REAKSI REDOKS
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![Page 1: REAKSI REDOKS](https://reader035.fdokumen.com/reader035/viewer/2022081421/5681324a550346895d98bf3d/html5/thumbnails/1.jpg)
REAKSI REDOKS
![Page 2: REAKSI REDOKS](https://reader035.fdokumen.com/reader035/viewer/2022081421/5681324a550346895d98bf3d/html5/thumbnails/2.jpg)
Reaksi Reduksi1. Reaksi yang mengalami
penurunan bilangan oksidasi2. Oksidator3. Reaksi yang menerima
elektron4. Reaksi pelepasan oksigen
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Reaksi Oksidasi1. Reaksi yang mengalami
kenaikan bilangan oksidasi2. Reduktor3. Reaksi yang menyerahkan
elektron4. Reaksi pengikatan oksigen
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ATURAN BILOKS
1. Unsur bebas dan molekul unsur biloksnya = 0
→
→
Contoh :Contoh :
4K
+
O2 2 K2O
P4
6I2
4PI3
+ → Ba
Cl2
BaCl2 +
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2. Logam dalam senyawa biloksnya selalu positif (+)
Contoh Contoh ::2K + Cl2K + Cl2 2
→22KKClCl
Mg + ClMg + Cl2 2
→3Fe + 3Cl3Fe + 3Cl2 2
→2Fe + 3Cl2Fe + 3Cl2 2
→
MgMgClCl22
33FeFeClCl
2222FeFeClCl
33
KK = +1= +1
MgMg =+2=+2FeFe = = +2+2Fe Fe = = +3+3
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3. Hidrogen dalam senyawa biloksnya = +1
Contoh Contoh :: 2H2H22 + +
OO22 → 2H2
OHH22 + + 2K 2K
→ 2KH
MgH2
HH22 + + Mg Mg
→
H=+1H=-1 H=-1
Kecuali : bersenyawa langsung dengan logam, hidrogen
= -1
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4. Oksigen dalam senyawa biloksnya=-2
Contoh Contoh :: 2H2H22 + +
OO22
→ 2H2
OHH22 + O + O22
→ H2O2
KO2K + K + OO22
→
O=-2
O=-1 O=-1/2
Kecuali : a. peroksida = -1 b. superoksida = -1/2 c. OF2 = +2
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5. Biloks ion = muatannya
Contoh Contoh ::O2- = -2
Na+ = +1
Cl- = -1
Al3+ = +3
NH4+ =
+1NO3- = -1
SO42- = -
2Cr2O72- = -2
a. monoatomik b. poliatomik
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6. Total biloks senyawa = 0
Contoh Contoh ::KClOKClO44 = 0 = 0
H2O = 0
MnOMnO22 = 0 = 0
KO2 = 0
HH22SOSO44 = 0 = 0
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CONTOH SOAL:
1.1. Tentukan biloks Cl dalam Tentukan biloks Cl dalam senyawa KClOsenyawa KClO44
1K + 1Cl + 4O = 01K + 1Cl + 4O = 0
+1 + Cl - 8 = 0Cl = +7
1(+1) + 1(Cl) + 4(-2) = 0 1(+1) + 1(Cl) + 4(-2) = 0
Jawab :Jawab :
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2. Tentukan biloks Cr dalam 2. Tentukan biloks Cr dalam senyawa Crsenyawa Cr22OO77
2-2-
2Cr + 7O = -22Cr + 7O = -2
2Cr - 14 = -22Cr = +12
2(Cr) + 7(-2) = -2(Cr) + 7(-2) = -2 2
Jawab : Jawab :
Cr = +6
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A. Penyetaraan Persamaan Reaksi Redoks
1. Metode Bilangan Oksidasi2. Metode Ion Elektron
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1. Tuliskan persamaan reaksi ion yang terjadi antara Al dan NO3
- dalam suasana basa dengan cara perubahan biloks jika diketahui perubahan yang terjadi Al/AlO2
- dan NO3-/NH3
Contoh:
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Jawab:Al
0
+
NO3- → AlO2
- + NH
3
+5
3e+3
8e-3
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8 Al + 3 NO3- → 8 AlO2
- + 3NH3
8Al + 3NO3- + 5OH- + 2H2O
+ 5 OH- + 2 H2O
8AlO2 + 3NH3→
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Contoh:
2. Tuliskan pers. reaksi ion yang terjadi antara KI + KOH + KMnO4 jika diketahui I-/I2 dan MnO4
-
/MnO2
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Jawab:
Langkah 1:
Langkah 2:
oksidasi : I- → I2
reduksi : MnO4- → MnO2
oksidasi : 2 I- → I2
reduksi : MnO4- → MnO2
+ 4 OH-
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Langkah 3:
Oks : 2 I- → I2 + 2e
Red : MnO4- +2H2O + 3e→MnO2
+4OH-
) x3
) x2
Langkah 4:
oksidasi : 6I- → 3 I2 + 6e
reduksi : 2MnO4
- +4H2O + 6e→ 2MnO2 +
8OH-
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Langkah 5:
Reaksi Lengkap:
6KI + 2KMnO4 + 4H2O →
6I- + 2MnO4- +4 H2O →
3 I2 +2MnO2 +
8OH-
3I2 + 2MnO2 +
8KOH
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Pertanyaan:1. Reaksi: 2K + ClK + Cl22 → 2KCl
Bilangan oksidasi K dalam senyawa KCl reaksi adalah …
+1
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2. Zat yang mengalami penurunan bilangan oksidasi dalam reaksi redoks disebut …
oksidator
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