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Química Geral · Entropy, free energy and electrochemistry

Pilhas galvânicas e a equação de Nernst

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O que você vai aprender

  • Identify the anode, cathode, electrode signs, electron flow and ion migration in a galvanic cell and write or read its line notation.
  • Use a table of standard reduction potentials to rank oxidizing and reducing agents, calculate E°cell and predict spontaneity.
  • Interconvert E°cell, ΔG° and K with ΔG° = −nFE° and log K = nE°/(0.0592 V), including equilibrium constants such as Ksp built from two half-reactions.
  • Calculate cell and electrode potentials away from standard conditions with the Nernst equation, and work backwards from a measured potential to a concentration.
  • Explain and calculate the potentials of concentration cells and of half-reactions that consume H⁺, whose potentials depend on pH.

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Inside a galvanic cell

A galvanic cell keeps the oxidizing agent and the reducing agent apart so that the electrons they exchange must travel through an external wire. Oxidation happens at the anode and reduction at the cathode. In a galvanic cell the anode is the negative terminal, because the oxidation pushes electrons into the wire, and the cathode is positive. Take nickel in 1 M Ni²⁺ joined to silver in 1 M Ag⁺ (drawn): nickel is oxidized (Ni → Ni²⁺ + 2e⁻), electrons flow through the wire to the silver, and Ag⁺ is reduced there (Ag⁺ + e⁻ → Ag), so the nickel electrode loses mass and the silver electrode gains it. A salt bridge of an unreactive electrolyte such as KNO₃ completes the circuit: anions drift toward the anode compartment, where cations are being produced, and cations toward the cathode compartment, keeping both solutions neutral. Line notation writes the anode on the left and the cathode on the right, | for a phase boundary and || for the salt bridge: Ni(s) | Ni²⁺(aq) || Ag⁺(aq) | Ag(s). When both members of a couple are dissolved ions, as for Fe³⁺/Fe²⁺, an inert platinum electrode carries the electrons: … || Fe³⁺(aq), Fe²⁺(aq) | Pt(s). If the two reagents touch, as when an iron nail sits in CuSO₄ solution, the electrons transfer directly and the free energy appears only as heat.

Standard reduction potentials and E°cell

Only differences in potential can be measured, so every half-reaction is written as a reduction and measured against the standard hydrogen electrode, 2H⁺(1 M) + 2e⁻ → H₂(1 bar), defined as 0.00 V. A negative E°, such as Zn²⁺/Zn −0.76 V, means that zinc is the anode when paired with the hydrogen electrode. The higher E°, the stronger the oxidizing agent on the left of the half-reaction (Ag⁺ +0.80 V is stronger than Cu²⁺ +0.34 V); the lower E°, the stronger the reducing agent on the right (Al, from Al³⁺/Al −1.66 V, is a strong reducing agent, while Al³⁺ is a very weak oxidizing agent). For any cell, E°cell = E°(cathode) − E°(anode), using both values as reduction potentials from the table. For Al(s) | Al³⁺ || Ag⁺ | Ag(s), E°cell = 0.80 − (−1.66) = +2.46 V. E° is energy per unit charge, an intensive quantity, so it is never multiplied when a half-reaction is multiplied to balance electrons. A positive E°cell means the reaction as written is spontaneous under standard conditions; a negative value means the reverse reaction is. The same test decides whether an ion is stable in water: Cu⁺ disproportionates, 2Cu⁺ → Cu + Cu²⁺, because E° = 0.52 − 0.15 = +0.37 V.

Free energy and equilibrium constants

Sign convention: E°cell > 0 ⇔ ΔG° < 0 ⇔ K > 1. The three are linked by ΔG° = −nFE° = −RT ln K, where n is the number of electrons transferred in the balanced equation as written and F = 96 485 C/mol. Because 1 C × 1 V = 1 J, −nFE° comes out in J/mol; divide by 1000 for kJ/mol. For 2Al + 3Ni²⁺ → 2Al³⁺ + 3Ni, n = 6 and E° = −0.26 − (−1.66) = 1.40 V, so ΔG° = −6 × 96 485 × 1.40 = −8.10 × 10⁵ J/mol = −810 kJ/mol. At 25 °C, 2.303RT/F = 0.0592 V, so log K = nE°/(0.0592 V): for 2Fe³⁺ + Sn²⁺ → 2Fe²⁺ + Sn⁴⁺, E° = 0.77 − 0.15 = 0.62 V and log K = 2 × 0.62/0.0592 = 20.9. Reversing a reaction changes the sign of E° and of ΔG° and turns K into 1/K. Any equilibrium that can be written as the sum of two half-reactions has a K that follows from potentials, even if nothing is oxidized overall: AgBr(s) + e⁻ → Ag + Br⁻ (+0.071 V) minus Ag⁺ + e⁻ → Ag (+0.800 V) gives AgBr(s) ⇌ Ag⁺ + Br⁻ with E° = −0.729 V and log Ksp = −0.729/0.0592 = −12.3. A larger E° does not always mean a more negative ΔG°: ΔG° also scales with n.

The Nernst equation

Away from standard conditions, E = E° − (RT/nF) ln Q, which at 25 °C becomes E = E° − (0.0592 V/n) log Q. Q is written from the balanced equation, products over reactants, each concentration raised to its coefficient, with pure solids and liquids omitted. For Ni(s) | Ni²⁺(0.10 M) || Ag⁺(0.010 M) | Ag(s), the reaction is Ni + 2Ag⁺ → Ni²⁺ + 2Ag, so Q = [Ni²⁺]/[Ag⁺]² = 0.10/(0.010)² = 1000 and E = 1.06 − (0.0592/2)(3) = 0.971 V. Each tenfold rise in Q lowers E by 0.0592/n V, 29.6 mV when n = 2. The equation also works for one electrode: for Ag⁺ + e⁻ → Ag at [Ag⁺] = 0.010 M, E = 0.80 − 0.0592 log(1/0.010) = 0.68 V. Removing a reactant (for example precipitating Ag⁺ as AgCl) raises Q and lowers E. As a cell discharges, the concentrations change together, so recompute Q from the stoichiometry. When E° is small, a modest concentration ratio can make E negative, and then the spontaneous direction, and the anode, are reversed. Rearranged, the equation turns a measured potential into an unknown concentration.

Concentration cells and pH

A concentration cell has the same couple in both compartments, so E° = 0 and the whole potential is the Nernst term. The electrode in the dilute solution is the anode: for Ag | Ag⁺(0.020 M) || Ag⁺(0.80 M) | Ag, E = 0 − 0.0592 log(0.020/0.80) = 0.0592 × log 40 = 0.095 V, and the cell runs until the two concentrations are equal. A half-reaction that consumes m H⁺ per n electrons has a potential that falls by 0.0592 × m/n V for every unit rise in pH. For the hydrogen electrode (m = n = 2) at 1 bar H₂, E = −0.0592 pH, so a hydrogen concentration cell is a pH meter: pH = E/0.0592 when the cathode is at pH 0. For permanganate (drawn), MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O with E° = +1.51 V, the slope is 0.0592 × 8/5 = 0.0947 V per pH unit; with [MnO₄⁻] = [Mn²⁺], E = 1.51 − 0.0947 × 3.00 = 1.23 V at pH 3.00. A couple with no H⁺ in its half-reaction, such as Fe³⁺/Fe²⁺, does not change with pH as long as its species stay as written.

What to carry forward

Oxidation at the anode (negative in a galvanic cell), reduction at the cathode, electrons through the wire and ions through the salt bridge. Read potentials as reductions and use E°cell = E°cathode − E°anode without scaling E°. Convert with ΔG° = −nFE° (J/mol) and log K = nE°/0.0592 V, and correct for concentration with E = E° − (0.0592 V/n) log Q, writing Q with every coefficient as an exponent. Concentration cells run on the Nernst term alone, and every H⁺ in a half-reaction makes its potential fall with pH by 0.0592 m/n V per unit.

Problemas de prática (40)

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Preveja o produto 8

Concentration cells and pH

  • Which side of a concentration cell is the anode?Predict the anode of a concentration cell and how the two concentrations change as it runs.

Standard potentials and spontaneity

  • Is Cu⁺ stable in water?Predict whether an ion in an intermediate oxidation state disproportionates by combining its two reduction potentials.
  • Reduce Ni²⁺ but not Zn²⁺Use the position of couples in the potential table to pick a metal that reduces one metal ion but not another.

Inside a galvanic cell

  • Electron flow and changing concentrationsPredict the direction of electron flow in the external circuit and the resulting changes in ion concentrations as a galvanic cell runs.
  • Write the line notationWrite the line notation of a galvanic cell with the anode on the left and each electrode at the outer end.

The Nernst equation

  • Precipitating the cathode ionPredict the effect on a cell potential of removing a reactant ion by precipitation.
  • Write Q for the Nernst equationWrite the reaction quotient for a cell reaction with unequal coefficients, omitting pure solids.
  • When the concentrations reverse the cellPredict which electrode is the anode when the concentration term outweighs a small standard cell potential.

Classificar 4

Inside a galvanic cell

  • Which electrode is the anode?Identify the anode of a galvanic cell from standard reduction potentials and state its polarity.

Concentration cells and pH

  • Which potential ignores pH?Classify half-reactions by whether their potential depends on pH, from whether H⁺ appears in the half-reaction.

Free energy and equilibrium constants

  • Signs for the reverse reactionClassify E°, ΔG° and K for a reaction written in reverse from the forward standard cell potential.

Standard potentials and spontaneity

  • Pick the strongest reducing agentIdentify the strongest reducing agent from a table of standard reduction potentials, distinguishing metals from their ions.

Comparar 4

Standard potentials and spontaneity

  • Rank metals as reducing agentsRank metals by reducing strength from the standard reduction potentials of their ions.
  • Rank standard cell potentialsCalculate and rank the standard potentials of several galvanic cells from one table.

Free energy and equilibrium constants

  • Larger E°, smaller ΔG°?Compare two reactions to show that ΔG° depends on n as well as on E°cell.

Concentration cells and pH

  • Rank couples by pH sensitivityRank half-reactions by how steeply their potentials fall with pH using the ratio of H⁺ consumed to electrons transferred.

Calcular 14

Inside a galvanic cell

  • Mass lost by the cadmium anodeRelate the mass gained at the cathode to the mass lost at the anode through the moles of electrons transferred.
  • Electrons transferred in an aluminium–copper cellBalance the overall reaction of a cell whose half-reactions involve different numbers of electrons and find n.

The Nernst equation

  • Cell potential after partial dischargeUse reaction stoichiometry to update both concentrations in Q before applying the Nernst equation to a partly discharged cell.
  • Nickel–silver cell away from 1 MCalculate a nonstandard cell potential when one ion's concentration is squared in Q.
  • Potential of a silver electrode in dilute Ag⁺Apply the Nernst equation to a single half-reaction to find an electrode potential at a nonstandard concentration.
  • Effect of a tenfold change in QUse the Nernst equation to find how much a cell potential changes for a tenfold change in Q.

Standard potentials and spontaneity

  • E°cell of an aluminium–silver cellCalculate a standard cell potential from two reduction potentials without scaling either by its coefficient.

Free energy and equilibrium constants

  • ΔG° for a six-electron cell reactionConvert a standard cell potential into ΔG° for a reaction whose balanced equation transfers six electrons.
  • E° of the hydrogen–oxygen cell from ΔG°Calculate a standard cell potential from the standard free energy of the cell reaction.
  • log K for the iron(III)–tin(II) reactionCalculate log K of a redox reaction from E°cell when the balanced equation scales one half-reaction.
  • Ksp of silver bromide from two potentialsCombine two half-reactions into a solubility equilibrium and calculate its pKsp from E°.

Concentration cells and pH

  • Hydrogen electrode at pH 4Calculate the potential of a hydrogen electrode at a given pH from the Nernst equation.
  • Permanganate potential at pH 3Calculate the potential of a half-reaction that consumes many H⁺ per electron at a pH other than 0.
  • Potential of a silver concentration cellCalculate the potential of a concentration cell built on a one-electron couple.

Encontre o erro 1

  • A negative voltmeter readingDiagnose a wrong spontaneity conclusion drawn from a negative voltmeter reading caused by reversed leads.

Interprete as evidências 1

  • Overall reaction from line notationRead a cell's line notation and write its balanced overall reaction.

Trabalhe de trás para frente 3

The Nernst equation

  • Find [Cu²⁺] from a measured potentialRearrange the Nernst equation to find an unknown ion concentration from a measured cell potential.

Concentration cells and pH

  • pH from a hydrogen concentration cellWork backwards from the potential of a hydrogen concentration cell to the pH of the unknown solution.

Free energy and equilibrium constants

  • E° from an equilibrium constant near 1Work backwards from an equilibrium constant to the standard cell potential of a two-electron reaction.

Planeje uma síntese 1

  • Choose a salt bridgeChoose a salt-bridge electrolyte whose ions conduct charge without reacting with the ions in either half-cell.

Explique por quê 4

Standard potentials and spontaneity

  • Why not titrate in hydrochloric acid?Explain from standard potentials why an analyte's acid medium must not contain an ion the titrant can oxidize.
  • What a negative E° meansExplain the meaning of a negative standard reduction potential relative to the standard hydrogen electrode.

Inside a galvanic cell

  • Why a platinum electrode?Explain why a half-cell whose oxidized and reduced forms are both dissolved ions needs an inert electrode.
  • Why keep the half-reactions apart?Explain why a galvanic cell must keep its oxidizing and reducing agents in separate compartments to deliver electrical work.

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