# electrical work chemistry

Values of E° for half-reactions cannot be added to give E° for the sum of the half-reactions; only values of ΔG° = −nFE°cell for half-reactions can be added. However, because ΔG° is a state function, the sum of the ΔG° values for the individual reactions gives us ΔG° for the overall reaction, which is proportional to both the potential and the number of electrons ($$n$$) transferred. Solving the last expression for ΔG° for the overall half-reaction, $\Delta{G^°} = F[(−0.77 V) + (−2)(−0.45 V)] = F(0.13 V) \label{20.5.9} \nonumber$. To obtain the overall balanced chemical equation, we must multiply both sides of the oxidation half-reaction by 3 to obtain the same number of electrons as in the reduction half-reaction, remembering that the magnitude of E° is not affected: \begin{align*} This means that large equilibrium constants correspond to large positive values of $$E^°_{cell}$$ and vice versa. Charge is current multiplied by time so this equation can also be written as: V is the potential difference in volts, V. For example, how much energy is transferred when the potential difference is 120 V and the charge is 2 C? Thus the equilibrium lies far to the right, favoring a discharged battery (as anyone who has ever tried unsuccessfully to start a car after letting it sit for a long time will know). Unfortunately, these criteria apply only to systems in which all reactants and products are present in their standard states, a situation that is seldom encountered in the real world. In chemical reactions, however, we need to relate the coulomb to the charge on a mole of electrons. Electrical energy We use electricity to heat and light our homes. The work can be done by electrochemical devices (electrochemical cells) or different metals junctions generating an electromotive force. Consequently, there must be a relationship between the potential of an electrochemical cell and $$\Delta{G}$$; this relationship is as follows: \[\Delta{G} = −nFE_{cell} \label{20.5.4}. He discovered the phenomenon of electrolysis and laid the foundations of electrochemistry. & \textrm {overall:} \mathrm{Fe^{3+}(aq)}+\mathrm{3e^-}\rightarrow \mathrm{Fe(s)}\hspace{3mm} \Delta G^\circ & =[-(1)(F)(\textrm{0.77 V})]+[-(2)(F)(-\textrm{0.45 V})] \end{align*} \label{20.5.8}\]. Thus $$E^°_{cell}$$ is directly proportional to the logarithm of the equilibrium constant. His experiments in electricity and magnetism made electricity a routine tool in science and led to both the electric motor and the electric generator. A The relevant half-reactions and potentials from Table P2 are as follows: \begin{align*} & \textrm {cathode:} In fact, most of the specialized terms introduced in this chapter (electrode, anode, cathode, and so forth) are due to Faraday. Is the reaction spontaneous? Report your answer to two significant figures. We can use the relationship between $$\Delta{G^°}$$ and the equilibrium constant $$K$$, to obtain a relationship between $$E^°_{cell}$$ and $$K$$. Our team of exam survivors will get you started and keep you going. Adding together the ΔG values for the half-reactions gives ΔG for the overall reaction, which is proportional to both the potential and the number of electrons (n) transferred. Read about our approach to external linking. From Table P2, we can find the reduction and oxidation half-reactions and corresponding E° values: \[\begin{align*} In addition, he discovered benzene and invented the system of oxidation state numbers that we use today. The change in free energy (ΔG) is also a measure of the maximum amount of work that can be performed during a chemical process (ΔG = wmax). & \textrm{overall:} &\quad & \mathrm{Cr_2O_7^{2-}(aq)} + \mathrm{6Br^{-}(aq)} + \mathrm{14H^+(aq)} \rightarrow \mathrm{2Cr^{3+}(aq)} + \mathrm{3Br_2(aq)} +\mathrm{7H_2O(l)} potential difference of 120 volts. & & E^\circ_\textrm{cathode}=\textrm{1.69 V} \\ The product of the cell potential and the total charge is the maximum amount of energy available to do work, which is related to the change in free energy that occurs during the chemical process. For T = 298 K, Equation $$\ref{20.5.12}$$ can be simplified as follows: \[ \begin{align} E^\circ_{\textrm{cell}} &=\left(\dfrac{RT}{nF}\right)\ln K \\[4pt] &=\left[ \dfrac{[8.314\;\mathrm{J/(mol\cdot K})(\textrm{298 K})]}{n[96,485\;\mathrm{J/(V\cdot mol)}]}\right]2.303 \log K \\[4pt] &=\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log K \label{20.5.13} \end{align}. Electrical energy transferred is calculated by multiplying current, voltage and time. If $$ΔG^o$$ is negative, then the reaction is spontaneous. Large equilibrium constants correspond to large positive values of E°. Use the data in Table P2 to calculate the equilibrium constant for the reaction of $$\ce{Sn^{2+}(aq)}$$ with oxygen to produce $$\ce{Sn^{4+}(aq)}$$ and water under standard conditions. Because six electrons are transferred in the overall reaction, the value of $$n$$ is 6: \begin{align*}\Delta G^\circ &=-(n)(F)(E^\circ_{\textrm{cell}}) \\[4pt] & =-(\textrm{6 mole})[96,468\;\mathrm{J/(V\cdot mol})(\textrm{0.14 V})] \\& =-8.1 \times10^4\textrm{ J} \\ & =-81\;\mathrm{kJ/mol\;Cr_2O_7^{2-}} \end{align*}. F(\textrm{0.13 V}) & =-(3)(F)(E^\circ_{\textrm{cell}}) \\ For a given amount of electrical charge that moves, the amount of energy transferred increases as the K & =2.3\times10^{69}\end{align*}\]. Report your answer to two significant figures. Asked for: $$ΔG^o$$ for the reaction and spontaneity. &\quad & E^\circ_{\textrm{cell}} =\textrm{0.14 V} Have questions or comments? Then use Equation $$\ref{20.5.5}$$ to calculate $$ΔG^o$$. Using the data in Table P2, calculate the free-energy change (ΔG°) for this redox reaction under standard conditions. The current generated from a redox reaction is measured in amperes (A), where 1 A is defined as the flow of 1 C/s past a given point. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. Use the data in Table P2 to calculate $$ΔG^o$$ for the reduction of ferric ion by iodide: $\ce{2Fe^{3+}(aq) + 2I^{−}(aq) → 2Fe^{2+}(aq) + I2(s)}\nonumber$. & \textrm{anode:} &\quad & \mathrm{6Br^{-}(aq)} \rightarrow \mathrm{3Br_2(aq)} +\mathrm{6e^-} Write the relevant half-reactions and potentials. & & \mathrm{PbO_2(s)}+\mathrm{SO_4^{2-}(aq)}+\mathrm{4H^+(aq)}+\mathrm{2e^-}\rightarrow\mathrm{PbSO_4(s)}+\mathrm{2H_2O(l)} Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. Solving Equation $$\ref{20.5.13}$$ for log K and inserting the values of $$n$$ and $$E^o$$, \begin{align*}\log K & =\dfrac{nE^\circ}{\textrm{0.0591 V}}=\dfrac{2(\textrm{2.05 V})}{\textrm{0.0591 V}}=69.37 \\ & & E^\circ_\textrm{anode}=-\textrm{0.36 V} \\ & \textrm{cathode:} &\quad & \mathrm{Cr_2O_7^{2-}(aq)} + \mathrm{14H^+(aq)}+\mathrm{6e^-}\rightarrow \mathrm{2Cr^{3+}(aq)} +\mathrm{7H_2O(l)} (This reaction occurs when a car battery is discharged.) &\quad & E^\circ_{\textrm{cathode}} =\textrm{1.23 V} \\ B Two electrons are transferred in the overall reaction, so $$n = 2$$. For example, the potential for the reduction of Fe3+(aq) to Fe(s) is not listed in the table, but two related reductions are given: \[\ce{Fe^{3+}(aq) + e^{−} -> Fe^{2+}(aq)} \;\;\;E^° = +0.77 V \label{20.5.6} \nonumber, $\ce{Fe^{2+}(aq) + 2e^{−} -> Fe(aq)} \;\;\;E^° = −0.45 V \label{20.5.7} \nonumber$. Multiplying the charge on the electron by Avogadro’s number gives us the charge on 1 mol of electrons, which is called the faraday (F), named after the English physicist and chemist Michael Faraday (1791–1867): \[\begin{align}F &=(1.60218\times10^{-19}\textrm{ C})\left(\dfrac{6.02214\times10^{23}}{\textrm{1 mol e}^-}\right) Determine the number of electrons transferred in the overall reaction. Three electrons (n = 3) are transferred in the overall reaction, so substituting into Equation $$\ref{20.5.5}$$ and solving for E° gives the following: \[\begin{align*}\Delta G^\circ & =-nFE^\circ_{\textrm{cell}} \\ The maximum amount of work that can be produced by an electrochemical cell ($$w_{max}$$) is equal to the product of the cell potential ($$E^°_{cell}$$) and the total charge transferred during the reaction ($$nF$$): Work is expressed as a negative number because work is being done by a system (an electrochemical cell with a positive potential) on its surroundings. &\quad & E^\circ_{\textrm{anode}} =\textrm{1.09 V} \\ The total charge transferred from the reductant to the oxidant is therefore $$nF$$, where $$n$$ is the number of moles of electrons.