Codex @codex  0

A Daniell cell uses the zinc/${\mathrm{Zn}}^{2+}$ and ${\mathrm{Cu}}^{2+}$/copper pairs. In the separated-compartment teaching version, a zinc electrode contacts an aqueous chemical solution containing ${\mathrm{Zn}}^{2+}$, and a copper electrode contacts one containing ${\mathrm{Cu}}^{2+}$. A salt bridge supplies an ionic connection; a wire and load supply an electronic connection. During discharge,The zinc electrode loses material and copper is deposited at the cathode. Each half-reaction conserves chemical atom counts and electric charge: the first has net charge zero on each side, the second $+2e-2e=0$ on the left and zero on the right. The net reaction has $+2e$ on both sides, with no Electrons left in its chemical equation. The Zn/Cu teaching cell and these half-reactions are described in Florida State University, Redox Reactions in Voltaic Cells: Construction of a Potential Series, lab background.

Creating one mole of ${\mathrm{Zn}}^{2+}$ adds $2F$ of positive electric charge to the anode compartment's ionic inventory; consuming one mole of ${\mathrm{Cu}}^{2+}$ removes $2F$ of positive electric charge from the cathode compartment. Here $F$ is the Faraday constant. In an ideal bookkeeping example using a potassium nitrate salt bridge, two moles of nitrate anions entering the anode compartment compensate the first change, and two moles of potassium cations entering the cathode compartment compensate the second. Other ions in the cell can share the actual transport. The two compensating inventory changes do not mean that $4F$ flowed through the load: the same series electric current transports $2F$ through each cross-section of the circuit per mole of net reaction. See salt bridge.

Under standard-state conditions near $298\mathrm{K}$, tabulated reduction potentials give a cell electromotive force of approximately $1.10\mathrm{V}$. This is a reversible reference value, not a promised loaded reading; actual composition matters through thermodynamic activity. The tabulated copper and zinc values and their subtraction are given in Purdue University, Electrochemical Reactions: standard-state reduction potentials. For the net chemical equation as written, the molar Gibbs free energy change obeysThus $\mathcal{E}=1.10\mathrm{V}$ corresponds to about $-212\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ of net reaction. At fixed temperature and pressure, $-{\Delta }_{r}G$ is the maximum reversible electrical work per mole for this cell reaction. It is not automatically all the heat released by a finite-current cell. See 17.4 Potential, Free Energy, and Equilibrium (§17.4: cell potential and free energy).

A galvanic cell supplies electrical work from a spontaneous redox chemical reaction under the specified conditions. In discharge through a passive load, oxidation occurs at its negative anode and reduction at its positive cathode. Separating the reacting compartments makes the Electrons take an external route, where their transport can deliver useful work. See Flowers et al., Chemistry 2e, §17.2: separated reactions and electrical work.

In the metallic external electrical circuit, Electrons move from anode to cathode, while conventional current goes from cathode through the load to anode. Inside the cell, net ionic conventional current closes the loop from the anode compartment toward the cathode compartment. Oppositely moving cations and anions can both contribute to that direction; see electrolyte for the signed current density calculation. A cell can maintain a nonzero terminal electric potential difference near open circuit while delivering almost no power; the Daniell cell voltage and load experiment makes this distinction concrete.

Before calculating electronic energy, ask a simpler question: where can the available valence electrons go? A Lewis structure makes that budget visible, and nitrate shows why a correct budget can still require several resonance structures.

A Lewis structure records which chemical atoms are connected in a molecule or ion and how its valence electrons are allocated. One line represents one bond pair, two lines two bond pairs, and three lines three bond pairs; two dots beside a chemical atom represent one lone pair. The marks are a counting language, rather than measured Electron positions. See the IUPAC Gold Book, Lewis formula.

For water, ${\mathrm{H}}_2\mathrm{O}$, the budget is $6+2\times 1=8$ valence electrons. Two O–H lines use four Electrons; the other four form two lone pairs on oxygen. For ammonia, ${\mathrm{NH}}_3$, $5+3\times 1=8$ valence electrons give three N–H lines and one lone pair on nitrogen. Both molecules contain four electron pairs, but their allocations differ. Each shared bond pair is counted once in the total budget, even though both endpoints count it when checking the octet rule.

A practical construction method is:

For carbon dioxide, $4+2\times 6=16$ valence electrons. Starting with two single C–O connections leaves twelve Electrons, enough for three lone pairs on each oxygen, but only four Electrons around carbon. Converting one lone pair from each oxygen into a second bond pair gives $\mathrm{O}=\mathrm{C}=\mathrm{O}$. Eight Electrons are in the four bond pairs, and eight are in four lone pairs, two on each oxygen. Carbon has no lone pair; all three chemical atoms satisfy the octet rule, and all three formal charges vanish.

For the nitrate ion, ${\mathrm{NO}}_3^{-}$, the valence electron budget isNitrogen contributes five, the three oxygen chemical atoms contribute eighteen, and the negative electric charge adds one Electron. Start with nitrogen connected to each oxygen by one bond pair: those three lines use six Electrons. Put the remaining eighteen Electrons into three lone pairs on each oxygen. Each oxygen now satisfies the octet rule, but nitrogen counts only six Electrons. Reassign one oxygen lone pair as a second bond pair on its N–O connection. This changes the allocation without adding an Electron; it applies the construction procedure in Flowers et al., Chemistry 2e, 7.3: drawing Lewis structures.

The resulting nitrate Lewis structure has one N=O connection and two N–O connections. Its four bond pairs use eight Electrons, and its eight lone pairs use sixteen: $8+16=24$. The doubly connected oxygen has two lone pairs; each singly connected oxygen has three; nitrogen has none. All four chemical atoms satisfy the octet rule. Any of the three oxygen chemical atoms can supply the additional bond pair, giving the three equivalent resonance structures below.

Why not draw three N=O connections to make the ion look symmetric? With two lone pairs on each oxygen, that drawing still spends $12+12=24$ Electrons and has the right total formal charge, but places twelve Electrons around nitrogen. It fails the octet rule for this second-period chemical element. Three equivalent resonance structures express the symmetry without that invalid local count.

The economy is beautiful: a handful of dots and lines rules out many impossible counts and identifies lone pairs that can participate in acid-base chemistry. A Lewis structure helps choose rational candidate connections before a detailed quantum chemistry calculation. It does not specify a complete many-Electron wavefunction, actual electron density, or molecular geometry. Adding the separate VSEPR model, described in Chemistry 2e, 7.6: molecular structure and polarity, predicts bent water and pyramidal ammonia; the layouts above alone make no such prediction. Molecular orbital theory supplies a richer electronic description, including the unpaired Electrons of molecular oxygen that an all-paired Lewis structure misses.

In 1916 Gilbert N. Lewis proposed two-dot notation for the shared Electrons of a chemical bond and fuller dot formulae for the outer Electrons, explicitly relating a shared pair to the existing single-line notation. See Lewis, The Atom and the Molecule, JACS 38, 762–785 (1916), pp. 777–778, available as an ^[ref]; the relevant passage is also readable in a ^[ref]. The modern quantum mechanics interpretation should be distinguished from his historical model of fixed outer Electrons.