Chemistry part 31, Julia Burdge,2e (2009) doc

The following are examples of redox reactions: Equations for redox reactions, such as those shown here, can be balanced by inspection, the method of balancing introduced in Chapter 3 [ ~

N ow is a good t im e to revi e w how oxi dation

numbers a r e assigne d [ ~ Section 4.4)

•

Balancing Redox Reactions

In Chapter 4 we briefly discussed oxidation-reduction or "redox" reactions, those in which trons are transferred from one species to another In this section we review how to identify a reac-tion as a redox reaction and look more closely at how such reactions are balanced

elec-A redox reaction is one in which there are c hanges in oxidation states, which we identify

using the rules introduced in Chapter 4 The following are examples of redox reactions:

Equations for redox reactions, such as those shown here, can be balanced by inspection, the method

of balancing introduced in Chapter 3 [ ~~ Section 3.3], but remember that redox equations must

be balanced for mass (number of atoms) and for charge (number of electrons) [ ~~ Section 4.4]

In this section we introduce the half-reaction method to balance equations that cannot be balanced simply by inspection

Consider the aqueous reaction of the iron (II) ion (Fe2+) with the dichromate ion (Cr20~ - ):

Because there is no species containing oxygen on the product side of the equation, it would not

be possible to balance this equation simply by adjusting the coefficients of reactants and products However, there are two things about the reaction that make it possible to add species to the equa-tion to balance it without changing the chemical reaction it represents:

• The reaction takes place in aqueous solution, so we can add H20 as needed to balance the equation

• This particular reaction takes place in acidic solution, so we can add H+ as needed to ance the equation (Some reactions take place in basic solution, enabling us to add OH- as needed for balancing We will learn more about this shortly.)

bal-After writing the unbalanced equation, we balance it stepwise as follows :

1 Separate the unbalanced reaction into half-reactions A half-reaction is an oxidation or a

reduction that occurs as part of the overall redox reaction

Oxidation:

Reduction:

Fe2+ - _ FeH

Cr 2 0 ~ - - _ Cr3+

2 Balance each of the half-reactions with regard to atoms other than 0 and H In this case,

no change is required for the oxidation half-reaction We adjust the coefficient of the chromium(III) ion to balance the reduction half-reaction

Oxidation:

Reduction:

Fe2+ - _ Fe3+

Cr 2 0 ~ - - - _ 2Cr3+

3 Balance both half-reactions for 0 by adding H20 Again, the oxidation in this case requires

no change, but we must add seven water molecules to the product side of the reduction

Oxidation: Fe2+ + FeH

R e duction: Cr 2 0 ~ - - _ 2CrH + 7H20

4 Balance both half-reactions for H by adding H+ Once again, the oxidation in this case

requires no change, but we must add 14 hydrogen ions to the reactant side of the reduction

Oxidation: Fe2+ - _ Fe3+

Reduction: 14H+ + Cr 2 0 ~ - + 2Cr3+ + 7H20

SECTION 19.1 Balancing Redox Reactions 76

5 Balance both half-reactions for charge by adding electrons To do this, we determine the

total charge on each side and add electrons to make the total charges equal In the case of

the oxidation, there is a charge of +2 on the reactant side and a charge of +3 on the product side Adding one electron to the product side makes the charges equal

6 If the number of electrons in the balanced oxidation half-reaction is not the same as the

number of electrons in the balanced reduction half-reaction, multiply one or both of the reactions by the number(s) required to make the number of electrons the same in both In

half-this case, with one electron in the oxidation and six in the reduction, multiplying the

oxida-tion by six accomplishes this

6Fe2+ • 6Fe3+ + ~ e )

Reduction: ~ e ) + l4H+ + Cr20~ - • 2Cr3+ + 7H20

7 Finally, add the balanced half-reactions back together and cancel the electrons, in addition to

any other identical terms that appear on both sides

6Fe2+ • 6Fe3+ + W

A final check shows that the resulting equation is balanced both for mass and for charge

Some redox reactions occur in basic solution When this is the case, balancing by the reaction method is done exactly as described for reactions in acidic solution, but it requires two

half-additional steps:

8 For each H+ ion in the final equation, add one OH- ion to each side of the equation,

combin-ing the H + and OH- ions to produce H20

9 Make any additional cancellations made necessary by the new H20 molecules

Sample Problem 19.1 shows how to use the half-reaction method to balance a reaction that takes place in basic solution

Pennanganate ion and iodide ion react in ba s ic s olution to produce mangane s e(IV) oxide and

molecular iodine Use the half-reaction method to balance the equation:

Mn0 4 + r - Mn0 2 + 1 2

Strategy The reaction takes place in ba s ic solution, so apply steps 1 through 9 to balance for ma s s

and for charge

Setup Identify the oxidation and reduction half-reaction s by assigning oxidation numbers

I

Think About It Verify that the

final equation is balanced for mas s

and for charge Remember that

electrons cannot appear in the

overall balanced equation

Step 2 Balance each half-reaction for mass, excluding 0 and H

Step 8 For each H + ion in the final equation, add one OH - ion to each s id e of the equation,

combining the H + and OH - ion s to produce H20

19.1.1 Which of the following equations does

not represent a redox reaction? ( Select

all that apply.)

19.1.2 Mn04 and C 2 0 ~ - re ac t in basic

solution to form Mn0 2 and CO ~-

What are the coefficients of MnO 4 and

C20~ - in the balanced equation?

a) NH3 + HCI • NH4Cl b) 2H20 2 • 2H20 + O2 a) I and 1

d) 3N02 + H20 • NO + 2HN03 c) 2 and 3 e) LiCl Li + + Cl - d) 2 and 6

e ) 2 and 2

SECTION 19.2 Galvanic Cells 1 53

Galvanic Cells

When zinc metal is Qlaced in a solution containing copper(II) ions, Zn is oxidized to Zn2+ ions

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Electrochemistry- Operatle- :; voltaic ce ll

- Zn(s) + Cu 2 + (aq) -+ Zn2 + (aq) + Cu(s)

The electrons are transferred directly from the reducing agent, Zn, to the oxidizing agent, Cu2+ , in

solution However, if we physically separate two half-reactions from each other, we can arrange it

such that the electrons must travel through a wire in order to pass from the Zn atoms to the Cu2+

gen-erates electricity

reac-This reaction is shown in Figure 4.5

an aqueous CUS04 solution in another container The cell operates on the principle that the

in separate locations with the transfer of electrons between them occurring through an external

particu-lar arrangement of electrodes and electrolytes is called a Daniell cell

-R e duction: Cu ?+ (aq) + 2e- -+ Cu(s)

To complete the electric circuit, and allow electrons to flow through the external wire, the solutions

one half-cell to the other This requirement is satisfied by a salt bridge, which, in its simplest form,

is an inverted U tube containing an inert electrolyte solution, such as KCI or NH4N03 The ions

o

2e - lost perZn atom oxidized

A galvanic cell can also be called a voltaic C?

B o th t erms refer to a cell in which a spoma"!?_ , s

chem i cal reaction generates a f l o w of elee O"S

The salt bridge (an inverted U tube )

containing an Na 2 S04 solution

provides an electrically conducting medium between two solutions Th e

openings of the U tube are loosel y

plugged with cotton balls to pre vent me

flow externally from the Zn electr ode

This is the origin of the terms cathode and

anode:

• Cations move toward the ca t hode

• Anions move toward the anode

The volt is a derived SI unit: 1 V = 1 J/1 C

,

The terms cell potential, cell voltage, cell

electromotive force, and cell emf are used

i nterchangeably and are all symbolized the same

way with f eell

·

== _ "'" Multimedia

Electrochemistry- an interactive voltaic cell

A cell in which different half - reactions take

place wi ll have a different emf

Figure 19.2 The galvanic cell

described in Figure 19.1 Note the U

tube (salt bridge) connecting the two

beakers When the concentrations of

Zn2+ and Cu2+ are 1 molar (1 M) at

25 ° C, the cell voltage is 1.10 V

• •

in the salt bridge must not react with the other ions in solution or with the electrodes (see Figure

19.1) During the course of the redox reaction, electrons flow through the external wire from the

anode (Zn electrode) to the cathode (Cu electrode) In the solution, the cations (Zn2+, Cu2+ , and

+ - ~ " ' ? ' , • • • • •

K ) move toward the cathode, while the anions (S04 - and Cl- ) move toward the anode Without

the salt bridge connecting the two solutions, the buildup of positive charge in the anode

compart-ment (due to the departure of electrons and the resulting formation of Zn2+ ) and the buildup of

negative charge in the cathode compartment (created by the arrival of electrons and the reduction

of Cu2+ ions to Cu) would quickly prevent the cell from operating

An electric current flows from the anode to the cathode because there is a difference in trical potential energy between the electrodes This flow of electric current is analogous to the flow

elec-of water down a waterfall, which occurs because there is a difference in the gravitational potential

energy, or the flow of gas from a high-pressure region to a low-pressure region Experimentally

the difference in - electrical potential between the anode and the cathode is measured by a voltmeter

(Figure 19.2) and the reading (in volts) is called the cell potential (Eceu) The potential of a cell depends not only on the nature of the electrodes and the ions in solution,: but also on the concentra-

•

tions of the ion s and the temper a ture at which the cell i s operated :

The conventional notation for representing galvanic cells is the cell diagram For the cell

shown in Figure 19.1 , if we assume that the concentrations of Zn2+ and Cu2+ ions are 1 M, the cell diagram is

Zn(s) I Zn?+ (1 M) II Cu2+ (1 M) I Cu(s)

The single vertical line represents a phase boundary For example, the zinc electrode is a solid and the Zn2+ ions are in solution Thus, we draw a line between Zn and Zn2+ to show the phase bound-

ary The double vertical lines denote the salt bridge By convention, the anode is written first,

to the left of the double lines, and the other components appear in the order in which we would

encounter them in moving from the anode to the cathode (from left to right in the cell diagram)

Standard Reduction Potentials

When the concentrations of the Cu2+ and Zn2+ ions are both 1.0 M, the cell potential of the cell

described in Section 19.2 is 1.10 V at 25°C (see Figure 19.2) This measured potential is related to

the half-reactions that take place in the anode and cathode compartments The overall cell poten

-tial is the difference between the electric potentials at the Zn and Cu electrodes-known as the

half-cell potentials Just as it is impossible for only one of the half-reactions to occur dently, it is impossible to measure the potential of just a single half-cell However, if we arbitrarily define the potential value of a particular half-cell as zero, we can use it to determine the r e lativ e

indepen-potentials of other half-cells Then we can use these relative potentials to determine overall ce ll

potentials The hydrogen electrode, shown in Figure 19.3, serves as the reference for this purpose

Hydrogen gas is bubbled into a hydrochloric acid solution at 25°C The platinum electrode has

two functions First, it provides a surface on which the dissociation (and oxidation) of hydrogen

molecules can take place:

Second, it serves as an electrical conductor to the external circuit

Salt bridge

+

As before, the ° superscript denotes standard-state conditions, and E O is the standard reduction

potential; that is, EO is the potential associated with a reduction half-reaction at an electrode when

the ion concentration is 1 M and the gas pressure is 1 atm Because the hydrogen electrode is used

to determine all other electrode potentials, it is called the standard hydrogen electrode (SHE)

A half-cell potential is measured by constructing a galvanic cell in which one of the electrodes

is the SHE The measured voltage is then used to determine the potential for the other half-cell

Figure 19.4(a) shows a galvanic cell with a zinc electrode and a SHE When the circuit is completed

in this cell, + electrons flow from the zinc electrode to the SHE, thereby oxidizing Zn to Zn 2+ and

reducing H ions to H2 In this case, therefore, the zinc electrode is the anode (where oxidation

takes place) and the SHE is the cathode (where reduction takes place) The cell diagram is

Zn(s) I Zn2+(l M) II H+(1 M) I HzCl atm) I Pt(s)

The measured potential for this cell is 0.76 V at 25°C We can write the half-cell reactions as

follows:

-Cathode (reduction): 2H + (aq)(1 M) + 2e - • H 2 (g)(1 atrn)

Overall: Zn(s) + 2H + (aq)(l M) - Zn2+(aq)(l M) + H 2(g) (l atm)

Figure 19.3 A hydrogen ele ctr ode

operating under standard-state

conditions Hydrogen gas at 1 atrn

bubbled through aIM Hel soluti on

The electrode it se lf is made of

platinum

One i ndication of the direction of electro n fla

i s the fact that as the reaction proceeds , the mass

of the z in c electrode decreases as a result o f thE

Figure 19.4 (a) A cell consisting of a z inc electrode and a hydrogen electrode (b ) A cell consisting of a co pper electrode and a hydrogen ele c trode _

Both cells are operating under standard-state conditions Note that in (a) the SHE i s the cathode, but in ( b ) it i s the anode

I

Again, the direction of electron flow i s

determined in part by the mass of the electrode

Reduction of Cu '+ ions causes the mas s of the

Cu electrode to increase

The half-reaction with the greater reduction

potential has the greater potential to occur as

a r eduction

By convention, the standard cell potential, E~elj, which is composed of a contribution from the

anode and a contribution from the cathode, is given by

Equation 19.1 E o cell -- E O cathode - EO anode

where E ~athode and E~ode are the standard reduction potentials of the cathode and anode, tively For the Zn-SHE cell, we write

respec-0.76 V = 0 - Ezn 2+/ zn

where the subscript "H+ 1H 2" means "2H+ + 2e - • H2" and the subscript "Zn2+ IZn" means

"Zn2+ + 2e - • Zn" Thus, the standard reduction potential of zinc, E z n 2+/z n is - 0.76 V

The standard reduction potential of copper can be determined in a similar fashion, by using a

cell with a copper electrode and a SHE [Figure 19.4(b)] In this case, electrons flow from the SHE

to the copper electrode when the circuit is completed; that is, the co pper electrode is the cathode ,

and the SHE is the anode

The cell diagram is

Pt(s) H2(1 atm) I H +(l M) I Cu2+(l M) I Cu(s)

and the half-cell reactions are

Anode (ox idati on) : H 2(g)(1 atm) - 2H+(aq)(l M) + 2e

-Cathode ( r eduction) : Cu 2+ (aq)(1 M) + 2e - • Cu(s)

O verall : H 2(g)( 1 atm) + Cu 2 + ( aq)(l M) • 2H + (aq)(1 M) + Cu(s)

Under standard-state conditions and at 25°C, the measured potential for the cell is 0.34 V, so we

Thus, the standard reduction potential of copper, ECu 2+/ c u, is 0.34 V, where the subscript "Cu 2 + ICu"

Having determined the standard reduction potentials of Zn and Cu , we can use Equation

19.1 to calculate the cell potential for the Daniell cell described in ection 19.2:

of products at equilibrium Conversely, a negative E O indicates that reactants will be favored at

equilibrium We will examine how E ~ elj, /1G o , and K are related in Section 19.4 ~

Table 19.1 lists standard reduction potentials, in order of decreasing reduction potential, for

a number of half-cell reactions To avoid ambiguity, all half-cell reactions are shown as redud ions

A galvanic cell is composed of two half-cells, and therefore two half-cell reactions Whether not

a particular half-cell reaction occurs as a reduction in a galvanic cell depends on how its red ction

potential compares to that of the other half-cell reaction If it has the greater (or more positive

-H a lf - Reaction

Co3+(aq) + e - • Co 2+ (aq)

HzOz(aq) + 2H+(aq) + 2e - • 2Hz0(l)

Ce4 + (aq) + e- • Ce 3 + (aq)

Mn04(aq) + 8H + (aq) + 5e - • Mn 2+ (aq) + 4H zO(l)

Au 3 + (aq) + 3e - • Au(s)

CrzO~ - (aq) + 14H+(aq) + 6e - - _ 2Cr 3+ (aq) + 7Hz0(l)

Mn04(aq) + 2Hz0(l) + 3e - • MnO z (s) + 40H - (aq)

Al 3+ (aq) + 3e- • Al(s)

Be 2+ (aq) + 2e- • Be(s)

Mg z+ (aq) + 2e- • Mg(s)

Na + (aq) + e- • Na(s)

Ca 2+ (aq) + 2e - • Ca(s)

S? + (aq) + 2e- • Sr(s) Ba2+(aq) + 2e- • Ba(s)

K+(aq) + e- • K(s)

Li + (aq) + e- • Li(s)

' F or a ll h a l f-react i o n s the concen t ra t ion is 1 M for d i ssolved species and the pressure is 1 atm for gases These are

the sta n dard-state va lu es

The electrode where reduction occurs is t h e

cathode Theref or e, E ~a thode = 0.34 V

The ele ctro de w here oxidat ion occ ur s is th e

anode Therefore, E ~node = - 0.76 V

•

, The Cu half-reaction, having a greater (more positive) reduction potential, is the half-reaction that

Consider, however, what would happen if we were to construct a galvanic eell combining t 'C

half-cell with an Mn half-cell The reduction potential ofMn is - 1.18 V

The overall cell reaction is

E O ce ll - E- O c a thod e - E O a nod e

= (-0.76V) - (-U8V)

= 0.42 V

Mn + Zn2+ - _ I Mn2+ + Zn

By using Equation 19.1, we can predict the direction of an overall cell reaction

It is important to understand that the standard reduction potential is an intensive property

(like temperature and density), not an extensive property (like mass and volume) [I ~~ Section 1.4]

This means that the value of the standard reduction potential does not depend on the amount of a substance involved Therefore, when it is necessary to multiply one of the half-reactions by a coef-

ficient in order to balance the overall equation, the value of E O for the half-reaction remains the

same Consider a galvanic cell made up of a Zn half-cell and an Ag half-cell:

The half-cell reactions are

The standard cell potential can be calculated using Equation 19,1:

E O cell -- E O cathode - E O anode

= 0.80 V - (-0.76 V)

= 1.56 V

,

SECTION 19.3 Standard Reduction Potentials 769

Although we !:l:~.1.t!p.~~~4 ~~~ ~g ~~~~~ r.c:a:-;~~,?~ ?y' ?" .~~ ~i.d I! ? t !:l:~.l.t~p~y ~~~ ~.t.~~~.~~ ~~.~~~~!?~

potential by 2

Table 19.1 is really a more extensive and quantitative version of the activity s eries

[ ~~ Section 4.4] Standard reduction potentials can be used to determine what, if any, redox

reac-tion will take place in solureac-tion

Sample Problems 19.2 and 19.3 illustrate how to use standard reduction potential s to predict the direction of the overall reaction in a galvanic cell and to predict what, if any, redox reaction

will take place in solution

A galvanic cell consists of an Mg electrode in a 1.0 M Mg(N03)2 s olution and a Cd electrode in a

at 25 ° C

the anode, combine cathode and anode half-cell reaction s to get the o ve rall cell reaction, and use

Because the Cd half-cell reaction has the greater (less negati ve) s tandard reduction potential , it will

occur as the reduction The Mg half-cell reaction will occur a s the oxidation Therefore , E ~a ' hode =

-OAO V and E~nod e = -2.37 V

The standard cell potential is

Overall: Mg + Cd2+ - _ Mg 2+ + Cd

EO cell -- E O cathode - E O anode

= ( - OAO V ) - ( -2.3 7 V )

= 1.97 V

of a Cd electrode in a 1.0 M Cd(N03)2 solution and a Pb electrode in a 1.0 M Pb (N 0 3)2 so lution

an Al electrode in a 1.0 M AI(N03)3 s olution and a Cu electrode in a 1.0 M CU (N 0 3)2 so lution

Predict what redox reaction will take place, if any, when molecular bromine ( Br 2) i s added to (a) a

1 M solution of NaI and (b) aIM solution of NaCI (Assume a temperature of 25 ° C )

Br iZ) + 2e- - _ 2Br-(aq)

It is a common error to m ulti ply E" values by

the same number as the half-cell equ atio n

Th i nk of the reductio n potentia l as the he i g ht

of a waterfall Just as water fal l s ' from a hig her

elevat i on to a lower elevat i on, electrons m o e

from an electrode with a hig h er poten tia l t o :.3:::

one with a lower potential The amoun t o f

w ater falling does not affect the net cha nge II'

its elevation Likewise, the nu mbe r of elect rons moving from higher potential to lower pote ntic

does not aff ect the si ze of the change in

potential

Think About It If you ever

done so mething wrong-check

condition s, the o v erall cell reaction

gives a positive E ~ell '

!

,

Think About It We can u s e

Equation 19.1 and treat problem s

of this type like galvanic cell

problems Write the proposed redox

reaction , and identify the " cathode "

and the "anode " If the calculated

E ~ ell is positi v e , the reaction w ill

occur If the calculated E ~e ll i s

negative, the reaction w ill not •

occur

Solution ( a ) If a redox reaction i s to occur, it will be the oxidation of r ions by Br 2 :

Becau s e the reduction potential of Br 2 is gr e ater than that of 12, Br 2 will be reduced to Br - and 1 - will

be oxidi z ed to 12, Thu s, the preceding reaction will occur

( b ) In thi s ca s e , the propo s ed reaction i s the reduction of Br 2 b y CI - ions:

H ow e v er, becau s e th e reduction potential of Br 2 i s s maller than that of C12, thi s reaction will not occur CI2 i s more readil y reduced than Br 2, so Br 2 i s not reduced b y Cl -

Practice Problem A Determine w hat redox reaction, if any , occurs ( at 25 ° C) when l ead metal (Pb) is added to ( a ) a 1.0 M s olution of N iCI2 and (b ) a 1.0 M s o l ution of HCl

Practice Problem B Determine w hat redo x reaction , if any , occurs (at 25 ° C) when Zn(s) is added to

( a ) a 1.0 M s oluti o n of Mn (N 0 3h and ( b ) a 1.0 M s olution of HCl

19.3 1

19.3.2

C al c ul a te E ~ e ll at 2 5 ° C for a gal v anic ce ll 19.3.3 What redox reaction, if any , will occur made of a Cr ele c trode in a s olution that at 2YC when Al metal is placed in a

i s 1.0 M in Cr 3+ and an Au electrod e in a s o l ution that is 1.0 M in Cr 3+ ?

s olution that i s 1.0 Min Au3+

electric energy = volts X coulombs

= joules

The total charge is determined by the number of moles of electrons en) that pass through the cuit By definition,

cir-total charge = nF

SECTION 19.4 Spontaneity of Redox Reactions Under Standard-State Conditions 77

equivalent to 96,485.3 C, although we usually round the number to three significant figures Thus,

reac-tion This energy is used to do electrical work (Welectr ical), so

W max = W elec tri ca l

= -nFEc e ll

where W rrtax is the maximum amount of work that can be done The negative sign on the right-hand

definedJree energy as the energy available to do work Specifically, the change in free energy, D.G,

represents the maximum amount of useful work that can be obtained from a reaction:

D.G = W max

Therefore, we can write

Both nand F are positive quantities and D.G is negative for a spontaneous process, so E ce ll must

be positive for a spontaneous process For reactions in which reactants and products are in their

Equation 19.3 makes it possible to relate E ~e ll to the equilibrium constant, K, of a redox reaction

equilibrium constant as follows [ ~~ Section 18.5, Equation 18 .15] :

When T = 298 K and n moles of electrons are transferred per mole ofreaction, Equation 19.4 can

I Michael Faraday ( 1791 - 1867 ) English c h e mi st and phy s ici s t Farad ay i s regarded by man y as th e greate st ex perim e ntal

scientist of the n i n eteen th century He s tarted as an apprentice to a bookbinder at the age of 1 3, but became interested in

sc ience after readin g a book on chem i s try Farada y in ve nted the electri c motor and was the fir st person to demon s trate the

prin c ipl e go v erning electric generators B es id es making n o table co ntributions to the field s of e l ec tricity a nd ma g neti sm,

Farada y also worked o n optical a c tivity and di scov ered and named b e n ze n e

•

,

•

Think About It Th e large positive

value of I1Go indicates that

reactant s are favored at equilibrium,

which i s consistent with the fact

that E O for the reaction is negative

•

And, by converting to the base-lO logarithm of K, we get

Thus, if we know anyone of the three quantities !1G o , K, or E ge1), we can convert to the others by

using Equations 18.15, 19.3, and 19.5

Sample Problems 19.4 and 19.5 demonstrate the interconversions among !1G o, K, and Egell•

For simplicity, the subscript "cell" is not shown

Sample Problem 19.4

Calculate the sta ndard free-energy change for the following reaction at 25 ° C:

2Au(s) + 3Ca 2+( aq)(l.0M) :;:.=:!:' 2Au 3+(a q)(1.0M) + 3Ca(s)

Strategy Use E O values from Table 19 1 to calculate E O for the reaction, and then u se Equation 19.3

to calculate the standar d free-energy change

Setup The half-cell reactions are

Cathode (reduction) : 3Ca 2+ (aq) + 6e - -+ 3Ca(s)

Anode (ox idation ) : 2Au(s) -+ 2Au 3+( aq) +

6e-From Table 19.1, E '(j+/ ca = -2 87 V and E 'j, u'+/ Au = l.SO V

Solution

E O cell -- E O cathode - E O anode

= -2.87 V - 1.50 V

= -4.37V Next, substitu te thi s va lue of E O into Equation 19.3 to obtain I1G o :

Practice Problem A Calculate I1G o for the following reaction at 25 ° C:

3Mg (s) + 2AI 3+ (aq) :;:.=:!:' 3Mg2 + (aq) + 2AI(s)

Practice Problem B Calculate I1Go for the following reaction at 25 ° C:

Pb (s) + N e+(a q) • • Pb 2+( aq) + Ni(s)

Sample Problem 19.5

Calculate the equilibrium con s tant for the following reaction at 25 ° C:

Strategy Use E O va lues from Table 19.1 to calculate E O for the reaction, and then calculate the

equilibrium constant u s ing Equation 19.5 ( rearranged to solve for K)

Setup The half-cell reactions are

Cathode (reduction): 2Cu 2+ (aq) + 2e- -+ 2Cu+(aq) Anode (ox idation): Sn(s) -+ Sn 2+ (aq) + 2e-