activity series lithiumpotassiummagnesiumaluminumzincironnickelleadhydrogencoppersilverplatinumgold...

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Activity Series Activity Series lithium lithium potassium potassium magnesium magnesium aluminum aluminum zinc zinc iron iron nickel nickel lead lead HYDROGEN HYDROGEN copper copper silver silver platinum platinum gold gold Oxidizes easily Reduces easily Less active More active

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Activity SeriesActivity Series

lithiumlithium

potassiumpotassiummagnesiummagnesiumaluminumaluminumzinczincironironnickelnickelleadleadHYDROGENHYDROGENcoppercoppersilversilverplatinumplatinum

goldgold

Oxidizes easily

Reduces easily

Less active

More active

Activity seriesActivity series

More active metals will displace less More active metals will displace less active metal ions.active metal ions.

Hydrogen is the reference point.Hydrogen is the reference point. Electrons will flow spontaneously Electrons will flow spontaneously

from the stronger reducing agent to from the stronger reducing agent to the stronger oxidizing agent.the stronger oxidizing agent.

lithiumlithium

potassiumpotassium

magnesiummagnesium

aluminumaluminum

zinczinc

ironiron

nickelnickel

leadlead

HYDROGENHYDROGEN

coppercopper

silversilver

platinumplatinum

goldgold

Will the following reactions Will the following reactions occur ?occur ?

AgAg++ + Cu + Cu

CuCu2+2+ + Ag + Ag

MgMg2+ 2+ + Zn+ Zn

ZnZn2+2+ + Mg + Mg

ZnZn2+2+ + Cu + Cu

MgMg2+2+ + Li + Li

No

No

Yes

The metal must be above (more active than) the ion for it to be a spontaneous reaction.

No

Yes

Yes

ElectrochemistryElectrochemistry

Chemical energy becoming Chemical energy becoming electrical energyelectrical energy

Current- the flow of electricity (amps)Current- the flow of electricity (amps) Potential difference- the force Potential difference- the force

pushing the current (volts)pushing the current (volts) Electrolyte- an ionic solution that can Electrolyte- an ionic solution that can

carry electric current.carry electric current. Spontaneous redox reactions Spontaneous redox reactions

produce this electric energy.produce this electric energy.

Voltaic CellVoltaic CellAnodeAnode

attracts anionsattracts anions where oxidation occurswhere oxidation occurs

CathodeCathode attracts cationsattracts cations where reduction occurswhere reduction occurs

Salt bridge Salt bridge connects the two half cells connects the two half cells contains a strong electrolyte contains a strong electrolyte

Zn Zn+ Cu2+ Cu in shorthand

Two half cells connected by a salt Two half cells connected by a salt bridgebridge

Reduction half reactions

F2 is the strongest oxidizing agent

Li is the strongest reducing agent

Reduction Potentials (E)Reduction Potentials (E) EEoo is the standard electrode potential is the standard electrode potential

all ions are 1M and gases are 1 atmall ions are 1M and gases are 1 atm The net EThe net Eo o is the sum of the Eis the sum of the Eoo of the half of the half

reactionsreactions The stronger oxidizing agent reduces.The stronger oxidizing agent reduces. Reverse the sign of the substance oxidized.Reverse the sign of the substance oxidized. If E is positive, the reaction is spontaneous.If E is positive, the reaction is spontaneous. If E is negative, the reverse reaction is If E is negative, the reverse reaction is

spontaneous.spontaneous.

What is the voltage produced from the What is the voltage produced from the reaction of Zn metal with Cureaction of Zn metal with Cu2+2+ ions? ions?

Zn(Zn(ss) + Cu) + Cu2+2+((aqaq) ) Zn Zn2+2+((aqaq) + Cu() + Cu(ss))

ZnZn2+2+ + 2e + 2e-- Zn -0.7628 Zn -0.7628

CuCu2+2+ + 2e + 2e-- Cu 0.3402 Cu 0.3402

-(-0.7628) + 0.3402 = 1.103 volts-(-0.7628) + 0.3402 = 1.103 volts

Will happen spontaneouslyWill happen spontaneously

ZnZn2+2+ + Ni( + Ni(ss) ) Zn( Zn(ss) + Ni) + Ni2+2+

-(-0.23) + (-0.7628) = -.53-(-0.23) + (-0.7628) = -.53

Will not occur spontaneouslyWill not occur spontaneously

CuCu2+2+ will reduce!

Faraday’s LawFaraday’s Law

Coulombs = amperes x secondsCoulombs = amperes x seconds

1 C = 1amp1 C = 1amp·1·1secsec

96,485 coulombs = 1 mole e96,485 coulombs = 1 mole e--

What mass of copper will be What mass of copper will be deposited by a current of 7.89 deposited by a current of 7.89

amps flowing for 1200 amps flowing for 1200 seconds?seconds? CuCu2+2+ + 2e + 2e-- Cu at the cathode Cu at the cathode

7.89A x 1200s x 7.89A x 1200s x 1C1C x x 1 mol e1 mol e-- == .0981 mol .0981 mol

ee--

AA·s 96,485C·s 96,485C

.0981 mol e.0981 mol e-- x x 1 mol Cu1 mol Cu xx 63.5g Cu 63.5g Cu = 3.1g = 3.1g

Cu Cu

2 mol e2 mol e- 1- 1 mole Cu mole Cu

What mass of CrWhat mass of Cr3+3+ is produced is produced by a current of 0.713A flowing by a current of 0.713A flowing

for 12,800 seconds?for 12,800 seconds?14H14H33OO+++ 6Fe+ 6Fe2+2++ Cr+ Cr22OO77

2-2- 6Fe 6Fe3+3++ 2Cr+ 2Cr3-3-+21H+21H22OO

.713A x 12,800s x .713A x 12,800s x 1C1C x x 1 mol e1 mol e-- == .0946 mol e .0946 mol e--

AA·s 96,485C·s 96,485C

.0946 mol e.0946 mol e-- xx 2mol Cr2mol Cr3+3+ x x 52.0g Cr52.0g Cr3+3+ = 1.64gCr = 1.64gCr3+3+

6mol e6mol e- - 1mol Cr1mol Cr3+3+

Nernst EquationNernst Equation

EEo Voltage under standard conditions o Voltage under standard conditions

(1M solutions at 25°C and 101.3kPa)(1M solutions at 25°C and 101.3kPa)

At non-standard conditions, use Nernst equationAt non-standard conditions, use Nernst equation

E = E° - E = E° - 0.059160.05916 log [ log [products]products]

n [reactants]n [reactants]

n = no. of electrons transferredn = no. of electrons transferredCoefficients in front of reactants or products are used as powers of their concentrations.Coefficients in front of reactants or products are used as powers of their concentrations.

Determine the voltage of a Determine the voltage of a

Zn ZnZn Zn2+ 2+ (.75M) Ni(.75M) Ni2+2+(2.0M) Ni cell at (2.0M) Ni cell at standard temperature and pressure. standard temperature and pressure.

EEoo = +0.7626V + -.257V = .506V = +0.7626V + -.257V = .506V

E = .506V - E = .506V - .0592 .0592 log log .75M.75M

2 2.0M2 2.0M

E = .506V – 0.0296 (log 0.375)E = .506V – 0.0296 (log 0.375)

E = .506V – (-0.0126)E = .506V – (-0.0126)

E = .519VE = .519V