfequation:
Tabulated values of standard reduction potentials, E^{o}, are given so that the stoichiometric ratio (mole ratio) of electrons to reductant (species being oxidised) is the lowest whole number possible in order to balance the reduction equation, so in order to make it easy to find our two halfequations in the tables we need to write both reactions:
 with the lowest whole number stoichiometric ratio of electrons to reductant (species that gets oxidised)
 and as reduction reactions
For our two halfequations:
 Write both equations with lowest whole number stoichiometric ratio (mole ratio) of electrons to reductant:
(a) Reduction reaction: Y_{(s)} is the reductant
Y^{2+}_{(aq)} + 2e^{} → Y_{(s)}
stoichiometric ratio e^{}:Y_{(s)} is 2:1
This is the lowest whole number ratio we can have so we don′t need to do anything to this equation.
(b) For the oxidation reaction: X_{(s)} is the reductant
2X_{(s)} → 2X^{+}_{(aq)} + 2e^{}
stoichiometric ratio e^{}:X_{(s)} is 2:2
Clearly we could divide the coefficients by 2 to arrive at a stoichiometric ratio of 1:1
So we need to divide the coefficient for each reactant and product species in our equation by 2:
2X_{(s)} 
→ 
2X^{+}_{(aq)} 
+ 
2e^{} 
2 

2 

2 
So, X_{(s)} → X^{+}_{(aq)} + e^{}
is the equation we will look up the value of E^{o} for.
 Write BOTH equations as if they are reduction reactions:
(a) Reduction reaction is already written as a reduction reaction so we don′t need to change it:
Y^{2+}_{(aq)} + 2e^{} → Y_{(s)}
We can look up this equation and find the value for E^{o} in the table of standard reduction potentials:
Y^{2+}_{(aq)} + 2e^{} → Y_{(s)} E^{o}_{(reduction)}
(b) Oxidation reaction needs to be rewritten as a reduction reaction, that is, the equation is reversed:
oxidation reaction: X_{(s)} → X^{+}_{(aq)} + e^{}
reduction reaction (reverse of oxidation equation): X^{+}_{(aq)} + e^{} → X_{(s)}
This equation, the reduction equation, is the one that we can look up in the table of standard reduction potentials BUT when we find the value of E^{o} it will be the value for the reduction reaction!
X^{+}_{(aq)} + e^{} → X_{(s)} E^{o}_{(rev)}
We will need to remember to reverse the sign for the value of E^{o} in order to find the E^{o} value for the oxidation reaction:
for our oxidation reaction: E^{o}_{(oxidation)} = E^{o}_{(rev)}
Step 3: Add together the standard electrode potentials for the two halfequations
E^{o}_{(redox)} = E^{o}_{(reduction)} + E^{o}_{(oxidation)}
Table of Standard Electrode Potentials
If you have been asked to find the value of E^{o} for a redox reaction, you will need to have access a table of standard electrode potentials.
This will often be on a separate data sheet, or, incorporated into a data booklet.
Note:
The table usually lists E^{o} values for reduction reactions.
In this case we refer to it as a table of Standard Reduction Potentials.
The reduction equations are given so that the stoichiometric ratio (mole ratio) of electrons to reductant is the lowest whole number ratio.
Reactants and products are given in their standard states:
For elements this means their state at 100 kPa
For aqueous solutions, the concentration of ions is 1.0 mol L^{1}
Some standard reduction potentials are given in the table below:
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 
↑ 
↓ 
SO_{4}^{2} + 4H^{+} + 2e^{} 
⇋ 
SO_{2}(aq) + 2H_{2}O 
0.16 
↑ 
↓ 
Cu^{2+} + 2e^{} 
⇋ 
Cu(s) 
0.34 
↑ 
↓ 
½O_{2}(g) + H_{2}O + 2e^{} 
⇋ 
2OH^{} 
0.40 
↑ 
↓ 
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 
↑ 
↓ 
½O_{2}(g) + 2H^{+} + 2e^{} 
⇋ 
H_{2}O 
1.23 
↑ 
↓ 
½Cl_{2}(g) + e^{} 
⇋ 
Cl^{} 
1.36 
↑ 
↓ 
½Cl_{2}(aq) + e^{} 
⇋ 
Cl^{} 
1.40 
↑ 
↓ 
MnO_{4}^{} + 8H^{+} + 5e^{} 
⇋ 
Mn^{2+} + 4H_{2}O 
1.51 
↑ 
Strongest Oxidant 
½F_{2}(g) + e^{} 
⇋ 
F^{} 
2.89 
Weakest Reductant 
Worked Examples of Calculating E^{o} for Redox Reactions
Question 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+}_{(aq)} + Pb_{(s)} → Zn_{(s)} + Pb^{2+}_{(aq)}
The relevant halfequations are:
Zn^{2+}_{(aq)} + 2e^{} → Zn_{(s)}
Pb_{(s)} → Pb^{2+}_{(aq)}+ 2e^{}
Solution:
(Based on the StoPGoPS approach to problem solving.)
 What is the question asking you to do?
Calculate E^{o} for the redox reaction
E^{o}_{(redox)} = ? V
 What data (information) have you been given in the question?
Extract the data from the question:
Reduction reaction: Zn^{2+}_{(aq)} + 2e^{} → Zn_{(s)}
Oxidation reaction: Pb_{(s)} → Pb^{2+}_{(aq)}+ 2e^{}
Reactants and products in their standard states
(this allows us to use tabulated values of standard reduction potentials)
 What is the relationship between what you know and what you need to find out?
Step 1: Write the two balanced halfreaction equations (given in the question):
Reduction reaction: Zn^{2+}_{(aq)} + 2e^{} → Zn_{(s)}
Oxidation reaction: Pb_{(s)} → Pb^{2+}_{(aq)}+ 2e^{}
Step 2: Use a table of Standard Electrode Potentials (Standard Reduction Potentials) to find the value of E^{o} for both reactions.
(a) Reduction reaction: Zn^{2+}_{(aq)} + 2e^{} → Zn_{(s)} E^{o}_{(reduction)} = 0.76 V
(b) Reverse the oxidation reaction to write it as a reduction reaction:
Pb^{2+}_{(aq)}+ 2e^{} → Pb_{(s)}
Look up the value of E^{o} for this reversed reaction:
Pb^{2+}_{(aq)}+ 2e^{} → Pb_{(s)} E^{o}_{(rev)} = 0.13 V
E^{o} for the oxidation reaction = E^{o} for the reversed reaction:
Pb_{(s)} → Pb^{2+}_{(aq)}+ 2e^{} E^{o}_{(oxidation)} = (0.13 V) = +0.13 V
Step 3: Add E^{o}_{(reduction)} to E^{o}_{(oxidation)} to determine the standard electrode potential (emf or voltage) for the redox reaction (E^{o}_{(redox)}):
E^{o}_{(redox)} = E^{o}_{(reduction)} + E^{o}_{(oxidation)}
 Substitute the values into the expression for E^{o}_{(redox)} and solve:
E^{o}_{(redox)} = E^{o}_{(reduction)} + E^{o}_{(oxidation)}
E^{o}_{(redox)} = 0.76 + (+0.13)
E^{o}_{(redox)} = 0.63 V
 Is your answer plausible?
Work backwards: use your calculated value of E^{o}_{(redox)} and the tabulated value for the reduction of zinc ions to zinc, to see if you arrive at a value of +0.13 V for the oxidation of metallic lead, and hence 0.13 V for reduction of lead(II) ions as tabulated.
E^{o}_{(redox)} = E^{o}_{(reduction of Zn2+)} + E^{o}_{(oxidation of Pb)}
0.63 = 0.76 + E^{o}_{(oxidation)}
E^{o}_{(oxidation of Pb)} = 0.63 + 0.76 = +0.13 V
E^{o}_{(reduction of Pb2+)} = E^{o}_{(oxidation of Pb)} = 0.13 V
which is the same as the value tabulated so we are reasonably confident that our answer is correct.
 State your solution to the problem "calculate E^{o}_{(redox)}":
E^{o}_{(redox)} = 0.63 V
Question 2. Calculate the standard electrode potential, E^{o}, for the following redox reaction in which all species are present in their standard states:
2Ag^{+}_{(aq)} + Cu_{(s)} → 2Ag_{(s)} + Cu^{2+}_{(aq)}
Solution:
(Based on the StoPGoPS approach to problem solving.)
 What is the question asking you to do?
Calculate E^{o} for redox reaction
E^{o}_{(redox)} = ? V
 What data (information) have you been given in the question?
Extract the data from the question:
redox reaction: 2Ag^{+}_{(aq)} + Cu_{(s)} → 2Ag_{(s)} + Cu^{2+}_{(aq)}
Reactants and products in their standard states
(this allows us to use tabulated values of standard reduction potentials)
 What is the relationship between what you know and what you need to find out?
Step 1: Write the two balanced halfreaction equations (given in the question):
(a) 2Ag^{+}_{(aq)} is reduced to 2Ag_{(s)}:
reduction reaction: 2Ag^{+}_{(aq)} + 2e^{} → 2Ag_{(s)}
(b) Cu_{(s)} is being oxidised to Cu^{2+}_{(aq)}:
oxidation reaction: Cu_{(s)} → Cu^{2+}_{(aq)} + 2e^{}
Step 2: Use a table of Standard Electrode Potentials (Standard Reduction Potentials) to find the value of E^{o} for both reactions.
(a) reduction reaction: 2Ag^{+}_{(aq)} + 2e^{} → 2Ag_{(s)}
stoichiometric ratio (mole ratio) e^{}:Ag_{(s)} is 2:2
divide the stoichiometric coefficient of all species by 2 to reduce the ratio to 1:1
2Ag^{+}_{(aq)} 
+ 
2e^{} 
→ 
2Ag_{(s)} 
2 

2 

2 
Ag^{+}_{(aq)} 
+ 
e^{} 
→ 
Ag_{(s)} 
Look up the E^{o} value for this reduction reaction in the tables:
Ag^{+}_{(aq)} + e^{} → Ag_{(s)} E^{o}_{(reduction)} = +0.80 V
(b) oxidation equation: Cu_{(s)} → Cu^{2+}_{(aq)} + 2e^{}
Reverse the oxidation equation to turn it into a reduction reaction (remember the table contains E^{o} values for the reduction reaction!)
reverse oxidation: Cu^{2+}_{(aq)} + 2e^{} → Cu_{(s)}
Look up the value of E^{o} for this reversed reaction:
Cu^{2+}_{(aq)} + 2e^{} → Cu_{(s)} E^{o}_{(rev)} = +0.34 V
Remember that the E^{o} value for the oxidation reaction is the same as for the reduction reaction BUT the sign is changed, so we change the + to a 
Cu_{(s)} → Cu^{2+}_{(aq)} + 2e^{} E^{o}_{(oxidation} = E^{o}_{(rev)} = (+0.34 V) = 0.34 V
Step 3: Add E^{o}_{(reduction)} to E^{o}_{(oxidation)} to determine the standard electrode potential (emf or voltage) for the redox reaction (E^{o}_{(redox)}):
E^{o}_{(redox)} = E^{o}_{(reduction)} + E^{o}_{(oxidation)}
 Substitute the values into the equation to calculate E^{o}_{(redox)}
E^{o}_{(redox)} = E^{o}_{(reduction)} + E^{o}_{(oxidation)}
E^{o}_{(redox)} = +0.80 + (0.34)
E^{o}_{(redox)} = +0.80  0.34
E^{o}_{(redox)} = +0.46 V
 Is your answer plausible?
Work backwards: use your calculated value of E^{o}_{(redox)} and the value tabulated for E^{o} for the reduction of silver ions to metallic silver to find the E^{o} value for the oxidation of copper metal, and hence the E^{o} value for the reduction of copper(II) ions which you can check against the tabulated values.
E^{o}_{(redox)} = E^{o}_{(reduction of Ag+)} + E^{o}_{(oxidation of Cu)}
+0.46 = +0.80 + E^{o}_{(oxidation of Cu)}
E^{o}_{(oxidation of Cu)} = +0.46  0.80 = 0.34 V
E^{o}_{(reduction of Cu2+)} = E^{o}_{(oxidation of Cu)} = (0.34) = +0.34 V
which is the same as the tabulated value so we are reasonably confident our answer is correct.
 State your solution to the problem "calculate the value of E^{o} for the redox reaction":
E^{o}_{(redox)} = +0.46 V
1. If species are not in their standard states, you can use the Nernst Equation to calculate the electrode potentials.