The connection between cell potential, Gibbs free energy and constant equilibrium are directly related in the following multi-part equation:
\[ \Delta G^o= -RT\ln K_{eq} = -nFE^o_{cell} \]
∆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 E_{a}(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 }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} or E_{cell} = E_{cathode} + E_{anode}
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.
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, Eº_{cell}, the E_{cell} is the non-standard state cell potential, which means that it is not determined under a concentration of 1 Molar 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.
\[ E_{cell}= E^o_{cell} -\dfrac{RT}{nF} ln\; Q \]
Other simplified forms of the equation that we typically see:
\[ E_{cell}= E^o_{cell} -\dfrac{0.0257}{n} ln \; Q \]
or in terms of log base 10, instead of the natural logarithm (base e)
\[ E_{cell}= E^o_{cell} - \dfrac{0.0592}{n} log_{10}\; Q \]
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 |
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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) E_{cell}= ? SOLUTION To use the Nersnt 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 = E^{o}_{Ag}_{/Ag } - E^{o}_{Fe3}_{+/Fe2+ } = 0.800V-0.771V = 0.029V Now to determine Ecell for the reaction Fe^{2}^{+}(0.1M) + Ag^{+}(1.0M) → Fe^{3}^{+}(0.20M) + Ag(s) Use the Nernst equation E_{cell}= 0.029V -(0.0592V/1)log [Fe^{3+}]/[Fe^{2}^{+}][Ag] =0.029V - 0.0592V*log [0.2]/[0.1]*[1.0] =0.011V |
K is the equilibrium constant of a reaction and is given by the reaction quotient:
\[ aA + bB \leftrightharpoons cC + dD \]
\[ K= \dfrac{[A]^a[B]^b}{[C]^c[D]^d} \]
The relationship between ∆G, K, and E° cell can be represented by the following diagram.
where
E°_{cell }can be calculated using the following formula:
E°_{cell }= E° (cathode) – E° (anode) = E°(Reduction) – E°(Oxidation)
Example 2 |
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Question Find the E° cell for the following coupled half-reactions SOLUTION 1. Determine the cathode and anode in the reaction Zn_{(s) }↔ Zn^{2}^{+}_{(aq)} + 2e^{- }Anode, Oxidation (since Zn_{(s)} increase oxidation state from 0 to +2) Cu^{2}^{+}_{(aq)} + 2e^{- }↔ Cu_{(s) }Cathode, Reduction (since Cu^{2}^{+}_{(aq)} decreases oxidation state from +2 to 0) 2. Determine the E° cell values using the standard reduction potential table Zn_{(s) }↔ Zn^{2}^{+}_{(aq)} + 2e^{- } -0.763 Cu^{2}^{+}_{(aq)} + 2e^{- }↔ Cu_{(s) }+0.340 3. Use E° cell = E°cathode - E°anode = 0.340 - (-0.763) = 1.103 V |
1. Find E° cell for 2Br^{-}_{(aq)} + I_{2}_{(s) }↔ Br_{2}_{(s)} + 2I^{-}_{(aq)}
2. Find E° for Sn_{(s) }↔ Sn^{2}^{+}_{(aq) }+ 2e^{-}
3. Find E° cell for Zn_{(s)} | Zn^{2}^{+}_{(aq)} || Cr^{3}^{+}_{(aq)} , Cr^{2}^{+}_{(aq)}
(See answer key for solutions)
*If the E° values of the reaction is negative, then the reaction is NOT spontaneous and therefore the reverse reaction is occuring and the electrons are flowing in the opposite direction.
* These values are tabulated in the standard reduction potential table.
E^{o} cell | Delta G | Q & K Relationship | Reaction Direction | Spontaneity (as written) |
>0 | - | Q<K | Forward | Spontaneous |
<0 | + | Q>K | Backward | Non-spontaneous |
=0 | =0 | Q=K | No Reaction | N/A |
Example: Using ∆G=-RTlnK |
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Given K = 2.81x10^{-16} for a following reaction Cu^{2}^{+}_{(aq)} + Ag_{(s)} ↔ Cu_{(s)} + 2Ag^{+, } find ∆G. Solution: Use the following formula: ∆G=-RTlnK = 8.314 x 298 x ln(2.81x10^{-16}) = -8.87x10^{5} = 8.871 kJ |
Example: Using ∆G=-nFE°_{cell} |
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Find ∆G for the following reaction: 2Al_{(s)} +3Br_{2}_{(l)} ↔ 2Al^{3}^{+}_{(aq, 1M)} + 6Br^{-}_{(aq, 1M)}
Solution: Step 1: Separate the reaction into its two half reactions 2Al_{(s)} ↔ 2Al^{3}^{+(aq)} 3Br^{2}_{(l)} ↔ 6Br^{-}_{(aq, 1M)} Step 2: Balance the half equations using O, H, and charge using e- 2Al_{(s)} ↔ 2Al^{3}^{+(aq)} +6e^{-} 6e^{-} + 3Br^{2}_{(l)} ↔ 6Br^{-}_{(aq)} Step 3: From the balanced half reactions, we can conclude the number of moles of e^{-} for use later in the calculation of ∆G. Determine the E° values using the standard reduction potentials, using the E° cell table. 2Al_{2}_{(s)} ↔ Al^{3}^{+}_{(aq)} +6e- -1.676V 3Br_{2}_{(l)} + 6e^{- }↔ 6Br^{-}_{(aq)} +1.065V Step 4: Determine E°_{cell }= E°_{cathode }- E°_{anode}. = 1.065 - (-1.676) = 2.741 V Step 5: Once E° cell has be calculated and the number of moles of electrons have been determined, we can use ∆G = -nFE°_{cell} = (-6 mol e^{-})(96458 C/mol e^{-})(2.741 V) = -1586kJ This equation can be used to calculate E° cell given K or K given E°_{cell}. If T=298 K, the RT is a constant then the following equation can be used: E°_{cell}= (0.025693V/n) ln K |
Example: Using E° cell=(RT/nF) lnK |
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Given the E° cell for the reaction Cu_{(s)} + 2H^{+}_{(aq)} ↔ Cu^{2}^{+}_{(aq)} +H_{2}_{(g)} is -0.34V, find the equilibrium constant (K) for the reaction. SOLUTION Step 1: Split into two half reaction Cu_{(s)} ↔ Cu^{2}^{+}_{(aq)} 2H^{+}_{(aq)}↔ H_{2}_{(g)} Step 2: Balance the half reactions with charges to determine n Cu_{(s)} ↔ Cu^{2}^{+}_{(aq)} + 2e^{-} 2H^{+}_{(aq)} +2e^{-} ↔ H_{2}_{(g)} Therefore n=2 Step 3: From the example above, E° cell = -0.34V -0.34 = (0.025693/2) lnK K = e^{(-0.34 x 2/0.025693) } K = 3.19 x 10^{-12} |
Example 1: I see you |
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Given the following reaction determine ∆G, K, and E^{o}_{cell}for the following reaction at standard conditions? Is this reaction spontaneous? Mn^{2}^{+}_{(aq)} + K_{(s)} ↔ MnO_{2}_{(s)} + K_{(aq)} SOLUTION Step 1: Separate and balance the half reactions. Label which one is reduction and which one is oxidation. Find the corresponding E° values for the half reactions. MnO_{2}_{(s)} + 4H^{+}_{(aq)} + 2e ↔ ^{-}Mn^{2}^{+}_{(aq)} + 2H_{2}O_{(l)} Reduction +1.23V K^{+}_{(aq)} + e^{-} ↔ K_{(s)} Oxidation -2.92V Step 2: Write net balanced reaction in acidic solution, and determine the E° cell. E° cell = E°_{cathode}- E°_{anode} = +1.23 - (-2.92) = 4.15 Mn^{2}^{+}_{(aq)} + 2K^{+}_{(aq)} + 2H_{2}O_{(l)} ↔ MnO_{2}_{(s)} + 4H^{+}_{(aq)} + K_{(s)} E° cell = 4.15V Step 3: Find ∆G for the reaction. Use ∆G = -nFE°_{cell} = -2 mol e^{-} x 96458 C x 4.15 = -800.60kJ Therefore, since E^{o}cell is positive and ∆G is negative, this reaction is spontaneous. |
4. Find ∆G for the following combined half reactions:
F^{2}_{(g)} + 2e^{-} ↔ 2F^{-}_{(aq)}
Li^{+}_{(aq)} + e^{-} ↔ Li_{(s)}
5. Find the equilibrium constant (K) for the following reaction: (Hint: Find E°_{Cell} first!)
Zn_{(s)} + 2H^{+}_{(aq)} ↔ Zn^{2}^{+}_{(aq)} + H_{2}_{(g)}
6. Find E°_{Cell }for the given reaction at standard conditions:
Cu^{+}_{(aq)} + e^{-} ↔ Cu_{(s)}
I_{2}_{(s)} + 2e^{-} ↔ 2I^{-}_{(aq)}
(See answer key for solutions)
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