Calculate Eo (standard cell potential), E (cell potential), and ΔG (free energy change) for the following cell reaction. Would the following reaction proceed spontaneously under the conditions given below? Please explain.
3Zn(s) + 2Cr3+(aq) → 3Zn2+ (aq) + 2Cr(s)
[Cr3+] = 0.010 M, [Zn2+] = 0.0085 M, T: 25oC
(Standart reduction potentials:
Eo (Zn2+/Zn): -0.76 V; Eo(Cr3+/Cr): -0.74 V )
(1 F = 96,500 C/mol (or J/V.mol); R = 8.314 J/mol.K)
Answers
Answer:
Learning Objectives
The connection between cell potential, Gibbs energy and constant equilibrium are directly related in the following multi-part equation:
\[ \Delta G^o= -RT\ln K_{eq} = -nFE^o_{cell} \]
∆G: Gibbs Energy
∆G is the change of Gibbs (free) energy for a system and ∆G° is the Gibbs energy change for a system under standard conditions (1 atm, 298K). On an energy diagram, ∆G can be represented as:
Where ∆G is the difference in the energy between reactants and products. In addition ∆G is unaffected by external factors that change the kinetics of the reaction. For example if Ea(activation energy) were to decrease in the presence of a catalyst or the kinetic energy of molecules increases due to a rise in temperature, the ∆G value would remain the same.
E°cell: Standard Cell Potential
E°cell is the electromotive force (also called cell voltage or cell potential) between two half-cells. The greater the E°cell of a reaction the greater the driving force of electrons through the system, the more likely the reaction will proceed (more spontaneous). E°cell is measured in volts (V). The overall voltage of the cell = the half-cell potential of the reduction reaction + the half-cell potential of the oxidation reaction. To simplify,
\[E_{cell} = E_{reduction} + E_{oxidation} \tag{1}\]
or
\[E_{cell} = E_{cathode} + E_{anode} \tag{2}\]
The potential of an oxidation reduction (loss of electron) is the negative of the potential for a reduction potential (gain of electron). Most tables only record the standard reduction half-reactions. In other words, most tables only record the standard reduction potential; so in order to find the standard oxidation potential, simply reverse the sing of the standard reduction potential.
Note
The more positive reduction potential of reduction reactions are more spontaneous. When viewing a cell reduction potential table, the higher the cell is on the table, the higher potential it has as an oxidizing agent.
Difference between Ecell and Eºcell
Eº cell is the standard state cell potential, which means that the value was determined under standard states. The standard states include a concentration of 1 Molar (mole per liter) and an atmospheric pressure of 1. Similar to the standard state cell potential, Eºcell, the Ecell is the non-standard state cell potential, which means that it is not determined under a concentration of 1 M and pressure of 1 atm. The two are closely related in the sense that the standard cell potential is used to calculate for the cell potential in many cases.
\[ E_{cell}= E^o_{cell} -\dfrac{RT}{nF} \ln\; Q \tag{3}\]
Other simplified forms of the equation that we typically see:
\[ E_{cell}= E^o_{cell} -\dfrac{0.0257}{n} \ln \; Q \tag{4}\]
or in terms of \(\log_{10}\) (base 10) instead of the natural logarithm (base e)
\[ E_{cell}= E^o_{cell} - \dfrac{0.0592}{n} \log_{10}\; Q \tag{5}\]
Both equations applies when the temperature is 25ºC. Deviations from 25ºC requires the use of the original equation. Essentially, Eº is E at standard conditions
Example 1
What is the value of Ecell for the voltaic cell below:
\[Pt_{(s)}|Fe^{2+}_{(0.1M)},Fe^{3+}_{(0.2M)}||Ag^+_{(0.1M)}|Ag_{(s)} \]
Solution
To use the Nernst equation, we need to establish \(E^o_{cell}\) and the reaction to which the cell diagram corresponds so that the form of the reaction quotient (Q) can be revealed. Once we have determined the form of the Nernst equation, we can insert the concentration of the species.
Solve:
\[ E^o_{cell} = E^o_{cathode}- E^o_{anode}\]
= EoAg/Ag - EoFe3+/Fe2+
= 0.800V-0.771V = 0.029V
Now to determine Ecell for the reaction
Fe2+(0.1M) + Ag+(1.0M) → Fe3+(0.20M) + Ag(s)
Use the Nernst equation
Ecell= 0.029V -(0.0592V/1)log [Fe3+]/[Fe2+][Ag]
=0.029V - 0.0592V*log [0.2]/[0.1]*[1.0]
=0.011V