!"#$%&'"()*)+,-./ #%0""4&5(+ 6708%#&+9,.,+:+9;.9++ non-spontaneous reactions by applying electrical energy.

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An electrolysis cell consists of two electrodes in either aqueous solution (of ions)

1 !"#$%&'"()*),-./ 012 3#%0""4&5( 6708%#&9,.,:9;.9 3<='>01 non-spontaneous reactions by applying electrical energy. An electrolysis cell consists of two electrodes in either aqueous solution (of ions) or in a molten salt e.g. molten NaCl. The anode The cathode 3<='>01 3#%0""4&5(012#"#$%&'"()*) 9

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4 Y10"(h#%7##"#$%&'"()*)'T0O4#'4)V06"4)*15 %7#)0G#;$'1)*2#&0J'1).,. Q70%*)*1%7#)'"4J'1R 9. Q70%0&#%7#70"T&#0$J'1)012%7#*&! ' R e'))*s"#$0%7'2#&#0$j'1)f! ' e'))*s"#01'2#&#0$j'1)f! ' 3<='>01 3#%0""4&5(012#"#$%&'"()*) / Y10"(h*15%7##"#$%&'"()*)'T0O4#'4)V06"g I$'1J14#2K. ;.U)#%7#V#&1)%!O40J'1%'0$$'41%T'& bl M c\bbl [ c\,i,] [/ 3*184&#X0%#&I$7015#)70"T$#"" 8'%#1J0")KX7#102>4)%#2%'8L/. Cathode! $#"" Na (aq) e!! Na(s)!2.71 V 2H 2 O(l) 2e!! H 2 (g) 2OH! (aq)!0.42 V Anode! $#"" 2Cl [ (aq)! Cl 2 (g) 2e!!1.36 V 2H 2 O(l)! O 2 (g) 4H (aq) 4e!!0.83 V Which reaction will occur at the electrodes? 3<='>01 3#%0""4&5(012#"#$%&'"()*) ^

5 Y10"(h*15#"#$%&'"()*)'T0O4#'4)V06"g I$'1J14#2K Z.Q70%0&#%7#&#0$J'1)0%%7##"#$%&'2#)R 60%7'2#F Y1'2#F B_#&0""F 3<='>01 3#%0""4&5(012#"#$%&'"()*) - Y10"(h*15#"#$%&'"()*)'T0O4#'4)V06"g I$'1J14#2K. f. Q7(*)01'2#70"T&#0$J'11'%8&#2*$%#2T&'G! ' R 0. =#$04)#6" [ I0OK*)0j&0$%#2%'%7#01'2#)'%7#"'$0" $'1$#1%&0J'1*)kk,3. What is left behind? 3<='>01 3#%0""4&5(012#"#$%&'"()*),]

6 9 DB Z What are the reactions at the electrodes? Cathode Anode Overall 3<='>01 3#%0""4&5(012#"#$%&'"()*),, L'XG4$7#"#$%&*$*%(*)1##2#2*1#"#$%&'"()*)R IL'XG01(#"#$%&'1)RK For example use the electrolysis of CuSO 4 Cu 2 2e!! Cu 1. How many moles of electrons are required to produce a mole of Cu(s)? 2. What is the charge on a mole of electrons? 3. How many Coulombs is this? 4. How long does it take for a mole of electrons to pass through a circuit? 3<='>01 3#%0""4&5(012#"#$%&'"()*),9

$Y@EBVD 64&&#1%012JG# YG8\6n) m401j%('t $70&5#I6K$ 3'"#)'T#"#$%&'1) IT0&020(K 3'"#)'T)4S)%01$#

of e! 1 F = 96,500 coulombs/mol o&0g)'t)4s)%01$#

force that causes electrons to flow (voltage) 1

Amp-sec-Volt = 1 Watt-sec 1 kw-hour = (1000

!Y@EBV Electrolysis of CuSO 4 gives 1.00g of Cu.

7 64&&#1%012JG# YG8\6n) m401j%('t $70&5#I6K 3'"#)'T#"#$%&'1) IT0&020(K 3'"#)'T)4S)%01$# 'i*2*h#2'&&#24$#2 1 Ampere = 1 coulomb/second 1 coulomb = 1 Amp-sec 1 mole of e! = 1 Faraday = 96,500 Coulombs = charge on 1 mole of e! 1 F = 96,500 coulombs/mol o&0g)'t)4s)%01$# 'i*2*h#2'&&#24$#2 Electromotive Force (EMF) force that causes electrons to flow (voltage) 1 Watt = 1 Amp-Volt 1 Joule = 1 coul-volt = 1 Amp-sec-Volt = 1 Watt-sec 1 kw-hour = (1000 Watt)(3600 sec) = 3.6 x 10 6 Watt-sec = 3.6 x 10 6 Joules! 3<='>01 3#%0""4&5(012#"#$%&'"()*),; DY3e?!!?!6@AB?CDED6Y?6U?Y@EBV Electrolysis of CuSO 4 gives 1.00g of Cu. Reaction is: Cu 2 2e!! Cu 1. How many Faradays (F) of charge are required? 2. How many Coulombs is this? 3<='>01 3#%0""4&5(012#"#$%&'"()*),Z

DY3e?!!?!6@AB?CDED6Y?6U?Y@EBVI$'1J14#2K 3. If 1.

However they are not the most abundant elements, and they are not usually

8 3. If 1.00g of Cu is obtained in 1 hour, how many amps are required? 4. If 2 amps were used, how long would it take to produce 5 g of Cu?! 3<='>01 3#%0""4&5(012#"#$%&'"()*),f The majority of elements are metals. However they are not the most abundant elements, and they are not usually found as pure metals in nature. Y88"*$0J'1'T#"#$%&'"()*)F%' #i%&0$%g#%0")t&'g%7#*&'&#). 3<='>01 3#%0""4&5(012#"#$%&'"()*),`

,-./0.12#34#25262.78#3.#2097:;# (<#!"<#5<#=2<#>0 10 (####)*&*'!

9 ,-./0.12#34# #3.#2097:;# (<#!"<#5<#=2<#>0 10 (####)*&*'!"##$%&%$'# % by weight H B C N F Na Mg Al P O S Cl K Ca Ti V Cr Mn Fe Co Cu Zn Si Se Rb Mo Sn I 3<='>01 3#%0""4&5(012#"#$%&'"()*),/ L'X2'G#%0")'$$4&*110%4&#RD'G#0&#T'4120) 84&#G#%0")S4%G')%0&#*13*1#&0") 3#%0")$'GG'1"(T'412*1%7#*&84&##"#G#1%0"T'&GF 3*1#&0")0&#)'"*2*1'&501*$$'G8'412). 3*1#&0")0&#10G#2S($'GG'1g1'%$7#G*$0"g10G#). 3<='>01 3#%0""4&5(012#"#$%&'"()*),^

3')%*G8'&%01%G#%0")0&#T'412*1G*1#&0")0) 'i*2#)g)4"p2#)g'&$0&s'10%#). Aluminosilicates and silicates Metal Al, Si, O!

CuFeS 2 Carbonates CaCO 3 3<='>01 3#%0""4&5(012#"#$%&'"()*) Salts Active metals- Group I, II Not found in nature as pure metals,

salts in the ocean, minerals),- 3#%0""4&5(Fis the science and technology of extracting metals from their natural sources and

10 3')%*G8'&%01%G#%0")0&#T'412*1G*1#&0")0) 'i*2#)g)4"p2#)g'&$0&s'10%#). Aluminosilicates and silicates Metal Al, Si, O! difficult to extract metals (Beryl = Be 3 Al 2 Si 6 O 18 ) Nonsilicate minerals Oxides Al 2 O 3, TiO 2, Fe 2 O 3 Sulfides PbS, ZnS, CuFeS 2 Carbonates CaCO 3 3<='>01 3#%0""4&5(012#"#$%&'"()*) Salts Active metals- Group I, II Not found in nature as pure metals, always combined with other elements (e.g. salts in the ocean, minerals),- 3#%0""4&5(Fis the science and technology of extracting metals from their natural sources and preparing them for practical use E%*1_'"_#) 3*1*15. 6'1$#1%&0J15'&#). A#24$*15'&#)%''S%0*1T&##G#%0").!"#$%&'G#%0""4&5(. e(&'g#%0""4&5( L(2&'G#%0""4&5( e4&*t(*15g#%0"). 3*i*15G#%0")%'T'&G0""'()%70%70_#%7#8&'8#&J#)2#)*&#2. 3<='>01 3#%0""4&5(012#"#$%&'"()*) 9]

Electrometallurgy is the process of obtaining metals through electrolysis.

ELECTROLYSIS Production of metals Na, Al. Purification of Metals Cu. Electroplating.

V06"*)#"#$%&'"(h#2*10 q'x1)$#"".?*o4*2v0 o0)#'4)6" 9 *) 0""'X#2%'2*)8#&)#. 3'"%#1V0*) )*87'1#2'r.

11 Electrometallurgy is the process of obtaining metals through electrolysis. Two different starting materials: -molten salt -aqueous solution COMMERCIAL APPLICATIONS OF ELECTROLYSIS Production of metals Na, Al. Purification of Metals Cu. Electroplating. 3<='>01 3#%0""4&5(012#"#$%&'"()*) 9, D'2*4GG#%0"*)'S%0*1#2_*0#"#$%&'"()*)'TG'"%#1 )'2*4G$7"'&*2#. V06"*)#"#$%&'"(h#2*10 q'x1)$#"".?*o4*2v0 o0)#'4)6" 9 *) 0""'X#2%'2*)8#&)#. 3'"%#1V0*) )*87'1#2'r.!"#$%&'"()*)'TG'"%#1)0"%)*)%7#BV?CX0(%'5#%%7#0$J_#G#%0")*184&#T'&G. Io&'48Ego&'48EEgY"4G*14GK 3<='>01 3#%0""4&5(012#"#$%&'"()*) 99

12 @7*)*)01#1#&5(*1%#1)*_#8&'$#)). What are the reactions at the electrodes? Cathode Anode Overall PYROMETALLURGY: using high temperatures to obtain the free metal. Depending on the mineral/metal, some of these steps are employed: Calcination is heating of ore to cause decomposition and elimination of a volatile product: PbCO 3 (s)! PbO(s) CO 2 (g) Roasting is heating which causes chemical reactions between the ore and the furnace atmosphere: 1. Burns off organic matter. 2. Converts carbonates and sulfides to oxides: 2 ZnS(s) 3O 2 (g)!2zno(s) SO 2 (g) 3. Less active metals are often reduced HgS(s) O 2 (g)! Hg(l) SO 2 (g) Smeltingis a melting process in which materials formed during reactions separate into two or more layers. Refining is the treatment of a crude, relatively impure metal to improve its purity and better define its composition. 3<='>01 3#%0""4&5(012#"#$%&'"()*) 9Z

DG#"J15'T*&'1*)01#i0G8"#'Te(&'G#%0""4&5( sources of iron: hematite Fe 2 O 3 and

Iron Ore: Fe 2 O 3 and SiO 2 Add limestone and coke Coke is coal that has been

3<='>01 3#%0""4&5(012#"#$%&'"()*) 9f @7#)#0&#%7#$7#G*$0"&#0$J'1)'$$4&&*15*1S"0)%

2CO(g) heat heat C(s) H 2 O(g)! CO(g) H 2 (g) Fe 3 O 4 (s) 4CO(g)!

13 DG#"J15'T*&'1*)01#i0G8"#'Te(&'G#%0""4&5( sources of iron: hematite Fe 2 O 3 and magnetite Fe 3 O 4. Iron Ore: Fe 2 O 3 and SiO 2 Add limestone and coke Coke is coal that has been heated to drive off the volatile components. e4&*p#2*&'1#i*%)%7#t4&10$#0%%7# S'j'G. 3<='>01 3#%0""4&5(012#"#$%&'"()*) T4&10$#.E15&#2*#1%)F*&'1'&#g"*G#)%'1#g$'l# Reactions 2C(s) O 2 (g)! 2CO(g) heat heat C(s) H 2 O(g)! CO(g) H 2 (g) Fe 3 O 4 (s) 4CO(g)! 3Fe(l) 4CO 2 (g) Fe 3 O 4 (s) 4H 2 (g)! 3Fe(l) 4H 2 O(g) Coke: 1) heats furnace 2) reduces iron Why is water added? Why is limestone CaCO 3 added? 3<='>01 3#%0""4&5(012#"#$%&'"()*) 9`

CaCO 3 does not participate in the reduction of Fe. Why is limestone CaCO 3 added? At high T CaCO 3! CaO CO 2 CaO SiO 2!

slag oxide oxide basic acidic Limestone (CaCO 3 ) removes SiO 2 (and other) impurities slag floats on Fe(l); protects it from

pig iron, is brittle; not strong. Some &#p1*15't%7#8&'24$%1##2)%'s#2'1#.

14 CaCO 3 does not participate in the reduction of Fe. Why is limestone CaCO 3 added? At high T CaCO 3! CaO CO 2 CaO SiO 2! CaSiO 3 (l) Metal nonmetal! slag oxide oxide basic acidic Limestone (CaCO 3 ) removes SiO 2 (and other) impurities slag floats on Fe(l); protects it from oxidation by O 2 Slag: cement cinder block building materials 3<='>01 3#%0""4&5(012#"#$%&'"()*) 9/ Product in blast furnace, pig iron, is brittle; not strong. Some &#p1*15't%7#8&'24$%1##2)%'s#2'1#. A#p1*15F2'1#*1 Bessemer Converter O 2 (g) bubbled through molten iron to oxidize remaining impurities CaO slag still present to remove impurities Alloying elements added as liquid iron is being removed. Purified molten steel is poured into molds. 3<='>01 3#%0""4&5(012#"#$%&'"()*) 9^

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15 L(2&'G#%0""4&5(F0&#%#$71*O4#)*1X7*$70G#%0"*) #i%&0$%#2t&'g'&#_*0%7#4)#'t0o4#'4)&#0$j'1).?#0$7*15*)08&'$#))*1x7*$70g#%0"w$'1%0*1*15$'g8'412*) )#"#$J_#"(2*))'"_#2. 3#%0")"*l#Y4gY50&#T'412*110%4&#0)84&##"#G#1%)g7'X#_#&g "0&5#2#8')*%)'T%7#)#G#%0")0&#&0&#. Q#$01&#G'_#)G0""0G'41%)T&'G&'$lS("#0$7*15. B1#$014)#X0%#&*T%7#G#%0"W$'1%0*1*15$'G8'412*)X0%#&)'"4S"#g S4%G'&#'s#1'1#G4)%4)#0$*2gS0)#g'&0)0"%)'"4J'1.!i0G8"#F2*))'"_#5'"2S($'G8"#i0J'1X*%76V [ F ZY4I"KM^6V [ I#$KMB 9 I%KM9L 9 BI!K!ZY4I6VK 9[ I#$KMZBL [ I#$K t T by4i6vk 9 c [ \9i,] ;^ e4&#5'"2*)%7#1's%0*1#2s(&#24$j'1f 9Y4I6VK 9[ I#$KMN1I"K!N1I6VK Z 9[ I#$KM9Y4I"K 3<='>01 3#%0""4&5(012#"#$%&'"()*) 9- Y"4G*14G'&#*)'S%0*1#2*1$'1$#1%&0%#2T'&G_*0 L(2&'G#%0""4&5(.@7*)*124)%&*0"8&'$#))*)$0""#2%7#=0(#&e&'$#)) Y"4G*14G*))#$'12G')% 4)#T4"G#%0". Bauxite: Al 2 O 3.xH 2 O. primary ore for Al uf]vy" 9 B ; impurities: SiO 2 Fe 2 O 3 =0(#&e&'$#))FS04i*%#*)$'1$#1%&0%#2%'8&'24$#0"4G*14G'i*2#. Y" 9 B ; L 9 BI"KM9L 9 BI!KM9BL [ I#$K!9Y"IBLK Z[ I#$K hydrated metal complex q*))'"_#s04i*%#*1)%&'15s0)#iv0blk0%7*57@ge(al 2 O 3 dissolves) a*"%#&'4%)'"*2)g0"4g*10%#*'1)%0()2*))'"_#2. a# 9 B ; gd*b 9 2'1'%2*))'"_#?'X#&8LgY"IBLK ; 3<='>01 3#%0""4&5(012#"#$%&'"()*) ;]

Problem: Al 2 O 3 melts at 2000 C and it is impractical to perform electrolysis on the molten salt.

16 The only way to get Aluminum metal is via electrolysis (electrometallurgy). In the Hall process, an electrolysis cell is used to produce aluminum. Problem: Al 2 O 3 melts at 2000 C and it is impractical to perform electrolysis on the molten salt. Hall: use purified Al 2 O 3 in molten cryolite (Na 3 AlF 6, melting point 1012 C). Anode: C(s) 2O 2! (l)! CO 2 (g) 4e [ Cathode: 3e [ Al 3 (l)! Al(l) The graphite rods are consumed in the reaction. 3<='>01 3#%0""4&5(012#"#$%&'"()*) ;, The cost of purifying aluminum is high. 3<='>01 3#%0""4&5(012#"#$%&'"()*) ;9

$2Cu 2 (aq) 2SO 4 2- (aq) Fe 2 O 3 (s) 3S(s) H 2 O \ / 2CuSO 4 (aq) " CuSO 4 (aq) is electrolyzed to Cu: but still not pure Because of its good conductivity, Cu is used to make electrical wiring.$

17 !i%&0$j'101284&*p$0j'1't640")'&#o4*&#) %X')%#8)F7(2&'G#%0""4&5(012#"#$%&'"()*). Hydrometallurgy: Copper containing ore (CuFeS 2 ) is stirred with aqueous H 2 SO 4 O 2 Redox Reaction 2CuFeS 2 (s)2h (aq)mso 4 2! (aq) 4O 2 (g)! 2Cu 2 (aq) 2SO 4 2- (aq) Fe 2 O 3 (s) 3S(s) H 2 O \ / 2CuSO 4 (aq) " CuSO 4 (aq) is electrolyzed to Cu: but still not pure Because of its good conductivity, Cu is used to make electrical wiring. Impurities reduce conductivity, therefore pure copper is required in the electronics industry. 3<='>01 3#%0""4&5(012#"#$%&'"()*) ;; A second electrolysis process is done to purify copper What are the reactions at the electrodes? Cathode thin sheet of pure copper Anode impure copper As the reaction proceeds, what happens to Cu? 3<='>01 3#%0""4&5(012#"#$%&'"()*) ;Z

Chapter 20 CHEMISTRY. Metallurgy and the Chemistry of Metals. Dr. Ibrahim Suleiman

Chapter 20 CHEMISTRY. Metallurgy and the Chemistry of Metals. Dr. Ibrahim Suleiman CHEMISTRY Chapter 20 Metallurgy and the Chemistry of Metals Dr. Ibrahim Suleiman GENERAL PROPERTIES AND STRUCTURE OF METALS opaque good conductors of heat and electricity high malleability and ductility