1.11 electrode potentials and cells, commercial electrochemical cells

oxidation vs reduction

oxidation

• loss of electrons / loss of hydrogen

reduction

• gain of electrons / gain of hydrogen

oxidising agent vs reducing agent

oxidising agent

• electron acceptor

• gets reduced

reducing agent

• electron donor

• gets oxidised

half equations

• what they show

• where we find e^- in a half equation for reduction and oxidation

• either show reduction OR oxidation

• represent a dynamic equilibrium as reduction and oxidation are reversible processes

• reduction: e^- on LHS of half equation

• oxidation: e^- on RHS of half equation

redox equations

• what they show

• where we find e^- in a redox equation for reduction and oxidation

• show both reduction AND oxidation, is a combination of 2 half equations

• e^- cancel out on both sides so not shown

anode vs cathode

anode = positive electrode

cathode = negative electrode

electrode potential, E^Ɵ

a number that indicates how readily a substance gains electrons

• i.e. how well a substance acts as an oxidising agent

electrochemical series

• a list of standard electrode potentials for half equations in numerical order

• always shows the half equations as reversible reduction reactions

  • oxidising agents on LHS of half equation in series

  • reducing agents on RHS of half equation in series

largest electrode potential means…

best oxidising agent

• forward reaction (where e^- is gained) is very likely to occur

• reverse reaction (where e^- is lost) is very unlikely to occur

smallest electrode potential means…

best reducing agent

• the reverse reaction (where e- is lost) is very likely to occur

• forward reaction (where e- is gained) is very unlikely to occur

components of a standard electrochemical cell

• 2 half cells

  • different solutions and electrodes

• voltmeter

  • measures EMF (pd)

  • high resistance to prevent current flow so that current is zero, so max EMF accurately measured

• salt bridge

  • ions are able to move through it

  • it completes the circuit (provides an electrical connection between the two solutions)

standard electrode potentials, E^Ɵ, refers to what conditions?

• 298 K

• 1 moldm^-3 solution of ions (aqueous ionic compound, e.g. a chloride)

• 100 kPa of gases

what substance is used as the salt bridge and why?

saturated potassium nitrate solution, KNO₃ (aq)

• soluble ionic solution

• does not react with chemicals in the half cells

can also use KCl

why and when platinum may be used as an electrode

why:

• it is chemically inert

• it is an electrical conductor and provides a surface for electron transfer

• it stays stable in acidic conditions

when:

• used for standard electrochemical cells

• used for a mixture of two solutions

• used for a solution and a gas

types of half cells

solid metal with metal ions

• electrode: piece of pure metal, e.g. Cu and Zn

• electrolyte: aqueous ions from 1 moldm^-3 chloride solution, e.g. CuCl₂ (Cu²⁺) and ZnCl₂ (Zn²⁺)

mixture of two solutions, or one solution one gas

• electrode: piece of platinum, Pt

• electrolyte: aqueous ions from 1 moldm^-3 chloride solution, e.g. FeCl₂ (Fe³⁺) and FeCl₃ (Fe²⁺)

LHS half cell vs RHS half cell

LHS

• R → O

• gets oxidsed

• smaller E^Ɵ

• more reactive metal

RHS

• O → R

• gets reduced

• larger E^Ɵ

• less reactive metal

conventional cell representation

• R O | | O R

  • LHS = gets oxidised, smaller E^Ɵ

  • RHS = gets reduced, larger E^Ɵ

  • | | in the middle for salt bridge

  • oxidising agents are closest to the salt bridge

~

• always include state symbols

• phase boundaries

  • comma if in same phase

  • vertical line if in different phases

~

• if platinum electrode used, place on outermost point and use vertical line for phase boundary

• include H⁺ but not water

give the conventional cell representation

Zn _(s) | Zn²⁺_(aq) | | Cu²⁺_(aq) | Cu _(s)

give the conventional cell representation

Fe _(s) | Fe²⁺_(aq) | | VO²⁺ _(aq) , H⁺_(aq) , V³⁺_(aq) | Pt_(s)

electromotive force (EMF)

potential difference

• the difference between the electrode potential of 2 half cells or 2 half equations

• the energy that is transferred per unit charge/per mole of electrons flowing between 2 points

E_cell or EMF = ?

E^Ɵ RHS - E^Ɵ LHS

larger E^Ɵ - smaller E^Ɵ

if E^Ɵ a > E^Ɵ b, then…

• a can oxidise b, because a is a better oxidising agent

• b can reduce a, because b is a better reducing agent

to oxidise another species, i need to have…

a larger E^Ɵ than that species

to reduce another species, i need to have…

a smaller E^Ɵ than that species

Identify the species that can be used to reduce VO₂⁺ ions to VO²⁺ ions in aqueous solution and no further.

Explain your answer.

(2 marks)

• Fe²⁺

  • (NOT Fe³⁺, remember reducing agents are on the RHS)

• E^Ɵ VO₂⁺ > E^Ɵ Fe³⁺ > E^Ɵ VO²⁺

Au⁺ + e^- → Au E^Ɵ = +1.68 V

O₂ + 4H⁺ + 4e^- → 2H₂O E^Ɵ = +1.23 V

Explain what happens when a soluble gold (I) compound containing Au⁺ ions is added to water.

State what you would observe.

Write an equation for the reaction that occurs.

E^Ɵ Au⁺ > E^Ɵ O₂ + H⁺

so Au⁺ oxidses H₂O to form O₂

observation: effervescence, solid gold forms

equation: 4 Au⁺ + 2 H₂O → 4 Au⁺ + 4 H⁺ + O₂

when would a redox reaction not occur?

E_cell or EMF is negative

the standard hydrogen electrode

how to set it up

is a reference used to measure unknown electrode potentials

has E^Ɵ = 0.0 V

set up:

• solution of aqueous HCl, with H₂ gas pumped in from the side to react with the H⁺ on a Pt electrode

• this half cell is placed on LHS and connected to the standard cell of another species, using a voltmeter and a salt bridge

  • voltmeter has high resistance → no current flowing, so voltmeter reading accurately indicates max E^Ɵ of the unknown

conditions of standard hydrogen electrode

conventional cell representation for standard hydrogen electrode

conditions:

• 298 K and Pt electrode

• 1 moldm^-3 HCl solution (or HNO₃)

• H₂ gas at 100 kPa

representation:

Pt (s) | H₂ (g) | H⁺ (aq) | | other standard cell

how do reaction conditions affect EMF?

changes in concentration of ions or gas pressure shift equilibrium

• e.g. conc is not / is less than 1 moldm^-3

• E^Ɵ increases if conditions shift equilibrium to RHS of the reduction reaction

• E^Ɵ decreases if conditions shift equilibrium to LHS of the reduction reaction

  • more electrons formed = more negative E^Ɵ

effects of replacing the voltmeter with a lamp / wire / ammeter

• chemical reaction occurs

• conc of reactants will decrease as they are being used up

• cell reaction goes to completion

• current will flow through circuit

• EMF decreases to 0 V

what doesn’t affect electrode potentials?

changes that don’t affect equilibrium

• e.g. changing the surface area of electrodes

commercial electrochemical cells

electrochemical cells used as a commercial source of electrical energy

negative and positive cathode in commercial electrochemical cells

negative terminal / negative electrode:

• half equation with smaller E^Ɵ

• oxidation, as reducing agent loses e^-

positive terminal / positive electrode:

• half equation with larger E^Ɵ

• reduction, as oxidising agent gains e^-

movement of electrons and ions in commercial electrochemical cells

• reducing agent releases e^- released at negative electrode

• e^- move through an external circuit

• e^- transfer their energy to a lamp/bulb/etc.

• oxidising agent gains e^- at positive cathode

• ions move through salt bridge (electrolyte) from positive cathode to negative cathode, to complete the circuit and balance overall charge

• potential difference created that can drive an electrical current to do work

why could the EMF in commercial electrochemical cells be different than expected?

non-standard conditions

rechargeable cell

the reactions are reversible

to recharge, apply a reverse potential to force electrons to move in the opposite direction, causing the redox reaction to reverse

pros

• minimises waste and use of natural resources

cons

• more expensive to make and buy

• if electricity used to recharge the cell is generated by burning fuels, CO₂ is released so contributes to global warming

• reactants are used up, so conc of reactants decreases, and evenutally conc of ions in the two solutions become equal, so EMV decreases to 0 V

rechargeable cell example and uses

• lithium ion cells

• used in phones, laptops and cars

rechargeable cell example - electrodes and half equations at these electrodes

negative electrode - graphite (oxidation)

• Li → Li⁺ + e^–

positive electrode - lithium cobalt oxide (reduction)

• Li⁺ + CoO₂ + e^– → Li⁺ [CoO₂ ]^–

overall

• Li + CoO₂ → Li⁺ [CoO₂]^–

rechargeable cell example - why an aqueous electrolyte is not used

• Li would react (violently) with water / electrolyte

• E^Ɵ H₂O > E^Ɵ Li, so water will oxidise Li to Li⁺

non-rechargeable cells

one or both of the reactions are not reversible

pros

• cheaper to produce and buy

cons

• single use

  • reactants are used up, so conc of reactants decreases, and eventually conc of ions in the two solutions become equal, so EMV decreases to 0 V

• use finite resources

• release toxic chemicals as they corrode so pollute environment

porous separator

acts as a salt bridge:

• ions are able to move through it

• it completes the circuit

fuel cells

• generate an electric current without needing to be electrically recharged

• EMF does not change with use because:

  • concentrations of reactants remain constant

  • reactants / fuel can be continuously supplied

• more energy efficient than combusion reactions so less energy is wasted as heat, because energy gets converted into a useful form such as kinetic energy for vehicles

which type of fuel is stored and transported more easily:

liquid fuel or gas fuel?

liquid

hydrogen-oxygen fuel cells

• uses a continuous supply of H₂ (fuel) and O₂ from air, to generate a continuous current

• pros:

  • water is the only waste product, no CO₂ or SO₂ released

  • same overall reaction as combusion of hydrogen but is much more energy efficient than combustion → less energy wasted as heat

• cons:

  • expensive to produce as H₂ is very flammable

  • use of finite resources to generate H₂, e.g. using methane and reacting with steam

    • CH₄ (g) + 2H₂O (g) → 4H₂ (g) + CO₂ (g)

  • burning fossil fuels to generate electricity to generate H₂

    • use renewable energy resources to power electrolysis of water

alkaline hydrogen-oxygen fuel cell half equations

KOH electrolyte between electrodes (OH^- moves through salt bridge)

negative electrode (oxidation)

• 2 OH^- + H₂ → 2 H₂O + 2 e^-

positive electrode (reduction)

• O₂ + 2 H₂O + 4 e^- → 4 OH^-

overall

• 2 H₂ (g) + O₂ (g) → 2 H₂O (l)

why are the EMF values of acidic and alkaline hydrogen-oxygen fuel cells are the same?

because same overall reaction:

2 H₂ (g) + O₂ (g) → 2 H₂O (l)