Electrochemical Potential and Free Energy We will now extend electrochemistry by determining the relationship between \(E^\circ_\ce{cell}\) and the thermodynamics quantities such as Δ G ° (Gibbs free energy) and K (the equilibrium constant). In spontaneous redox reactions, chemical energy is converted into electrical energy, which can do work. The electrical work is the product of the charge transferred multiplied by the potential difference (voltage): \[\mathrm{electrical\: work=volts \times (charge\: in\: coulombs)=J} \nonumber \] The charge on 1 mole of electrons is given by Faraday’s constant ( F ) \[ \begin{align*} F &=\dfrac{6.022 \times 10^{23}\:e^-}{mol} \times \dfrac{1.602 \times 10^{−19}\:C}{e^-} \\[4pt] &=9.648 \times 10^4\:\dfrac{C}{mol} \\[4pt] &=9.648 \times 10^4\:\dfrac{J}{V⋅mol} \end{align*} \nonumber \] Therefore \[\mathrm{total\: charge=(number\: of\: moles\: of\: e^-)} \times F=nF \nonumber \] In this equation, \(n\) is the number of moles of electrons for the balanced oxidation-reduction reaction. The measured cell potential is the maximum potential the cell can produce and is related to the electrical work ( w ele ) by \[E_\ce{cell}=\dfrac{−w_\ce{ele}}{nF}\hspace{40px}\ce{or}\hspace{40px}w_\ce{ele}=−nFE_\ce{cell} \nonumber \] The negative sign for the work indicates that the electrical work is done by the system (the galvanic cell) on the surroundings. Free energy is defined as the energy that is available to do work. In particular, the change in free energy is defined in terms of the maximum work (\(w_{max}\)), which is \(w_{ele}\) for electrochemical systems. \[\begin{align*} ΔG&=w_\ce{max}=w_\ce{ele} \\[4pt] &=−nFE_\ce{cell} \end{align*} \nonumber \] We can verify the signs are correct when we realize that \(n\) and \(F\) are positive constants and that galvanic cells, which have positive cell potentials, involve spontaneous reactions. Thus, spontaneous reactions, which have \(ΔG < 0\), must have \(E_{cell} > 0\). If all the reactants and products are in their standard states, this becomes \[ΔG°=−nFE^\circ_\ce{cell} \nonumber \] Although the sign of \(E^o_{cell}\) tells us whether a particular redox reaction will occur spontaneously under standard conditions, it does not tell us to what extent the reaction proceeds, and it does not tell us what will happen under nonstandard conditions. To answer these questions requires a more quantitative understanding of the relationship between electrochemical cell potential and equilibrium constants. Since \[ΔG°=−RT\ln K \nonumber \] \[−nFE^\circ_\ce{cell}=−RT\ln K \nonumber \] or \[ E^\circ_\ce{cell}=\dfrac{RT}{nF}\ln K \nonumber \] Most of the time, the electrochemical reactions are run at standard temperature (298.15 K). Collecting terms at this temperature yields \[\begin{align*} E^\circ_\ce{cell}&=\dfrac{RT}{nF}\:\ln K \\[4pt] &=\dfrac{\left(8.314\:\dfrac{\ce{J}}{\textrm{K⋅mol}}\right)(298.15\:K)}{n \times 96,485\: \textrm{C/V⋅mol}}\:\ln K \\[4pt] &=\dfrac{\mathrm{0.0257\: V}}{n}\:\ln K \end{align*} \nonumber \] where \(n\) is the number of moles of electrons. The logarithm in equations involving cell potentials is often expressed using base 10 logarithms (i.e., \(\log_{10}\) or just \(\log\)), which changes the constant by a factor of 2.303: \[E^\circ_\ce{cell}=\dfrac{\mathrm{0.0592\: V}}{n}\:\log K \nonumber \] Thus, if Δ G °, K , or \(E^\circ_\ce{cell}\) is known or can be calculated, the other two quantities can be readily determined. The relationships are shown graphically in Figure \(\PageIndex{1}\). Given any one of the quantities, the other two can be calculated. Example \(\PageIndex{1}\): Equilibrium Constants, Potentials, & Free Energy Changes What is the standard Gibbs free energy change and equilibrium constant for the following reaction at 25 °C? \[\ce{2Ag+}(aq)+\ce{Fe}(s)⇌\ce{2Ag}(s)+\ce{Fe^2+}(aq) \nonumber \] So lution The reaction involves an oxidation-reduction reaction, so the standard cell potential can be calculated using the data in Table P1. \[\begin{align*} &\textrm{anode (oxidation): }\ce{Fe}(s)⟶\ce{Fe^2+}(aq)+\ce{2e-} \hspace{40px} E^\circ_{\ce{Fe^2+/Fe}}=\mathrm{−0.447\: V} \nonumber\\ &\textrm{cathode (reduction): }2 \times (\ce{Ag+}(aq)+\ce{e-}⟶\ce{Ag}(s)) \hspace{40px} E^\circ_{\ce{Ag+/Ag}}=\mathrm{0.7996\: V} \nonumber\\ &E^\circ_\ce{cell}=E^\circ_\ce{cathode}−E^\circ_\ce{anode}=E^\circ_{\ce{Ag+/Ag}}−E^\circ_{\ce{Fe^2+/Fe}}=\mathrm{+1.247\: V} \nonumber \end{align*} \nonumber \] Remember that the cell potential for the cathode is not multiplied by two when determining the standard cell potential. With n = 2, the equilibrium constant is then \[E^\circ_\ce{cell}=\dfrac{\mathrm{0.0592\: V}}{n}\:\log K \nonumber \] \[\begin{align*} K&=10^{n \times E^\circ_\ce{cell}/\mathrm{0.0592\: V}} \\[4pt] &=10^{2 \times \mathrm{1.247\: V/0.0592\: V}} \\[4pt] &=10^{42.128} \\[4pt] &=1.3 \times 10^{42}\end{align*} \nonumber \] The standard free energy is then \[ΔG°=−nFE^\circ_\ce{cell} \nonumber \] \[ΔG°=\mathrm{−2 \times 96,485\:\dfrac{J}{\textrm{V⋅mol}} \times 1.247\: V=−240.6\:\dfrac{kJ}{mol}} \nonumber \] Check your answer: A positive standard cell potential means a spontaneous reaction, so the standard free energy change should be negative, and an equilibrium constant should be >1. The Nernst Equation Thus far, we have addressed electrochemical reactions that occur under standard conditions. The Nernst Equation will be introduced to address reactions that occur under nonstandard Conditions. Recall that \[ΔG=ΔG°+RT\ln Q \nonumber \] where \(Q\) is the reaction quotient (see the chapter on equilibrium fundamentals). Converting to cell potentials: \[−nFE_\ce{cell}=−nFE^\circ_\ce{cell}+RT\ln Q \label{nernst1A} \] or \[E_\ce{cell}=E^\circ_\ce{cell}−\dfrac{RT}{nF}\:\ln Q \label{nernst1B} \] Equation \ref{nernst1B} is the generalized Nernst equation that is applicable at any temperature . However, is can be simplified for reactions occuring at 25 °C (298.15 K) by rewriting it as \[E_\ce{cell}=E^\circ_\ce{cell}−\dfrac{\mathrm{0.0257\: V}}{n}\:\ln Q \label{nernst2A} \] or \[ E_\ce{cell}=E^\circ_\ce{cell}−\dfrac{\mathrm{0.0592\: V}}{n}\log_{10} Q \label{nernst2B} \] If the temperature is not 298.15 K, it is necessary to recalculate the potential with Equation\ref{nernst1B}. With the Nernst equation, it is possible to calculate the cell potential at nonstandard conditions. This adjustment is necessary because potentials determined under different conditions will have different values. Example \(\PageIndex{2}\): Cell Potentials at Nonstandard Conditions Consider the following reaction at room temperature: \[\ce{Co}(s)+\ce{Fe^2+}(aq,\:1.94\:M)⟶\ce{Co^2+}(aq,\: 0.15\:M)+\ce{Fe}(s) \nonumber \] Is the process spontaneous? Solution There are two ways to solve the problem. If the thermodynamic information in Table T1 were available, you could calculate the free energy change. If the free energy change is negative, the process is spontaneous. The other approach, which we will use, requires information like that given in Table P1. Using those data, the cell potential can be determined. If the cell potential is positive, the process is spontaneous. Collecting information from Table P1 and the problem, \[\begin{align*} &\textrm{Anode (oxidation): }\ce{Co}(s)⟶\ce{Co^2+}(aq)+\ce{2e-} \hspace{40px} E^\circ_{\ce{Co^2+/Co}}=\mathrm{−0.28\: V}\\ &\textrm{Cathode (reduction): }\ce{Fe^2+}(aq)+\ce{2e-}⟶\ce{Fe}(s) \hspace{40px} E^\circ_{\ce{Fe^2+/Fe}}=\mathrm{−0.447\: V}\\ &E^\circ_\ce{cell}=E^\circ_\ce{cathode}−E^\circ_\ce{anode}=\mathrm{−0.447\: V−(−0.28\: V)=−0.17\: V} \end{align*} \nonumber \] The process is not spontaneous under standard conditions. Using the Nernst equation and the concentrations stated in the problem and \(n = 2\), \[Q=\ce{\dfrac{[Co^2+]}{[Fe^2+]}}=\dfrac{0.15\:M}{1.94\:M}=0.077 \nonumber \] Now we can insert these into the Nernst Equation at room temperature (Equation \ref{nernst2B}) \[\begin{align*} E_\ce{cell} &=E^\circ_\ce{cell}−\dfrac{\mathrm{0.0592\: V}}{n}\:\log Q \\[4pt] &=\mathrm{−0.17\: V−\dfrac{0.0592\: V}{2}\:\log 0.077} \\[4pt] &=\mathrm{−0.17\: V+0.033\: V=−0.14\: V} \end{align*} \nonumber \] The process is (still) nonspontaneous. Finally, we will take a brief look at a special type of cell called a concentration cell . In a concentration cell, the electrodes are the same material and the half-cells differ only in concentration. Since one or both compartments is not standard, the cell potentials will be unequal; therefore, there will be a potential difference, which can be determined with the aid of the Nernst equation. Example \(\PageIndex{3}\): Concentration Cells What is the cell potential of the concentration cell described by \[\ce{Zn}(s)│\ce{Zn^2+}(aq,\: 0.10\:M)║\ce{Zn^2+}(aq,\: 0.50\:M)│\ce{Zn}(s) \nonumber \] So lution From the information given: \[\begin{align*} &\textrm{Anode: }\ce{Zn}(s)⟶\ce{Zn^2+}(aq,\: 0.10\:M)+\ce{2e-} \hspace{40px} E^\circ_\ce{anode}=\mathrm{−0.7618\: V}\\ &\textrm{Cathode: }\ce{Zn^2+}(aq,\: 0.50\:M)+\ce{2e-}⟶\ce{Zn}(s) \hspace{40px} E^\circ_\ce{cathode}=\mathrm{−0.7618\: V}\\ &\overline{\textrm{Overall: }\ce{Zn^2+}(aq,\: 0.50\:M)⟶\ce{Zn^2+}(aq,\: 0.10\:M) \hspace{40px} E^\circ_\ce{cell}=\mathrm{0.000\: V}} \end{align*} \nonumber \] The standard cell potential is zero because the anode and cathode involve the same reaction; only the concentration of Zn 2 + changes. Substituting into the Nernst equation, \[E_\ce{cell}=\mathrm{0.000\: V−\dfrac{0.0592\: V}{2}\:\log\dfrac{0.10}{0.50}=+0.021\: V} \nonumber \] and the process is spontaneous at these conditions. Check your answer: In a concentration cell, the standard cell potential will always be zero. To get a positive cell potential (spontaneous process) the reaction quotient Q must be <1. Q < 1 in this case, so the process is spontaneous. Problems Exercise \(\PageIndex{1}\) What is the standard Gibbs free energy change and the equilibrium constant for the following reaction at room temperature? Is the reaction spontaneous? \[\ce{Sn}(s)+\ce{2Cu^2+}(aq)⇌\ce{Sn^2+}(aq)+\ce{2Cu+}(aq) \nonumber \] Answer Spontaneous; n = 2; \(E^\circ_\ce{cell}=\mathrm{+0.291\: V}\); \(ΔG°=\mathrm{−56.2\:\dfrac{kJ}{mol}}\); \(K = 6.8 \times 10^9\). Exercise \(\PageIndex{2}\) What is the cell potential for the following reaction at room temperature? \[\ce{Al}(s)│\ce{Al^3+}(aq,\:0.15\:M)║\ce{Cu^2+}(aq,\:0.025\:M)│\ce{Cu}(s) \nonumber \] What are the values of n and Q for the overall reaction? Is the reaction spontaneous under these conditions? Answer n = 6; Q = 1440; E cell = +1.97 V, spontaneous. Exercise \(\PageIndex{3}\) What value of Q for the previous concentration cell would result in a voltage of 0.10 V? If the concentration of zinc ion at the cathode was 0.50 M , what was the concentration at the anode? Answer Q = 0.00042; [Zn 2 + ] cat = 2.1 \times 10 −4 M . Exercise \(\PageIndex{4}\): Silver Sulfide The black tarnish that forms on silver objects is primarily \(\ce{Ag2S}\). The half-reaction for reversing the tarnishing process is as follows: \[\ce{Ag2S(s) + 2e^{−} → 2Ag(s) + S^{2−} (aq)} \quad E°=−0.69\, V \nonumber \] Referring to Table \(\PageIndex{1}\), predict which species—H 2 O 2 (aq), Zn(s), I − (aq), Sn 2 + (aq)—can reduce Ag 2 S to Ag under standard conditions. Of these species—H 2 O 2 (aq), Zn(s), I − (aq), Sn 2 + (aq), identify which is the strongest reducing agent in aqueous solution and thus the best candidate for a commercial product. From the data in Table \(\PageIndex{1}\), suggest an alternative reducing agent that is readily available, inexpensive, and possibly more effective at removing tarnish. Answer We can solve the problem in one of two ways: (1) compare the relative positions of the four possible reductants with that of the Ag 2 S/Ag couple in Table \(\PageIndex{1}\) or (2) compare E° for each species with E° for the Ag 2 S/Ag couple (−0.69 V). The species in Table \(\PageIndex{1}\) are arranged from top to bottom in order of increasing reducing strength. Of the four species given in the problem, I − (aq), Sn 2 + (aq), and H 2 O 2 (aq) lie above Ag 2 S, and one [Zn(s)] lies below it. We can therefore conclude that Zn(s) can reduce Ag 2 S(s) under standard conditions, whereas I − (aq), Sn 2 + (aq), and H 2 O 2 (aq) cannot. Sn 2 + (aq) and H 2 O 2 (aq) appear twice in the table: on the left side (oxidant) in one half-reaction and on the right side (reductant) in another. The strongest reductant is Zn(s), the species on the right side of the half-reaction that lies closer to the bottom of Table \(\PageIndex{1}\) than the half-reactions involving I − (aq), Sn 2 + (aq), and H 2 O 2 (aq). (Commercial products that use a piece of zinc are often marketed as a “miracle product” for removing tarnish from silver. All that is required is to add warm water and salt for electrical conductivity.) Of the reductants that lie below Zn(s) in Table \(\PageIndex{1}\), and therefore are stronger reductants, only one is commonly available in household products: Al(s), which is sold as aluminum foil for wrapping foods. Exercise \(\PageIndex{5}\) Use the data in Table \(\PageIndex{1}\) to determine whether each reaction is likely to occur spontaneously under standard conditions: \(\ce{Sn(s) + Be^{2+}(aq) → Sn^{2+}(aq) + Be(s)}\) Answer Metallic tin is oxidized to \(\ce{Sn^{2+}(aq)}\), and \(\ce{Be^{2+}(aq)}\) is reduced to elemental beryllium. We can find the standard electrode potentials for the latter (reduction) half-reaction (−1.99 V) and for the former (oxidation) half-reaction (−0.14 V) directly from Table \(\PageIndex{1}\). Adding the two half-reactions gives the overall reaction: \[\begin{align*}\textrm{cathode} &: \ce{Be^{2+}(aq)} +\ce{2e^-} \rightarrow \ce{Be(s)} \\ \textrm{anode} &: \ce{Sn(s)} \rightarrow \ce{Sn^{2+}(s)} +\ce{2e^-} \\ \hline \textrm{overall} &: \ce{Sn(s)} + \ce{Be^{2+}(aq)} \rightarrow \ce{Sn^{2+}(aq)} + \ce{Be(s)} \end{align*} \] with \[\begin{align*} E^\circ_{\textrm{cathode}} &=\textrm{-1.99 V} \\[4pt] E^\circ_{\textrm{anode}} &=\textrm{-0.14 V} \\[4pt] E^\circ_{\textrm{cell}} &=E^\circ_{\textrm{cathode}}-E^\circ_{\textrm{anode}} \\[4pt] &=-\textrm{1.85 V} \end{align*} \] The standard cell potential is quite negative, so the reaction will not occur spontaneously as written. That is, metallic tin cannot reduce Be 2 + to beryllium metal under standard conditions. Instead, the reverse process, the reduction of stannous ions (Sn 2 + ) by metallic beryllium, which has a positive value of E° cell , will occur spontaneously. Exercise \(\PageIndex{6}\) Use the data in Table \(\PageIndex{1}\) to determine whether each reaction is likely to occur spontaneously under standard conditions: \(\ce{MnO2(s) + H2O2(aq) + 2H^{+}(aq) → O2(g) + Mn^{2+}(aq) + 2H2O(l)}\) Answer \(\ce{MnO2}\) is the oxidant (\(\ce{Mn^{4+}}\) is reduced to \(\ce{Mn^{2+}}\)), while \(\ce{H2O2}\) is the reductant (\(\ce{O^{2−}}\) is oxidized to \(\ce{O2}\)). We can obtain the standard electrode potentials for the reduction and oxidation half-reactions directly from Table \(\PageIndex{1}\). The two half-reactions and their corresponding potentials are as follows: \[\begin{align*}\textrm{cathode} &: \ce{MnO_2(s)}+\ce{4H^+(aq)}+\ce{2e^-} \rightarrow\ce{Mn^{2+}(aq)}+\ce{2H_2O(l)} \\ \ce{anode} &: \ce{H_2O_2(aq)}\rightarrow\ce{O_2(g)}+\ce{2H^+(aq)}+\ce{2e^-} \\\hline \textrm{overall} &: \ce{MnO_2(s)}+\ce{H_2O_2(aq)}+\ce{2H^+(aq)}\rightarrow\ce{O_2(g)}+\ce{Mn^{2+}(aq)}+\ce{2H_2O(l)} \end{align*} \] with \[\begin{align*} E^\circ_{\textrm{cathode}} &=\textrm{1.23 V} \\[4pt] E^\circ_{\textrm{anode}} &=\textrm{0.70 V} \\[4pt] E^\circ_{\textrm{cell}} &=E^\circ_{\textrm{cathode}}-E^\circ_{\textrm{anode}} \\[4pt] &=+\textrm{0.52 V} \end{align*} \] The standard potential for the reaction is positive, indicating that under standard conditions, it will occur spontaneously as written. Hydrogen peroxide will reduce \(\ce{MnO2}\), and oxygen gas will evolve from the solution. Exercise \(\PageIndex{7}\) Use the data in Table \(\PageIndex{1}\) to determine whether each reaction is likely to occur spontaneously under standard conditions: \(\ce{2Ce^{4+}(aq) + 2Cl^{−}(aq) → 2Ce^{3+}(aq) + Cl2(g)}\) \(\ce{4MnO2(s) + 3O2(g) + 4OH^{−}(aq) → 4MnO4^{−}(aq) + 2H2O}\) Answer a spontaneous with \(E^o_{cell} = 1.61\, V - 1.396\, V = 0.214\, V\) Answer b nonspontaneous with \(E^°_{cell} = −0.20\, V\)