In our last post, we looked at the overview of electrode potentials, where we discussed metal ions/metal systems or half-cells, standard electrode potential and electrochemical cells in depth. Here, we will focus on the calculations involving electrode potentials, which include calculations of the electromotive force (e.m.f) of electrochemical cells, the relationship between e.m.f & free energy and the relationship between e.m.f & equilibrium constant.
Half-Cell Reactions Std Reduction Potential, E° (V)
K+(aq) + e- <----> K(s) -2.92
Ca2+(aq) + 2e- <----> Ca(s) -2.87
Na+(aq) + e- <----> Na(s) -2.71
Mg2+(aq) + 2e- <----> Mg(s) -2.37
Al3+(aq) + 3e- <----> Al(s) -1.66
Zn2+(aq) + 2e- <----> Zn(s) -0.76
Fe2+(aq) + 2e- <----> Fe(s) -0.44
Sn2+(aq) + 2e- <----> Sn(s) -0.14
Pb2+(aq) + 2e- <----> Pb(s) -0.13
2H+(aq) + 2e- <----> H2(g) 0.00
Cu2+(aq) + 2e- <----> Cu(s) +0.34
Fe3+(aq) + e- <----> Fe2+(aq) +0.77
Hg2+(aq) + 2e- <----> Hg(s) +0.79
Ag+(aq) + e- <----> Ag(s) +0.80
Au3+(aq) + 3e- <----> Au(s) +1.50
Non-metals
I2(s) + 2e- <----> 2I-(aq) +0.54
Br2(l) + 2e- <----> 2Br-(aq) +1.09
Cl2(g) + 2e- <----> 2Cl-(aq) +1.36
F2(g) + 2e- <----> 2F-(aq) +2.87
The above table is the electrochemical or electromotive series of some elements showing their standard reduction potentials (SRP), E° in volts. The half-cell reactions represented above are the reduction half-reactions and the reverse reactions will give us the oxidation half-reactions. Similarly, the electrode potential of the reversed reaction will become the standard oxidation potential with an opposite sign.
If we take a closer look at the metallic series, it will be observed that the oxidizing ability or tendency of ions to gain electrons increases downward and decreases upward; while the reducing ability or the tendency of atoms to lose electrons decreases downward and increases upward. This implies that those elements lower in the series (especially those with positive SRP values) will make good cathodes, while those above (especially the ones with negative SRP values) can act as good anodes in electrochemical calls.
For the non-metallic series, the oxidizing ability or the tendency of atoms to gain electrons increases downward, while the reducing ability or the tendency of ions to lose electrons increases upward.
Electromotive Force (e.m.f)
The e.m.f of a cell, E°cell, is a measure of the driving force with which current flows out of the cell. It is the potential difference between the two electrodes when no current is flowing.
Recall the electrochemical cell made up of a Zn2+(aq)/Zn(s) system and a Cu2+(aq)/Cu(s) system, where the zinc electrode is the anode and the copper electrode, the cathode. The cell can be represented symbolically by:
Zn(s)/Zn2+(aq)//Cu2+(aq)/Cu(s)
An electrochemical cell can be generally represented by:
Anode(s)/Anode(aq)//Cathode(aq)/Cathode(s)
where the single slash (/) represents the interface between the electrode and its electrolyte, and the double slash (//) represents the salt-bridge. The left-hand side of the salt-bridge represents the oxidation half-reaction, while the right-hand side represents the reduction half-reaction.
Recall that the e.m.f of the cell is given as:
E°cell = E°cathode - E°anode ...................(i), or
E°cell = E°reduction - E°oxidation ...............(ii)
Equations (i) and (ii) are used if the standard electrode potentials of the electrode are provided as standard reduction potentials (as in the table above).
However, if the standard electrode potentials are given as standard oxidation potentials (SOPs), i.e, based on the reverse reactions of the half-cell reactions in the table, then equations (iii) and (iv) below are applied.
E°cell = E°anode - E°cathode ...................(iii), or
E°cell = E°oxidation - E°reduction ...............(iv)
Example 1
If the electrode potential of copper is +0.34V and that of zinc is -0.76V, calculate the e.m.f of the cell.
Answer
From the above question, zinc will always act as an anode in the presence of copper, and the values of the standard electrode potentials provided are the SRPs, so we apply equation (i) or (ii)
E°cell = E°cathode - E°anode
= +0.34 - (-0.76)
= +0.34 + 0.76
= +1.1V
It is important to note that an electrochemical cell with a large e.m.f can be produced by using metals that are far apart on the electrochemical series as the electrodes.
Example 2
Calculate the standard cell potential of each of the following cells:
a) Zn(s)/Zn2+(aq)//Cu2+(aq)/Cu(s)
b) Mg(s)/Mg2+(aq)//Ca2+(aq)/Ca(s)
Answer
a) Zn(s)/Zn2+(aq)//Cu2+(aq)/Cu(s)
From the above representation, we can infer that the zinc electrode is the anode (since its half-cell is on the left-hand side of the salt-bridge), while the copper electrode is the cathode.
Therefore, using the values of their respective SRPs (Zn = -0.76V, Cu = +0.34V), the solution will be the same as that for Example 1 above.
b) Mg(s)/Mg2+(aq)//Ca2+(aq)/Ca(s)
From the above representation, the magnesium electrode is the anode, while the calcium electrode is the cathode. So, using their SRPs and applying equation (i) or (ii), we will obtain:
E°cell = E°cathode - E°anode
= -2.87 - (-2.37)
= -2.87 + 2.37
= -0.5V
Example 3
Given the following reactions:
Fe(s) ----> Fe2+(aq) + 2e-; E° = +0.44V
Ag(s) ----> Ag+(aq) + e-; E° = -0.80V
Calculate the e.m.f of a cell that can be produced from the combination of both reactions.
Answer
The above reactions are oxidation reactions, hence, the standard electrode potentials provided are the SOPs. Also, based on their positions in the electromotive series, the iron electrode will act as the anode, while the silver electrode will serve as the cathode.
Hence, applying equation (iii) or (iv) gives:
E°cell = E°anode - E°cathode
= +0.44 - (-0.80)
= +0.44 + 0.80
= +1.24V
It is important to note that in problems involving two half-reactions of any kind, it is always necessary to determine the number of moles electrons involved in the redox reaction by balancing the electron loss and gain, though, it is does not affect the calculation of the e.m.f
So, considering the reactions above, we will have:
Fe(s) ----> Fe2+(aq) + 2e-; E° = +0.44V (oxidation half-equation)
2Ag+(aq) + 2e- ----> 2Ag(s); E° = +0.80V (reduction half-equation)
Fe(s) + 2Ag+(aq) ----> Fe2+(aq) + 2Ag(s) (overall/net reaction)
Given the above, the e.m.f of the cell can also be calculated based on the net reaction as:
E°cell = E°reduction + E°oxidation ..........(v)
= E°cathode + E°anode ..........(vi)
= +0.80 + 0.44
= +1.24V
From the above, it can be observed that the standard electrode potential for the anodic/oxidation half-reaction is the standard oxidation potential, while that of the cathodic/reduction half-reaction is the standard reduction potential.
Therefore, when the electrode potentials provided are for the respective redox processes, (i.e, SOP for the oxidation half-reaction, and SRP for the reduction half-reaction), we add both values to obtain the e.m.f of the cell.
Relationship between e.m.f of a Cell, E°cell, and Change in Gibb's Free Energy, DG
Gibb's Free Energy (G), is a measure of the spontaneity of a process. It is an indication that a process will proceed on its own without the interference of an external factor. Examples of spontaneous processes include:
a) the melting of ice
b) the crying of a new born baby
c) the flow of water from a highland to a lowland
d) the flow of electrons from the anode to the cathode
e) the flow of current from a region of higher potential to that of lower potential
To show that a process is spontaneous and feasible, the value of DG (read as delta G or change in free energy), must be negative (DG < 0).
The relationship between free energy and the e.m.f of a cell is given as:
DG° = -nFE°cell .................(vii)
where n = number of electrons transferred in the redox reaction
F = Faraday's constant
= 96500Cmol^-1
E°cell = standard e.m.f of the cell
From equation (vii) above, for the process to be spontaneous, the e.m.f of the cell must be positive.
Therefore, using the e.m.f values of the electrochemical cells in our examples above, let us see how to determine the spontaneity of a redox process.
Example 1.1
If the electrode potential of copper is +0.34V and that of zinc is -0.76V, calculate the e.m.f of the cell. Hence or otherwise, determine the spontaneity of the cell.
Answer
From our previous solution, the e.m.f, E°cell = 1.1V
Zn(s) ----> Zn2+(aq) + 2e- (oxidation half-equation)
Cu2+(aq) + 2e- ----> Cu(s) (reduction half-equation)
Zn(s) + Cu2+(aq) ----> Zn2+(aq) + Cu(s) (overall/net reaction)
From the above reactions, the number of moles of electrons transferred = 2 moles
Therefore, substituting these values into equation (vii), we will obtain:
DG° = -nFE°cell
DG° = -2 x 96500 x 1.1
= -212300 J
= -212.3 kJ
Since DG° < 0, it means the redox reaction is very spontaneous (due to the high negative value) and the cell will work as arranged. That is to say, electrons will flow from the zinc electrode (anode) to the copper electrode (cathode).
Example 2.1
Calculate the standard cell potential of the cell below and comment on the cell reaction:
Mg(s)/Mg2+(aq)//Ca2+(aq)/Ca(s)
Answer
From the earlier solution, the e.m.f, E°cell = -0.5V
Mg(s) ----> Mg2+(aq) + 2e- (oxidation half-equation)
Ca2+(aq) + 2e- ----> Ca(s) (reduction half-equation)
Mg(s) + Ca2+(aq) ----> Mg2+(aq) + Ca(s) (overall/net reaction)
Number of moles of electrons transferred = 2 moles
Therefore, substituting these values into equation (vii), we will obtain:
DG° = -nFE°cell
DG° = -2 x 96500 x -0.5
= +96500 J
= +96.5 kJ
From the positive value of DG (DG > 0), it means the redox reaction is not spontaneous, and the cell will not work. In other words, magnesium cannot act as an anode in the presence of calcium, because electrons will never flow from the former to the latter by virtue of their positions in the electromotive series. Therefore, to make the cell work, the electrodes must be rearranged.
Ca(s)/Ca2+(aq)//Mg2+(aq)/Mg(s)
E°cell = E°cathode - E°anode
= -2.37 - (-2.87)
= -2.37 + 2.87
= +0.5V
Hence,
DG° = -96.5kJ
Example 3.1
Given the following reactions:
Fe(s) ----> Fe2+(aq) + 2e-; E° = +0.44V
Ag+(aq) + e- ----> Ag(s); E° = -0.80V
Determine whether the electrochemical cell will work based on the above arrangement.
Answer
From the earlier solution,
the e.m.f of the cell, E°cell = +1.24V
the number of moles of electrons transferred = 2 moles
But,
DG° = -nFE°cell
Hence,
DG° = -2 x 96500 x 1.24
= -239320 J
= -239.3 kJ
Since DG < 0, it means the redox reaction is spontaneous, and the electrochemical cell will work based on the stated arrangement.
Relationship between the e.m.f of a Cell, E°cell & Equilibrium Constant, K
Consider a redox reaction that is at equilibrium and represented by the equation:
aA(s) + bB+(aq) <----> aA+(aq) + bB(s)
The equilibrium constant, Keq, of the reaction will be given as:
Keq = [Products]^n/[Reactants]^m
Keq = [A+(aq)]^a.[B(s)]^b/[A(s)]^a.[B+(aq)]^b
but,
The concentration of a solid substance at equilibrium is always taken as a unitary value(i.e, 1) in the presence of aqueous substances.
So, the above equation reduces to:
Keq = [A+(aq)]^a/[B+(aq)]^b ............(viii)
Keq = [Anode(aq)]^a/[Cathode(aq)]^b ............(ix)
The change in Gibb's free energy, DG, is related to the equilibrium constant, K, according to the equation:
DG° = -RT ln Keq .............................(x)
(but, ln (natural log) = 2.303 log)
DG° = - 2.303 RT log Keq ...............(xi)
Substitute equation (ix) into (xi)
DG° = - 2.303 RT log [Anode(aq)]^a/[Cathode(aq)]^b ............(xii)
Equate (vii) and (xii)
-nFE°cell = - 2.303 RT log [Anode(aq)]^a/[Cathode(aq)]^b
Divide both sides by -nF
E°cell = (2.303RT/nF) log [Anode(aq)]^a/[Cathode(aq)]^b ..........(xiii)
where R = gas constant (8.314 J mol^-1 K^-1)
T = standard temperature in Kelvin (25°C = 298K)
n = number of moles of electrons transferred
F = Faraday's constant
a = number of moles of the anodic electrolyte
b = number of moles of the cathodic electrolyte
[Anode(aq)] = concentration of the anodic electrolyte
[Cathode(aq)] = concentration of the cathodic electrolyte
Equation (xiii) gives the relationship between the e.m.f of a cell and the concentrations of the electrolytes at standard conditions.
Recall that in the course of using an electrochemical cell, its e.m.f decreases. This is due to a decrease in the concentrations of the electrolytes. Therefore, we can calculate the e.m.f of a cell (or the electrode potential of an electrode) at any concentration other than the standard concentration of 1.0 mol dm^-3, by using the equation:
Ecell = E°cell - (2.303RT/nF) log [Anode(aq)]^a/[Cathode(aq)]^b ...
.....(xiv)
The above equation is known as the Nernst equation, and it shows the relationship between the e.m.f of a cell (at non-standard concentration), Ecell, the standard e.m.f of the cell, E°cell and the concentrations of the electrolytes.
Example 4
Calculate the cell e.m.f of Al(s)/Al3+(aq)(0.010 mol dm^-3)//Cu2+(aq)/Cu(s)(0.100 mol dm^-3) at 25°C [E°al = +1.66V, E°cu = +0.34V]
Answer
From the arrangement, aluminum is oxidized (anode) and copper is reduced (cathode). So, the given standard electrode potential for Al is the SOP, while that for Cu is the SRP.
2Al(s) ----> 2Al3+(aq) + 6e- (oxidation half-reaction)
3Cu2+(aq) + 6e- ----> 3Cu(s) (reduction half-reaction)
2Al(s) + 3Cu2+(aq) ----> 2Al3+(aq) + 3Cu(s) (overall redox reaction)
Also, the concentrations of the electrolytes indicate that the required cell e.m.f is Ecell, and not the standard cell e.m.f, E°cell. Therefore, we apply the Nernst equation:
Ecell = E°cell - (2.303RT/nF) log [Al3+(aq)]^2/[Cu2+(aq)]^3
but,
E°cell = E°cathode + E°anode
= + 0.34 + 1.66
= + 2.00V
Number of moles of electrons transferred, n = 6 mol
Hence,
Ecell = 2.00 - (2.303x8.314x298/ 6x96500) log (0.010)^2/(0.100)^3
= 2.00 - (5705.8483/579000) log (1x10^-4/1x10^-3)
= 2.00 - 0.009855 log 10^-1
= 2.00 - 0.009855(-1)
= 2.00 + 0.009855
= 2.009855
~ 2.01V
Do These:
Question 1
Calculate the e.m.f of the cell:
Zn(s)/Zn2+(aq)(0.002 mol dm^-3)//Cu2+(aq)/Cu(s)(0.02 mol dm^-3) at 25°C [E°zn = +0.76V, E°cu = +0.34V]
Question 2
What will be the concentration of the electrolyte in the copper half-cell in Example 4 above, if the e.m.f of the cell, Ecell, is 2.06V and the concentration of the electrolyte in the aluminium half-cell is 0.001mol dm^-3 at 25°C?
Question 3
If the overall reaction in an electrochemical cell is given by:
Mg(s) + Fe2+(aq) ----> Mg2+(aq) + Fe(s), calculate the standard e.m.f of the cell.
Question 4
An electrochemical cell was constructed by combining the two half-cells below:
Cu2+(aq) + 2e- ----> Cu(s); E° = +0.34V
Fe2+(aq) + 2e- ----> Fe(s); E° = -0.44V
What is the standard e.m.f of the cell?
Question 5
Ca(s) ----> Ca2+(aq) + 2e-; E° = +2.87V
2Ag(s) ----> 2Ag+(aq) + 2e-; E° = -0.80V. Calculate the standard electrode potential of the calcium-silver cell represented by the half-cell reactions above.
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Half-Cell Reactions Std Reduction Potential, E° (V)
K+(aq) + e- <----> K(s) -2.92
Ca2+(aq) + 2e- <----> Ca(s) -2.87
Na+(aq) + e- <----> Na(s) -2.71
Mg2+(aq) + 2e- <----> Mg(s) -2.37
Al3+(aq) + 3e- <----> Al(s) -1.66
Zn2+(aq) + 2e- <----> Zn(s) -0.76
Fe2+(aq) + 2e- <----> Fe(s) -0.44
Sn2+(aq) + 2e- <----> Sn(s) -0.14
Pb2+(aq) + 2e- <----> Pb(s) -0.13
2H+(aq) + 2e- <----> H2(g) 0.00
Cu2+(aq) + 2e- <----> Cu(s) +0.34
Fe3+(aq) + e- <----> Fe2+(aq) +0.77
Hg2+(aq) + 2e- <----> Hg(s) +0.79
Ag+(aq) + e- <----> Ag(s) +0.80
Au3+(aq) + 3e- <----> Au(s) +1.50
Non-metals
I2(s) + 2e- <----> 2I-(aq) +0.54
Br2(l) + 2e- <----> 2Br-(aq) +1.09
Cl2(g) + 2e- <----> 2Cl-(aq) +1.36
F2(g) + 2e- <----> 2F-(aq) +2.87
The above table is the electrochemical or electromotive series of some elements showing their standard reduction potentials (SRP), E° in volts. The half-cell reactions represented above are the reduction half-reactions and the reverse reactions will give us the oxidation half-reactions. Similarly, the electrode potential of the reversed reaction will become the standard oxidation potential with an opposite sign.
If we take a closer look at the metallic series, it will be observed that the oxidizing ability or tendency of ions to gain electrons increases downward and decreases upward; while the reducing ability or the tendency of atoms to lose electrons decreases downward and increases upward. This implies that those elements lower in the series (especially those with positive SRP values) will make good cathodes, while those above (especially the ones with negative SRP values) can act as good anodes in electrochemical calls.
For the non-metallic series, the oxidizing ability or the tendency of atoms to gain electrons increases downward, while the reducing ability or the tendency of ions to lose electrons increases upward.
Electromotive Force (e.m.f)
The e.m.f of a cell, E°cell, is a measure of the driving force with which current flows out of the cell. It is the potential difference between the two electrodes when no current is flowing.
Recall the electrochemical cell made up of a Zn2+(aq)/Zn(s) system and a Cu2+(aq)/Cu(s) system, where the zinc electrode is the anode and the copper electrode, the cathode. The cell can be represented symbolically by:
Zn(s)/Zn2+(aq)//Cu2+(aq)/Cu(s)
An electrochemical cell can be generally represented by:
Anode(s)/Anode(aq)//Cathode(aq)/Cathode(s)
where the single slash (/) represents the interface between the electrode and its electrolyte, and the double slash (//) represents the salt-bridge. The left-hand side of the salt-bridge represents the oxidation half-reaction, while the right-hand side represents the reduction half-reaction.
Recall that the e.m.f of the cell is given as:
E°cell = E°cathode - E°anode ...................(i), or
E°cell = E°reduction - E°oxidation ...............(ii)
Equations (i) and (ii) are used if the standard electrode potentials of the electrode are provided as standard reduction potentials (as in the table above).
However, if the standard electrode potentials are given as standard oxidation potentials (SOPs), i.e, based on the reverse reactions of the half-cell reactions in the table, then equations (iii) and (iv) below are applied.
E°cell = E°anode - E°cathode ...................(iii), or
E°cell = E°oxidation - E°reduction ...............(iv)
Example 1
If the electrode potential of copper is +0.34V and that of zinc is -0.76V, calculate the e.m.f of the cell.
Answer
From the above question, zinc will always act as an anode in the presence of copper, and the values of the standard electrode potentials provided are the SRPs, so we apply equation (i) or (ii)
E°cell = E°cathode - E°anode
= +0.34 - (-0.76)
= +0.34 + 0.76
= +1.1V
It is important to note that an electrochemical cell with a large e.m.f can be produced by using metals that are far apart on the electrochemical series as the electrodes.
Example 2
Calculate the standard cell potential of each of the following cells:
a) Zn(s)/Zn2+(aq)//Cu2+(aq)/Cu(s)
b) Mg(s)/Mg2+(aq)//Ca2+(aq)/Ca(s)
Answer
a) Zn(s)/Zn2+(aq)//Cu2+(aq)/Cu(s)
From the above representation, we can infer that the zinc electrode is the anode (since its half-cell is on the left-hand side of the salt-bridge), while the copper electrode is the cathode.
Therefore, using the values of their respective SRPs (Zn = -0.76V, Cu = +0.34V), the solution will be the same as that for Example 1 above.
b) Mg(s)/Mg2+(aq)//Ca2+(aq)/Ca(s)
From the above representation, the magnesium electrode is the anode, while the calcium electrode is the cathode. So, using their SRPs and applying equation (i) or (ii), we will obtain:
E°cell = E°cathode - E°anode
= -2.87 - (-2.37)
= -2.87 + 2.37
= -0.5V
Example 3
Given the following reactions:
Fe(s) ----> Fe2+(aq) + 2e-; E° = +0.44V
Ag(s) ----> Ag+(aq) + e-; E° = -0.80V
Calculate the e.m.f of a cell that can be produced from the combination of both reactions.
Answer
The above reactions are oxidation reactions, hence, the standard electrode potentials provided are the SOPs. Also, based on their positions in the electromotive series, the iron electrode will act as the anode, while the silver electrode will serve as the cathode.
Hence, applying equation (iii) or (iv) gives:
E°cell = E°anode - E°cathode
= +0.44 - (-0.80)
= +0.44 + 0.80
= +1.24V
It is important to note that in problems involving two half-reactions of any kind, it is always necessary to determine the number of moles electrons involved in the redox reaction by balancing the electron loss and gain, though, it is does not affect the calculation of the e.m.f
So, considering the reactions above, we will have:
Fe(s) ----> Fe2+(aq) + 2e-; E° = +0.44V (oxidation half-equation)
2Ag+(aq) + 2e- ----> 2Ag(s); E° = +0.80V (reduction half-equation)
Fe(s) + 2Ag+(aq) ----> Fe2+(aq) + 2Ag(s) (overall/net reaction)
Given the above, the e.m.f of the cell can also be calculated based on the net reaction as:
E°cell = E°reduction + E°oxidation ..........(v)
= E°cathode + E°anode ..........(vi)
= +0.80 + 0.44
= +1.24V
From the above, it can be observed that the standard electrode potential for the anodic/oxidation half-reaction is the standard oxidation potential, while that of the cathodic/reduction half-reaction is the standard reduction potential.
Therefore, when the electrode potentials provided are for the respective redox processes, (i.e, SOP for the oxidation half-reaction, and SRP for the reduction half-reaction), we add both values to obtain the e.m.f of the cell.
Relationship between e.m.f of a Cell, E°cell, and Change in Gibb's Free Energy, DG
Gibb's Free Energy (G), is a measure of the spontaneity of a process. It is an indication that a process will proceed on its own without the interference of an external factor. Examples of spontaneous processes include:
a) the melting of ice
b) the crying of a new born baby
c) the flow of water from a highland to a lowland
d) the flow of electrons from the anode to the cathode
e) the flow of current from a region of higher potential to that of lower potential
To show that a process is spontaneous and feasible, the value of DG (read as delta G or change in free energy), must be negative (DG < 0).
The relationship between free energy and the e.m.f of a cell is given as:
DG° = -nFE°cell .................(vii)
where n = number of electrons transferred in the redox reaction
F = Faraday's constant
= 96500Cmol^-1
E°cell = standard e.m.f of the cell
From equation (vii) above, for the process to be spontaneous, the e.m.f of the cell must be positive.
Therefore, using the e.m.f values of the electrochemical cells in our examples above, let us see how to determine the spontaneity of a redox process.
Example 1.1
If the electrode potential of copper is +0.34V and that of zinc is -0.76V, calculate the e.m.f of the cell. Hence or otherwise, determine the spontaneity of the cell.
Answer
From our previous solution, the e.m.f, E°cell = 1.1V
Zn(s) ----> Zn2+(aq) + 2e- (oxidation half-equation)
Cu2+(aq) + 2e- ----> Cu(s) (reduction half-equation)
Zn(s) + Cu2+(aq) ----> Zn2+(aq) + Cu(s) (overall/net reaction)
From the above reactions, the number of moles of electrons transferred = 2 moles
Therefore, substituting these values into equation (vii), we will obtain:
DG° = -nFE°cell
DG° = -2 x 96500 x 1.1
= -212300 J
= -212.3 kJ
Since DG° < 0, it means the redox reaction is very spontaneous (due to the high negative value) and the cell will work as arranged. That is to say, electrons will flow from the zinc electrode (anode) to the copper electrode (cathode).
Example 2.1
Calculate the standard cell potential of the cell below and comment on the cell reaction:
Mg(s)/Mg2+(aq)//Ca2+(aq)/Ca(s)
Answer
From the earlier solution, the e.m.f, E°cell = -0.5V
Mg(s) ----> Mg2+(aq) + 2e- (oxidation half-equation)
Ca2+(aq) + 2e- ----> Ca(s) (reduction half-equation)
Mg(s) + Ca2+(aq) ----> Mg2+(aq) + Ca(s) (overall/net reaction)
Number of moles of electrons transferred = 2 moles
Therefore, substituting these values into equation (vii), we will obtain:
DG° = -nFE°cell
DG° = -2 x 96500 x -0.5
= +96500 J
= +96.5 kJ
From the positive value of DG (DG > 0), it means the redox reaction is not spontaneous, and the cell will not work. In other words, magnesium cannot act as an anode in the presence of calcium, because electrons will never flow from the former to the latter by virtue of their positions in the electromotive series. Therefore, to make the cell work, the electrodes must be rearranged.
Ca(s)/Ca2+(aq)//Mg2+(aq)/Mg(s)
E°cell = E°cathode - E°anode
= -2.37 - (-2.87)
= -2.37 + 2.87
= +0.5V
Hence,
DG° = -96.5kJ
Example 3.1
Given the following reactions:
Fe(s) ----> Fe2+(aq) + 2e-; E° = +0.44V
Ag+(aq) + e- ----> Ag(s); E° = -0.80V
Determine whether the electrochemical cell will work based on the above arrangement.
Answer
From the earlier solution,
the e.m.f of the cell, E°cell = +1.24V
the number of moles of electrons transferred = 2 moles
But,
DG° = -nFE°cell
Hence,
DG° = -2 x 96500 x 1.24
= -239320 J
= -239.3 kJ
Since DG < 0, it means the redox reaction is spontaneous, and the electrochemical cell will work based on the stated arrangement.
Relationship between the e.m.f of a Cell, E°cell & Equilibrium Constant, K
Consider a redox reaction that is at equilibrium and represented by the equation:
aA(s) + bB+(aq) <----> aA+(aq) + bB(s)
The equilibrium constant, Keq, of the reaction will be given as:
Keq = [Products]^n/[Reactants]^m
Keq = [A+(aq)]^a.[B(s)]^b/[A(s)]^a.[B+(aq)]^b
but,
The concentration of a solid substance at equilibrium is always taken as a unitary value(i.e, 1) in the presence of aqueous substances.
So, the above equation reduces to:
Keq = [A+(aq)]^a/[B+(aq)]^b ............(viii)
Keq = [Anode(aq)]^a/[Cathode(aq)]^b ............(ix)
The change in Gibb's free energy, DG, is related to the equilibrium constant, K, according to the equation:
DG° = -RT ln Keq .............................(x)
(but, ln (natural log) = 2.303 log)
DG° = - 2.303 RT log Keq ...............(xi)
Substitute equation (ix) into (xi)
DG° = - 2.303 RT log [Anode(aq)]^a/[Cathode(aq)]^b ............(xii)
Equate (vii) and (xii)
-nFE°cell = - 2.303 RT log [Anode(aq)]^a/[Cathode(aq)]^b
Divide both sides by -nF
E°cell = (2.303RT/nF) log [Anode(aq)]^a/[Cathode(aq)]^b ..........(xiii)
where R = gas constant (8.314 J mol^-1 K^-1)
T = standard temperature in Kelvin (25°C = 298K)
n = number of moles of electrons transferred
F = Faraday's constant
a = number of moles of the anodic electrolyte
b = number of moles of the cathodic electrolyte
[Anode(aq)] = concentration of the anodic electrolyte
[Cathode(aq)] = concentration of the cathodic electrolyte
Equation (xiii) gives the relationship between the e.m.f of a cell and the concentrations of the electrolytes at standard conditions.
Recall that in the course of using an electrochemical cell, its e.m.f decreases. This is due to a decrease in the concentrations of the electrolytes. Therefore, we can calculate the e.m.f of a cell (or the electrode potential of an electrode) at any concentration other than the standard concentration of 1.0 mol dm^-3, by using the equation:
Ecell = E°cell - (2.303RT/nF) log [Anode(aq)]^a/[Cathode(aq)]^b ...
.....(xiv)
The above equation is known as the Nernst equation, and it shows the relationship between the e.m.f of a cell (at non-standard concentration), Ecell, the standard e.m.f of the cell, E°cell and the concentrations of the electrolytes.
Example 4
Calculate the cell e.m.f of Al(s)/Al3+(aq)(0.010 mol dm^-3)//Cu2+(aq)/Cu(s)(0.100 mol dm^-3) at 25°C [E°al = +1.66V, E°cu = +0.34V]
Answer
From the arrangement, aluminum is oxidized (anode) and copper is reduced (cathode). So, the given standard electrode potential for Al is the SOP, while that for Cu is the SRP.
2Al(s) ----> 2Al3+(aq) + 6e- (oxidation half-reaction)
3Cu2+(aq) + 6e- ----> 3Cu(s) (reduction half-reaction)
2Al(s) + 3Cu2+(aq) ----> 2Al3+(aq) + 3Cu(s) (overall redox reaction)
Also, the concentrations of the electrolytes indicate that the required cell e.m.f is Ecell, and not the standard cell e.m.f, E°cell. Therefore, we apply the Nernst equation:
Ecell = E°cell - (2.303RT/nF) log [Al3+(aq)]^2/[Cu2+(aq)]^3
but,
E°cell = E°cathode + E°anode
= + 0.34 + 1.66
= + 2.00V
Number of moles of electrons transferred, n = 6 mol
Hence,
Ecell = 2.00 - (2.303x8.314x298/ 6x96500) log (0.010)^2/(0.100)^3
= 2.00 - (5705.8483/579000) log (1x10^-4/1x10^-3)
= 2.00 - 0.009855 log 10^-1
= 2.00 - 0.009855(-1)
= 2.00 + 0.009855
= 2.009855
~ 2.01V
Do These:
Question 1
Calculate the e.m.f of the cell:
Zn(s)/Zn2+(aq)(0.002 mol dm^-3)//Cu2+(aq)/Cu(s)(0.02 mol dm^-3) at 25°C [E°zn = +0.76V, E°cu = +0.34V]
Question 2
What will be the concentration of the electrolyte in the copper half-cell in Example 4 above, if the e.m.f of the cell, Ecell, is 2.06V and the concentration of the electrolyte in the aluminium half-cell is 0.001mol dm^-3 at 25°C?
Question 3
If the overall reaction in an electrochemical cell is given by:
Mg(s) + Fe2+(aq) ----> Mg2+(aq) + Fe(s), calculate the standard e.m.f of the cell.
Question 4
An electrochemical cell was constructed by combining the two half-cells below:
Cu2+(aq) + 2e- ----> Cu(s); E° = +0.34V
Fe2+(aq) + 2e- ----> Fe(s); E° = -0.44V
What is the standard e.m.f of the cell?
Question 5
Ca(s) ----> Ca2+(aq) + 2e-; E° = +2.87V
2Ag(s) ----> 2Ag+(aq) + 2e-; E° = -0.80V. Calculate the standard electrode potential of the calcium-silver cell represented by the half-cell reactions above.
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Explain the relationship between the electrode potential of a reference electrode and the ionic concentration of a solution in which is immersed. E=E=RT/f in c ( Nerst equation)
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