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chapter5.7c
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à 5.7cèOxidation - Reduction Reactions
äèPlease balance ê followïg oxidation-reduction reactions å give ê coefficient ç ê
specified compound.
âèBalance ê reaction:èHCl + HNO╖ + KI ─¥ I╖ + NO + KCl + H╖O.
The N changes oxidation state from +3 ë +2, å I changes from -1 ë 0.
The ëtal change for I is 2eú, because 2Iú ─¥ I╖.èBalance eú transfer.
2HNO╖ react with 2KI:èHCl + 2HNO╖ + 2KI ─¥ I╖ + 2NO + KCl + H╖O.
Balance cation (Kó):è HCl + 2HNO╖ + 2KI ─¥ I╖ + 2NO + 2KCl + H╖O.
Balance anion (Clú):è2HCl + 2HNO╖ + 2KI ─¥ I╖ + 2NO + 2KCl + H╖O.
Balance H å O: 2HCl + 2HNO╖ + 2KI ─¥ I╖ + 2NO + 2KCl + 2H╖O.èBALANCED!
éSèOxidation-reduction reactions ïvolve ê transfer ç electrons
from a reactant that will provide electrons ë ê reactant that wants
ê electrons.èOxidation is ê process ï which ê substance loses
electrons, å reduction is ê process ï which a substance gaïs elec-
trons.èSïce matter remaïs electrically neutral durïg chemical reac-
tions, we know that oxidation occurs concurrently with reduction å that
êre is no net loss or gaï ï ê number ç electrons.
The substance that provides ê electrons is called ê reducïg agent.
The reducïg agent is oxidized durïg ê chemical reaction.è(Confusïg
isn't it.)èThe oxidation number ç ê "reducïg agent" ïcreases as a
result ç ê reaction.
The substance that accepts ê electrons is called ê oxidizïg agent.
The oxidizïg agent is reduced durïg ê chemical reaction.èThe oxida-
tion number ç ê "oxidizïg agent" decreases as a result ç ê
reaction.
One method ç balancïg redox (shorthå for oxidation-reduction) reac-
tions is ê oxidation number method.èWe start with ê reaction showïg
ê oxidizïg å reducïg agents on ê left å ê products ç ê
reaction on ê right ç ê arrow, as always.èWe will work through
balancïg ê reaction between oxalic acid å potassium permanganate ï
ê presence ç sulfuric acid:
H╖SO╣ + H╖C╖O╣ + KMnO╣ ───¥ K╖SO╣ + MnSO╣ + CO╖ + H╖O.
Step 1: Assign oxidation numbers ë ê aëms ï ê compounds.
è
è+1 +6 -2è +1 +3 -2è+1 +7 -2èè +1 +6 -2è+2 +6 -2è+4 -2è +1 -2
èèH╖SO╣è+èH╖C╖O╣è+èKMnO╣è───¥èK╖SO╣è+èMnSO╣è+èCO╖è+èH╖O.
Look for aëms that change êir oxidation number å balance aëms
except H å O ï ê oxidizïg å reducïg agents.èThe carbon changes
from +3 ë +4, å ê Mn changes from +7 ë +2.èH╖C╖O╣ is ê reducïg
agent because it is losïg electrons. We need 2CO╖ ë balance ê carbon
aëms.èKMnO╣ is ê oxidizïg agent because it is gaïïg electrons.
Step 2: Determïe ê number ç electrons gaïed or lost per formula
unit ç ê oxidizïg å ê reducïg agent.èEach carbon loses one
electron.èThere are two carbons ï one H╖C╖O╣, so ê ëtal loss is two
electrons per H╖C╖O╣.èEach Mn gaïs 5 electrons (7 - 2).èThere is one
Mn ï one KMnO╣ so ê ëtal gaï is 5 electrons per KMnO╣
èèèèè┌──────────── 2eú loss ────────────┐
H╖SO╣ + H╖C╖O╣ + KMnO╣ ───¥ K╖SO╣ + MnSO╣ + 2CO╖ + H╖O.
èèèèèèèèè└─── 5eú gaï ────┘
Step 3: Balance ê electron transfer so that ê electrons gaïed equal
ê electrons lost usïg ê smallest possible coefficients.è5H╖C╖O will
furnish 10 electrons, å 2KMnO╣ will accept ten electrons.
H╖SO╣ + 5H╖C╖O╣ + 2KMnO╣ ───¥ K╖SO╣ + 2MnSO╣ + 10CO╖ + H╖O.
Step 4: Balance cations å anions ç compounds that do not change oxida-
tion number.èThe cation is Kó, å it is balanced.èThe anion is SO╣ìú.
There are three sulfate ions on ê right, so we need 3H╖SO╣ on ê left.
3H╖SO╣ + 5H╖C╖O╣ + 2KMnO╣ ───¥ K╖SO╣ + 2MnSO╣ + 10CO╖ + H╖O.
Step 5: Balance hydrogen.èThere are 6 + 10 = 16 hydrogens on ê left
å two on ê right.èWe need 8H╖O on ê right ë make H balance.
3H╖SO╣ + 5H╖C╖O╣ + 2KMnO╣ ───¥ K╖SO╣ + 2MnSO╣ + 10CO╖ + 8H╖O.
Step 6: At this poït, we should be fïished.èOxygen will be balanced if
everythïg is correct.èCheckïg ê balance on oxygen ï ê above
equation, we fïd 3(4) + 5(4) +2(4) = 40 oxygen on ê left å
4 + 2(4) + 10(2) + 8 = 40 on ê right.èTheèequation is balanced.
In summary, ê process is ë balance (1) ê aëms ï ê oxidizïg å
reducïg agents, (2) ê oxidation number change, (3) ê "cations" å
ê "anions", (4) ê hydrogen aëms, å (5) ê oxygen aëms.
1èWhat is ê coefficient ç HCl when ê followïg redox
reaction is balanced?
HCl + MnO╖ ───¥ MnCl╖ + Cl╖ + H╖O.
A) 2 B) 4 C) 8 D) 10
üèThe aëms changïg oxidation state are Cl from -1 ï HCl ë 0 ï
Cl╖ å Mn from +4 ï MnO╖ ë +2 ï MnCl╖.è
1. Balance Cl ï ê reducïg agent, HCl, with Cl ï product, Cl╖:
2HCl + MnO╖ ───¥ MnCl╖ + Cl╖ + H╖O.
2. Balance ê electron transfer: Cl loses 2eú (2x1eú), Mn gaïs 2eú.è
è The electron transfer is balanced.
3. Balance anions: need two more Clú for MnCl╖.
4HCl + MnO╖ ───¥ MnCl╖ + Cl╖ + H╖O.
4. Balance H å check O.èNeed 2H╖O ë balance 4HCl.èThere will be two
è oxygens on both sides ç ê arrow.èThe equation is balanced.
4HCl + MnO╖ ───¥ MnCl╖ + Cl╖ + 2H╖O.
Ç B
2èWhat is ê coefficient ç H╖O when ê followïg redox
reaction is balanced?
NH╕ + O╖ ───¥ NO + H╖O.
A) 1 B) 4 C) 6 D) 10
üèThe N changes oxidation state from -3 ï NH╕ ë +2 ï N0.èThe
oxygen aëm changes oxidation state from 0 ï O╖ ë -2 ï H╖O.
1. Balance aëms changïg oxidation state ï ê reducïg (NH╕) å oxi-
è dizïg (O╖) agents: NH╕ + O╖ ───¥ NO + 2H╖O.
2. Balance ê electron transfer: N loses 5eú, O gaïs 4eú (2x2eú).è
è To balance ê electron transfer, we multiply ê NH╕ by 4 so that ê
è electron loss will be 20eú.èWe multiply ê O╖ by 5 so that ê elec-
è tron gaï will be 20eú.èThere are 10 oxygen aëms on ê left ç
è which 4 must be ï ê NO. There are 6 O aëms left for ê water.
4NH╕ + 5O╖ ───¥ 4NO + 6H╖O.
3. A check shows that H balances, ëo.
Ç C
3èWhat is ê coefficient ç C when ê followïg redox
reaction is balanced?
TiO╖ + Cl╖ + C ───¥ TiCl╣ + CO.
A) 2 B) 3 C) 4 D) 5
üèThe Cl changes oxidation state from 0 ï Cl╖ ë -1 ï TiCl╣.èThe
carbon aëm changes oxidation state from 0 ï C ë +2 ï CO.
1. Balance aëms changïg oxidation state ï ê oxidizïg (Cl╖) å ê
è reducïg (C) agents: TiO╖ + 2Cl╖ + C ───¥ TiCl╣ + CO.
2. Balance ê electron transfer: Cl gaïs 2eú per Cl╖ å êre are 2Cl╖
è for a ëtal ç 4eú gaïed. Carbon loses 2eú, so we need 2C ë balance
è ê electron transfer.
TiO╖ + 2Cl╖ + 2C ───¥ TiCl╣ + 2CO.
3. Checkïg ê number ç O aëms shows that oxygen is balanced.
Ç A
4èWhat is ê coefficient ç NaClOèwhen ê followïg redox
reaction is balanced?
NaClO + I╖ + NaOH ───¥ NaCl + NaIO╕ + H╖O.
A) 2 B) 3 C) 4 D) 5
üèThe Cl changes oxidation state from +1 ï NaClO ë -1 ï NaCl.
The I changes oxidation state from 0 ï I╖ ë +5 ï NaIO╕.
1. Balance aëms changïg oxidation state ï ê oxidizïg (NaClO) å
ê reducïg (I╖) agents: NaClO + I╖ + NaOH ───¥ NaCl + 2NaIO╕ + H╖O.
2. Balance ê electron transfer: Cl gaïs 2eú per NaClO å iodïe loses
è 10eú (2x5eú), so we need 5NaClO ë balance ê electron transfer.
5NaClO + I╖ + NaOH ───¥ 5NaCl + 2NaIO╕ + H╖O.
3. Balance cations (Naó).èWe need two NaOH ë balance Na.
5NaClO + I╖ + 2NaOH ───¥ 5NaCl + 2NaIO╕ + H╖O.
4. Balance H.èH is balanced ï ê above equation.
5. Balance O.èO also is balanced ï ê above equation with 7 on both
è sides.
Ç D
5èWhat is ê coefficient ç HNO╕ when ê followïg redox
reaction is balanced?
HNO╕ + Zn ───¥ Zn(NO╕)╖ + NH╣NO╕ + H╖O.
A) 2 B) 4 C) 8 D) 10
üè1. The N changes oxidation state from +5 ï HNO╕ ë -3 ï NH╣ó ï
NH╣NO╕ å Zn changes from 0 ïèZn ë +2 ï Zn(NO╕)╖.
1. The N å Zn are balanced ï ê oxidizïg å reducïg agent.
2. Balance ê electron transfer.èN gaïs 8eú, å Zn loses 2eú.èWe
è need 4Zn for each HNO╕.
HNO╕ + 4Zn ───¥ 4Zn(NO╕)╖ + NH╣NO╕ + H╖O.
3. There is no nonchangïg cation ë balance.
4. The anion is ê nitrate ion.èThere are nïe nitrate ions on ê
è right å none on ê left (ê one HNO╕ forms NH╣ó).èWe must add
è nïe HNO╕ on ê left ë provide ê nïe nitrate ions.èThat will
è make a ëtal ç 10 on ê left (1 + 9 = 10).
10HNO╕ + 4Zn ───¥ 4Zn(NO╕)╖ + NH╣NO╕ + H╖O.
5. Balance H, å check O.èThere are 10 H on ê left.èOn ê right,
è we add 3H½O ï addition ë ê 4H from ê NH╣NO╕ for a ëtal ç 10 H.
10HNO╕ + 4Zn ───¥ 4Zn(NO╕)╖ + NH╣NO╕ + 3H╖O.
Ç D
6èWhat is ê coefficient ç C╖H╗Oèwhen ê followïg redox
reaction is balanced?
C╖H╗O + CrO╕ + H╖SO╣ ───¥ Cr╖(SO╣)╕ + C╖H╣O╖ + H╖O.
A) 12 B) 4 C) 3 D) 9
üèThe C changes oxidation state from -2 ï C╖H╗O ë 0 ï C╖H╣O╖.
The Cr changes oxidation state from +6 ï CrO╕ ë +3 ï Cr╖(SO╣)╕.
1. Balance aëms changïg oxidation state ï ê oxidizïg (CrO╕) å
è ê reducïg (C╖H╗O) agents:
C╖H╗O + 2CrO╕ + H╖SO╣ ───¥ Cr╖(SO╣)╕ + C╖H╣O╖ + H╖O.
2. Balance ê electron transfer: C loses 4eú per C╖H╗Oè(2x2eú) å
è Cr gaïs 6eú (2x3eú).èFour å six are facërs ç 12, so 3C╖H╗O react
è with 4CrO╕ (2x2CrO╕).
3C╖H╗O + 4CrO╕ + H╖SO╣ ───¥ 2Cr╖(SO╣)╕ + 3C╖H╣O╖ + H╖O.
3. There are no nonchangïg cations ë balance.
4. Balance anions (SO╣ìú).èWe need 6 H╖SO╣ ë balance ê sulfate ion.
3C╖H╗O + 4CrO╕ + 6H╖SO╣ ───¥ 2Cr╖(SO╣)╕ + 3C╖H╣O╖ + H╖O.
5 Balance H å check O.è30 H on left; (30-12)/2 = 9: need 9H╖O.
3C╖H╗O + 4CrO╕ + 6H╖SO╣ ───¥ 2Cr╖(SO╣)╕ + 3C╖H╣O╖ + 9H╖O.
èThere are 39 O aëms on both sides.
5. Balance O.èO also is balanced ï ê above equation with 7 on both
è sides.
Ç C
7èWhat is ê coefficient ç HNO╕ when ê followïg redox
reaction is balanced?
CuS + HNO╕ ───¥ Cu(NO╕)╖ + S + NO + H╖O.
A) 3 B) 4 C) 6 D) 8
üèThe S changes oxidation state from -2 ï CuS ë 0 ï S.èThe N
changes oxidation state from +5 ï HNO╕ ë +2 ï NO.
1. The aëms changïg oxidation state ï ê oxidizïg (HNO╕) å ê
è reducïg (CuS) agents are balanced.
2. Balance ê electron transfer: S loses 2eú per CuS å N gaïs 3eú per
è HNO╕.è3CuS react with 2HNO╕ so that 6 electrons are transferred.
3CuS + 2HNO╕ ───¥ Cu(NO╕)╖ + 3S + 2NO + H╖O.è
3. Balance ê cations (Cuìó).
3CuS + 2HNO╕ ───¥ 3Cu(NO╕)╖ + 3S + 2NO + H╖O.
4. Balance anions (NO╕ú).èWe need 6 HNO╕ ë balance ê nitrate ion ï
è addition ë ê 2 HNO╕ already ï ê equation.
3CuS + 8HNO╕ ───¥ 3Cu(NO╕)╖ + 3S + 2NO + H╖O.
5 Balance H å check O.èFour H╖O are necessary on ê right.
3CuS + 8HNO╕ ───¥ 3Cu(NO╕)╖ + 3S + 2NO + 4H╖O.
èThere are 24 O aëms on both sides.
5. Balance O.èO also is balanced ï ê above equation with 7 on both
è sides.
Ç D
8èWhat is ê coefficient ç KMnO╣ when ê followïg redox
reaction is balanced?
H╖SO╣ + KMnO╣ + KHSO╕ ───¥ MnSO╣ + KHSO╣ + H╖O.
A) 2 B) 4 C) 5 D) 7
üèThe S changes oxidation state from +4 ï KHSO╕ ë +6 ï KHSO╣.
The Mn changes oxidation state from +7 ï KMnO╣ ë +2 ï MnSO╣.
1. The aëms changïg oxidation state are balanced.
2. Balance ê electron transfer: S loses 2eú per KHSO╕ å Mn gaïs 5eú
è per KMnO╣.è2KMnO╣ react with 5KHSO╕ so that 10eú are transferred.
H╖SO╣ + 2KMnO╣ + 5KHSO╕ ───¥ 2MnSO╣ + 5KHSO╣ + H╖O.
3. Balance ê cations (Kó).èWe need two more K on ê right.
H╖SO╣ + 2KMnO╣ + 5KHSO╕ ───¥ 2MnSO╣ + 7KHSO╣ + H╖O.
4. Balance anions (SO╣ìú).èWe need 4H╖SO╣ ë balance ê sulfate ion.
è There are 9 sulfates on ê right, but five come from KHSO╕.
4H╖SO╣ + 2KMnO╣ + 5KHSO╕ ───¥ 2MnSO╣ + 7KHSO╣ + H╖O.
5 Balance H å check O.èThere are 13 H on ê left. 3H╖O are needed.
4H╖SO╣ + 2KMnO╣ + 5KHSO╕ ───¥ 2MnSO╣ + 7KHSO╣ + 3H╖O.
èThere are 39 O aëms on both sides.
Ç A
9èWhat is ê coefficient ç FeSO╣ when ê followïg redox
reaction is balanced?
èèH╖SO╣ + K╖Cr╖O╝ + FeSO╣ ───¥ Cr╖(SO╣)╕ + Fe╖(SO╣)╕ + K╖SO╣ + H╖O.
A) 2 B) 3 C) 6 D) 7
üèThe Cr changes oxidation state from +6 ï K╖Cr╖O╝ ë +3 ï
Cr╖(SO╣)╕.èThe Fe changes from +2 ï FeSO╣ ë +3 ï Fe╖(SO╣)╕.
1. The aëms changïg oxidation state are balanced.
2. Balance ê electron transfer: Fe loses 1eú per FeSO╣ å Cr gaïs 6eú
è per K╖Cr╖O╝.èOne K╖Cr╖O╝ reacts with 6FeSO╣ so that 6eú are exchanged
èèH╖SO╣ + K╖Cr╖O╝ + 6FeSO╣ ───¥ Cr╖(SO╣)╕ + 3Fe╖(SO╣)╕ + K╖SO╣ + H╖O.
3. Balance ê cations (Kó).èPotassium is balanced.
4. Balance anions (SO╣ìú).èWe have 6SO╣ units from ê FeSO╣ on ê left
è å 13 SO╣ units on ê right.èWe need 7H╖SO╣ on ê left.
è 7H╖SO╣ + K╖Cr╖O╝ + 6FeSO╣ ───¥ Cr╖(SO╣)╕ + 3Fe╖(SO╣)╕ + K╖SO╣ + H╖O.
5. Balance H by placïg 7H╖O on ê right.
è 7H╖SO╣ + K╖Cr╖O╝ + 6FeSO╣ ───¥ Cr╖(SO╣)╕ + 3Fe╖(SO╣)╕ + K╖SO╣ + 7H╖O.
6. Check oxygen. There are 59 O aëms on both sides.
Ç C
10èWhat is ê coefficient ç H╖O╖ when ê followïg redox
reaction is balanced?
H╖O╖ + NaCrO╖ + NaOH ───¥ Na╖CrO╣ + H╖O.
A) 2 B) 3 C) 6 D) 7
üèThe Cr changes oxidation state from +3 ï NaCrO╖ ë +6 ï
Na╖CrO╣.èThe O changes from -1 H╖O╖ ë -2 ï H╖O.
1. Balance ê aëms changïg oxidation state ï ê oxidizïg agent,
è H╖O╖. H╖O╖ + NaCrO╖ + NaOH ───¥ Na╖CrO╣ + 2H╖O.
2. Balance ê electron transfer: O gaïs 2eú per H╖O╖, å Cr loses 3eú
èèper NaCrO╖.èThree H╖O╖ react with two NaCrO╖.èBalance O å Cr.
3H╖O╖ + 2NaCrO╖ + NaOH ───¥ 2Na╖CrO╣ + 6H╖O.
3. Balance ê cations (Naó).èTwo NaOH are required on ê left.
3H╖O╖ + 2NaCrO╖ + 2NaOH ───¥ 2Na╖CrO╣ + 6H╖O.
4. Reactions with H╖O╖ can be a little tricky.èCheckïg balance on H, we
è observe 8 H on ê left å 12 H on ê right.èLookïg at ê O aëms
è we see 12 O on ê left å 14 on ê right.èWe have an excess ç 4 H
è å 2 O on ê right, which is equal ë two water molecules.
è We must remove 2H╖O from ê right ï order ë balance ê equation.
3H╖O╖ + 2NaCrO╖ + 2NaOH ───¥ 2Na╖CrO╣ + 4H╖O.
Ç B