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1

Electrochemistry & Redox

An oxidation-reduction

(redox) reaction involves the transfer of electrons from the reducing agent to the oxidising agent. 1

OXIDATION - is the LOSS of electrons Zn(s) Zn

2+ (aq) + 2e (aq)

REDUCTION - is the GAIN of electrons Cu2+

(aq) + 2e (aq) Cu(s)

These represents the redox HALF-EQUATIONS

Voltaic Cells

These cells are connected with a

wire, to allow electron flow and a lbd l h

A voltaic electrochemical cell involves two half cells one containing an oxidising agent and the other a reducing agent.

2 salt bridge to complete the circuit and maintain electrical neutrality.

The PULL or DRIVING FORCE

on the electrons is the cell potential ( E cell ) or the electromotive force (emf) of the cell, measured in volts.

Electrochemical Cells

Voltaic (Galvanic) cells are those

in which spontaneous chemical reactions produce electricity and supply it to other circuits. G< 0

An ox, red cat

Anode: oxidation

Reduction: cathode

3

Electrolytic cells are those in

which electrical energy causes non-spontaneous chemical reactions to occur. G> 0

Balancing Redox Equations

The concept of Oxidation Numberis artificial. In simple ions it is equivalent to the charge on the ion. Oxidation involves an increase in oxidation number

Reduction involves a decrease in oxidation number

4 As half cells, determine oxidation numbers and balance electrons. Combine half cells balancing gain/loss of electrons.

Balance with H

2

O and H+

or H 2

O and OH

Check charges balance.

Zn(s) Zn

2+ (aq) + 2e (aq) Cu 2+ (aq) + 2e (aq) Cu(s)

Zn(s) + Cu

2+ (aq) Zn 2+ (aq) + Cu(s) 0 +II +II 0

Standard Reduction Potentials

In data tables half cells are written as reductions.

Standard hydrogen electrode defined as E

= 0 V (1 atm H 2, [H ] = 1 M, at all temperatures). The more negative E, the greater the tendency to release electrons (t di t) 5 (act as a reducing agent). Fe 3+ + e Fe 2+ E = 0.77 V Cu 2+ + 2e Cu E = 0.34 V 2H + 2e H 2 (g) E = 0.00 V Zn 2+ + 2e Zn E = -0.76 VCalculating Cell Potential

When combining two half-reactions:

One half-cell reaction is reversed (thus the sign of the reduction potential is reversed). Number of electrons lost must equal the number gained. Note cell potential is an INTENSIVE PROPERTY i.e. when a half- reaction is multiplied by and integerE stays the SAME 6 reaction is multiplied by and integer E stays the SAME.

E.g. Fe

3+ + e Fe 2+ E = 0.77 V ...1 Cu 2+ + 2e Cu E = 0.34 V ...2 reverse ...2Cu Cu 2+ + 2e E = -0.34 V double ...1 2Fe 3+ + 2e 2Fe 2+ E = 0.77 V add equations2Fe 3+ + Cu 2Fe 2+ + Cu 2+ E

°cell

= 0.77+(-0.34) = 0.43 V 2

Question

Write a balanced equation for the oxidation of lactate to pyruvate and calculate the cell potential. O 2 + 4H + 4e 2H 2 OE = 0.82 V pyruvate+2H +2e lactateE =-019Vpyruvate+ 2H+ 2elactateE 0.19 V 7

G °= -nFE°

max Cu 2+ (aq) + Fe(s) Cu(s) + Fe 2+ (aq)

Is this a spontaneous reaction?

Cu 2+ (aq) + 2e

Cu(s) E° = 0.34 V

Fe 2+ (aq) + 2e

Fe(s)E°= -0.44 V

8 Cu 2+ (aq) + Fe(s) Cu(s) + Fe 2+ (aq) E° cell = 0.34 + (+0.44) = 0.78 V G ° = - 2 x 96485 x 0.78 (F=96485 C mol -1 = - 1.5 x 10 5 J

G °and the positive sign for

E° cell

Glucose metabolism

C 6 H 12 O 6 + 6O 2 6CO 2 + 6H 2 = -2875 kJ mol -1 If all this energy was released at once it would totally swamp the cell.

Instead energy in food used to create an electrochemical potential which is used to create the high energy molecule ATP.

The energy can be released in small steps rather than all at once. 9

Metabolism and energy storage

10 Redox reactions involved form part of the electron-transport chain (ETC). The ETC lies on the inner membrane of mitochondria. The ETC involves a series of large molecules, mostly proteins, that use Fe 2+ /Fe 3+ to pass electrons down the chain.

Electron transport chain ETC

The reaction that ultimately powers ETC is H

2 + ½O 2 H 2 O but do not have H 2 (g) but rather 2H + 2e in the presence of NADH. 2H + 2e + ½O 2 H 2 O 11

NADH(aq) + H

(aq) + ½O 2 (aq) NAD (aq) + H 2 O(l)

E °'

overall -1 At each of the points shown there is enough free energy released to drive ADP 3- + HPO 42-
+ H ATP 4- + H 2

O G ' = 30.5 kJ mol

-1

NADH + H

NAD + 2H + 2e

Concentration Cells

Imagine a cell: Cu(s)Cu

2+ (aq)Cu 2+ (aq)Cu(s)

Low [Cu

2+ ]High [Cu 2+

Cu(s) Cu

2+ (aq) + 2e Cu 2+ (aq) + 2e Cu(s) 12 3

Concentration Cells

Nerve cells operate as concentration cells.

Inside cell [Na

] low, [K ] high; outside cell [Na ] high, [K ] low. Outer cell membrane is positive. One third of our ATP is used to maintain this difference 13 difference.

On nerve stimulation, Na

enters cell, inner membrane becomes more positive, then K ions leave cell to re-establish positive potential on outside.

These changes occur on a millisecond timescale.

Large changes in charge in one region of the membrane stimulate the neighbouring region and the impulse moves down the length of the cell.

ATP Synthesis

So far looked at electrons,

but what about the protons?

NADH + H

NAD + 2H + 2e

As redox processes occur,

free energy used to force protons into the 14 protons into the intermembrane space.

This creates a concentration

cell across the membrane.

When [H

intermembrane /[H matrix ~ 2.5 a trigger allows protons to flow back across membrane and this spontaneous process drives the non-spontaneous formation of ATP.

Chemical Kinetics

The rate of a reaction is the speed with which the concentrations of the molecules present change. The rate is given by the gradient of a concentration vs time graph. 15

Decomposition of H

2 O 2 2H 2 O 2 (aq) 2H 2

O + O

2 (g)

Start 1.000 M H

2 O 2

After 10 s [H

2 O 2 ] = 0.983 M [H 2 O 2 ]=-0 017 MTime intervalt=10 0 s 16 [H 2 O 2 ] = 0.017 MTime interval t = 10.0 s

Rate of reaction = Rate of change of [H

2 O 2 = [ H 2 O 2 ]/ t = - 0.017/10.0= -1.7 x 10 -3 M s -1

To avoid negative rates....

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