electrical work definition

For example, under standard conditions, the reaction of Co(s) with NiElectrochemical cells convert chemical energy to electrical energy and vice versa. To obtain the value of E° for the overall half-reaction, we first must add the values of ΔG° (= −nFE°) for each individual half-reaction to obtain ΔG° for the overall half-reaction:\[\begin{align*}\mathrm{Fe^{3+}(aq)}+\mathrm{e^-} \rightarrow \mathrm{Fe^{2+}(aq)} \hspace{3mm} \Delta G^\circ &=-(1)(F)(\textrm{0.77 V})\\Solving the last expression for ΔG° for the overall half-reaction,\[\Delta{G^°} = F[(−0.77 V) + (−2)(−0.45 V)] = F(0.13 V) \label{20.5.9} \nonumber \]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}} \\This value of E° is very different from the value that is obtained by simply adding the potentials for the two half-reactions (0.32 V) and even has the opposite sign.Values of E° for half-reactions cannot be added to give E° for the sum of the half-reactions; only values of ΔG° = −nFE°We can use the relationship between \(\Delta{G^°}\) and the equilibrium constant \(K\), to obtain a relationship between \(E^°_{cell}\) and \(K\). The type of electrical system that they work on or with doesn’t matter, the work that they do does. Electric power flowing The relationship between real power, reactive power and apparent power can be expressed by representing the quantities as vectors. Recall that for a general reaction of the type \(aA + bB \rightarrow cC + dD\), the standard free-energy change and the equilibrium constant are related by the following equation:Given the relationship between the standard free-energy change and the standard cell potential (Equation \(\ref{20.5.5}\)), we can write\[E^\circ_{\textrm{cell}}= \left( \dfrac{RT}{nF} \right) \ln K \label{20.5.12B}\]For T = 298 K, Equation \(\ref{20.5.12}\) can be simplified as follows:\[ \begin{align} E^\circ_{\textrm{cell}} &=\left(\dfrac{RT}{nF}\right)\ln K \\[4pt] &=\left[ \dfrac{[8.314\;\mathrm{J/(mol\cdot K})(\textrm{298 K})]}{n[96,485\;\mathrm{J/(V\cdot mol)}]}\right]2.303 \log K \\[4pt] &=\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log K \label{20.5.13} \end{align}\]Thus \(E^°_{cell}\) is directly proportional to the logarithm of the equilibrium constant. Using the data in As always, the first step is to write the relevant half-reactions and use them to obtain the overall reaction and the magnitude of \(E^o\). Section Content Objective 1 Introduction to Electrical Safety Participants will be able to:

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