ELECTROCHEMISTRY_______________________________ Chapter 4.5-4.6, 21 Overview of an Electrochemical Process at Constant T and P

ΔG = ΔGo + RT ln Q = welec (maximum) = ‒ QEcell = ‒ ItEcell (exp – lab) (E, volt, intensive parameter, Q, charge is extensive) = ‒ nFEcell (theory)

or rearranging to the Nernst equation

Ecell = Eocell ‒ (RT/nF) ln Q where Eo

cell = ‒ ΔGo / nF

= Eo

cell ‒ (0.05916/n) log Q (at 25oC)

1. n determined by balancing redox equation 2. Eo

cell determined by Eo of species being oxidized and Eo of species being reduced (standard reduction potentials) 3. therefore only Q (reaction quotient) can change the voltage

Oxidation - Reduction Reactions (4.5-4.6)

oxidation

reduction

Balancing Redox Equations (half-reaction method, 21.1)

oxidation numbers, ON (in order of priority, 4.5)

1. Sum of the oxidation numbers of atoms in a neutral species is zero; in an ion the sum equals the ion charge. 2. Alkali metals (Group I) ON = 1, alkaline earth metals (Group II) ON = 2, Group III usually have ON = 3. 3. Fluorine ON = -1 always. Other halogens usually have ON = -1 except in compounds with oxygen or other halogens when the oxidation number can be positive (follows the trend in electronegativity). 4. Hydrogen has ON = 1 except in metal hydrides when the oxidation number is negative. 5. Oxygen usually has ON = -2 except in compounds with fluorine when the oxidation number can be positive and in compounds containing the O-O bond. For peroxides ON = -1.

André-Marie Ampère Michael Faraday

Note: I below stands for current measured in amperes

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Steps in Balancing Oxidation-Reduction Equations by Half-Reaction Method (21.1)

EX 1. Balance in acidic solution MnO4

-(aq) + Fe(s) → Mn2+(aq) + Fe2+(aq)

(use oxidation numbers and include electrons in half-reactions)

(omit step d if oxidation numbers used with electrons)

*for basic react-ions neutralize H+ with OH-

*for basic react-ions neutralize H+ with OH-

*include these steps for a basic reaction

Basically two ways to balance:

1. start with electrons in half-reactions determined from oxidation numbers

OR

2. use electrons to balance charge after all elements are balanced (our text)

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disproportionation (p. 609)

Electrochemical Cells (p. 944, 21.2, pp. 974-975; Figures 21.2, 21.26)

half-cells FIG I - Terms Used in a Galvanic (Voltaic) Cell electrodes

anode cathode

electrolyte salt bridge

EX 2. Balance in basic solution Ag(s) + CN-(aq) + O2(g) → Ag(CN)2(aq)

EX 3. Balance HNO2 → NO3

- + NO

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types of electrochemical cells, p 944

ΔG = welec(max) = ‒ nFEcell

1. spontaneous: ΔG < 0, Ecell> 0 2. nonspontaneous: ΔG > 0, Ecell < 0

FIG II- Electrochemical Cells (Figure 21.26)

a) spontaneous: galvanic (voltaic) b) nonspontaneous: electrolytic

electrochemical cell notation (cell diagram or line notation, p. 948)

Cell Diagram or Line Notation for Galvanic Cells half-cell notation

different phases separated by vertical lines species in same phase separated by commas

types of electrodes • active electrode – involved in electrode half-reaction (most metal electrodes)

ex: Zn2+/Zn metal electrode Zn(s) → Zn2+ (aq) + 2e- (oxidation)

notation: Zn|Zn2+

• inert electrode – not involved in electrode half-reaction (inert solid conductors which serve to contact solution with external electrical circuit)

ex: Pt electrode in Fe3+/Fe2+ solution Fe3+(aq) + e- → Fe2+(aq) (reduction)

notation: Fe3+, Fe2+|Pt • electrodes involving metals and their slightly soluble salts

ex: Ag/AgCl electrode AgCl(s) + e- → Ag(s) + Cl-(aq) (reduction)

notation: Cl-|AgCl|Ag

• electrodes involving gases – a gas is bubbled over an inert electrode

ex: H2 gas over Pt electrode H2(g) → 2H+(aq) + 2e- (oxidation)

notation: Pt|H2|H+ cell notation

anode half-cell written to left of cathode half-cell (ordered way spontaneous reaction goes) electrodes appear on the far left (anode) and far right (cathode) in notation salt bridges represented by double vertical lines

ex: Zn|Zn2+ and Fe3+, Fe2+|Pt half-cells

Zn(s) → Zn2+(aq) + 2e- (anode, oxidation) Fe3+(aq) + e- → Fe2+(aq) [×2] (cathode, reduction) Zn + 2Fe3+ → Zn2+ + 2Fe2+

notation: Zn|Zn2+ || Fe3+, Fe2+|Pt

OXIDATION half-reaction: Sn(s) → Sn2+(aq) + 2e- REDUCTION half-reaction: Cu2+(aq) + 2e- → Cu(s) OVERALL: Cu2+(aq) + Sn(s) → Cu(s) + Sn2+(aq)

OXIDATION half-reaction: Cu(s) → Cu2+(aq) + 2e- REDUCTION half-reaction: Sn2+(aq) + 2e- → Sn(s) OVERALL: Sn2+(aq) + Cu(s) → Sn(s) + Cu2+(aq)

Luigi Galvani Alessandro Volta

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Cell Potential, Electrical Work, and Gibbs Free Energy (p. 959)

electrical work

electric current

Electrolysis (pp. 980-981 − stoichiometry of spontaneous/nonspontaneous reactions)

EX 5. A galvanic (voltaic) electrochemical cell whose overall reaction is

Cu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s)

oxidizes metallic copper at the anode dissolving the metal

Cu(s) → Cu2+(aq) + 2 e-

and reduces silver ions in solution by plating them out as metallic silver at the cathode.

Ag+(aq) + e- → Ag(s)

a) The cell delivers 0.1 A. How long does it take to dissolve 5.00 g of copper at the anode? (MCu = 63.546)

b) An external power source is connected to the cell and the cell is run electrolytically. If the cell produces 0.500 A for 101 minutes, how much copper is deposited and how much silver dissolves? (MAg= 107.87 g/mol)

EX 4. A 6 volt battery delivers 1.25 amperes for 1.5 hours. How much work does the battery do?

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Standard Electrode Potentials (21.3)

standard states – stable form (allotrope) at P = 1 atm and specified T (usually 25°C) pure solid pure liquid gas – ideal gas behavior 1 molar (1 M) solution – ideal solution behavior

standard cell potential, Eo – there is no way to separately measure the potential of an oxidation or reduction half-reaction so conveniently define a zero for the scale standard reduction potential, Eo – by international con-vention, Eo for the reduction of H+(aq) to give H2(g) is taken to be zero at all temperatures:

2 H+(aq, 1 M) + 2 e- → H2(g, P = 1 atm)

RED: 2H+(aq) + 2e- → H2(g) => H+(aq, 1 M)|H2(g, 1 atm)|Pt OX: H2(g) → 2H+(aq) + 2e- => Pt|H2(g, 1 atm)|H+(aq, 1 M)

FIG IV – SHE/Zinc FIG V – SHE/Silver

EX 6. An aqueous solution of a palladium salt is electrolyzed by 1.50 A for one hour which produces 2.9779 g of Pd at the cathode. What is the oxidation state of Pd? (MPd = 106.42 g/mol)

We will not use activities A. Our standard state will have pressures and concentrations equal to one.

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Disproportionation (p. 609) and Oxidizing/Reducing Agents

strength of reducing agent stre

ngth

of o

xidi

zing

age

nt

∆G = ‒ nFEcell => if ∆G < 0 then Ecell > 0 for spontaneity

EX 8. Find Eo(Ag2 CrO4|Ag) for the electrode and determine Eo for the following reaction if Eo(Fe3+|Fe2+) = 0.770 V and ΔGo = -62.5 kJ for the reaction:

K2CrO4(aq) + 2 Ag(s) + 2 FeC13(aq) → Ag2CrO4(s) + 2 FeC12(aq) + 2 KCl(aq)

EX 7. Will Ag+ oxidize Zn metal or will Zn2+ oxidize metallic silver?

Eo Eo

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Cell Potential and Concentration (pp. 961-963)

The Nernst Equation Ecell = Eo

cell ‒ (RT/nF) ln Q = Eocell ‒ (0.05916/n) log Q (25oC); Eo

cell = Eocathode ‒ Eo

anode Half-Cells

EX 9. Determine E for the hydrogen electrode: 2 H+ + 2 e- → H2 a) [H+] = 0.00100 M, PH2 = 0.500 atm b) H+ + e- → 1/2 H2; [H+] = 0.00100 M, PH2 = 0.500 atm c) [H+] = 1.00 × 10-14 M, PH2 = 1.00 atm d) rewrite hydrogen electrode equation in base, [OH-] = 1.00 M, PH2 = 1.00 atm

EX 10. For the half reaction, Cu2+ + 2e- → Cu, where, Eo = 0.340 V what is the reduction potential of the half-cell when [Cu2+] = 0.10 M?

Walther Nernst

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Cells:

EX 11. If [Cu2+] = 0.10 M and [Ag+] = 0.20 M, what is the voltage of an electrochemical cell based upon the couple: Eo(Cu2+|Cu) = 0.340 V and Eo(Ag+|Ag) = 0.800 V?

EX 12. An electrochemical cell is based upon the couple: Eo (Cu2+|Cu) = 0.340 V and Eo (Ag+|Ag) = 0.800 V. If the voltage of the cell were 0.448 V when [Cu2+] = 0.10 M, what is the silver ion concentration?

EX 13. If the cell voltage is 0.473 V what is the pH of the cathode compartment for the following cell where Eo(Zn2+|Zn) = - 0.7618 V?

Zn(s)|Zn2+(aq, 1.00 M)||H+(aq, ? M)|H2(g, 1 atm)|Pt(s)

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Cell Potential, Free Energy, and the Equilibrium Constant (pp. 959-961)

solubility product, Ksp

Reaction Parameters at Standard State

K

E°cell

Reaction at standard-state conditions

< 0 = 0 > 0

> 1 = 1 < 1

> 0 = 0 < 0

spontaneous at equilibrium nonspontaneous

EX 15. If Eo(Cr3+|Cr) = -0.74 V and Eo(Zn2+|Zn) = -0.7628 V what is the equilibrium constant for

2 Cr3+(aq) + 3 Zn(s) → 2 Cr(s) + 3 Zn2+(aq)

EX 14. What is the equilibrium constant for the cell of example 11?

∆G°

E°cell K

E°cell = ln K RT nF

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EX 17. Determine the value of Ksp(AgSCN) given that

AgSCN(s) + e- → Ag(s) + SCN-(aq) Eo = 0.0898 V Ag+(aq) + e- → Ag(s) Eo = 0.7996 V

EX 16. From the following reaction cell where Ecell = 0.397 V

Pt|H2(1 atm)|H+(1.00 M)||Cl-(l.00 × 10-3 M), Ag+(? M)|AgCl(s)|Ag(s)

determine the value of Ksp(AgCl). Relevant standard reduction potentials are

2 H+(aq) + 2 e- → H2(g) Eo = 0.0000 V Ag+(aq) + e- → Ag(s) Eo = 0.7996 V

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Concentration Cells - Q wins! (pp. 964-967) Corrosion (21.6)

4 Fe2+(aq) + O2(g) + (4 + 2x) H2O(l) → 2 Fe2O3·xH2O(s) + 8 H+(aq)

OXIDATION (at anode) Fe(s) → Fe2+(aq) + 2e- -Eo = 0.447 V Cu(s) → Cu2+(aq) + 2e- -Eo = 0.345 V Zn(s) → Zn2+(aq) + 2e- -Eo = 0.762 V cathodic protection: use a “sacrificial” anode

REDUCTION (at cathode) O2(g) + 4 H+(aq) + 4e- → 2 H2O(l) Eo = 1.229 V acidic conditions O2(g) + 2 H2O (l) + 4e- → 4 OH-(aq) Eo = 0.401 V basic conditions

Ecell > 0 as long as the half-cell concentrations are different. The cell is no longer able to do work once the concentrations are equal.

rust

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Nerve Transmission

A nerve cell (neuron) can be likened to a receiver, transmitter, and transmission line between the two. Dendrites receive impulse signals from a neighboring neuron. Transmission of electrical signals within the receiving neuron proceeds from dendrites to synapse along the length of the axon. The signal is due to opening and closing of sodium channels in the axon’s membrane, allowing Na+ to enter or not enter the cell. Entering Na+ causes an increase in charge inside the cell. At rest, the inside of a neuron is slightly negative (~ -70 mV) due to a higher concentration of K+ ions inside (concentration gradient balanced by a potential difference across the membrane, Einside - Eoutside which can be calculated by modifying the Nernst equation)

E = Eo + (0.05916/n) log Q

where n is the charge of the ion. An impulse opens Na+ channels. When stimulated past a threshold Na+ rushes into the axon, causing a region of positive charge (depolarization) within the axon and a potential of ~30 mV across the membrane. The channels open in one direction

along the axon as an opened channel needs a rest period before it can reopen. The region of positive charge causes nearby Na+ channels to close. Just after the Na+ channels close, K+ channels open and K+ exits the axon (repolarization). The membrane is brought back to its negative resting potential by Na+/K+ pumps in the cell membrane. This process continues as a chain-reaction along the axon. The influx of sodium ions depolarises the axon, and the outflow of potassium ions repolarises it.

K+ Na+ inside 20 x y outside x 10 y