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Setarakan reaksi redoks berikut dengan cara setengah reaksi! CrI 3 ​ + KOH + Cl 2 ​ → K 2 ​ CrO 4 ​ + KIO 4 ​ + KCl + H 2 ​ O ( suasana basa )

Setarakan reaksi redoks berikut dengan cara setengah reaksi!

  

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L. Avicenna

Master Teacher

Mahasiswa/Alumni Institut Teknologi Bandung

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Pembahasan

Cara penyetaraan persamaan reaksi redoks dengan cara setengah reaksi,yaitu dengan melihat elektron yang diterima atau dilepaskan. Reaksi dilakukan pada suasana basa. 1. Tuliskan reaksi ionnya 2. Berikan biloks tiap unsur 3.Membuat reaksi setengah sel dari masing-masing reaksi 4.Menyetarakan atom yang mengalami kenaikan dan penurunan bilangan oksidasi 5. Menambahkan pada sisi yang kelebihanatom O 6.Setarakan unsur H dengan menambahkan pada suasana basa 7.Menyetarakan jumlah muatan dengan penambahan elektron 8.Menyetarakan jumlah elektron dikedua reaksi 9.Menjumlahkan kedua reaksi 10. Mengembalikan ke reaksi semula Jadi reaksi setaranya dalam suasana basa adalah:

Cara penyetaraan persamaan reaksi redoks dengan cara setengah reaksi,yaitu dengan melihat elektron yang diterima atau dilepaskan. Reaksi dilakukan pada suasana basa.

1. Tuliskan reaksi ionnya

Cr to the power of 3 plus sign and 3 I to the power of minus sign and K to the power of plus sign and O H to the power of minus sign and Cl subscript 2 yields 2 K to the power of plus sign and Cr O subscript 4 to the power of 2 minus sign end exponent plus K to the power of plus sign plus I O subscript 4 to the power of minus sign plus K to the power of plus sign and Cl to the power of minus sign and H subscript 2 O 
 

2. Berikan biloks tiap unsur 
 


 

3. Membuat reaksi setengah sel dari masing-masing reaksi

Reduksi space colon Cl subscript 2 yields Cl to the power of minus sign Oksidasi colon space Cr to the power of 3 plus sign yields Cr O subscript 4 to the power of 2 minus sign end exponent Oksidasi colon space 3 I to the power of minus sign yields I O subscript 4 to the power of minus sign  
 

4. Menyetarakan atom yang mengalami kenaikan dan penurunan bilangan oksidasi

Reduksi space colon Cl subscript 2 yields 2 Cl to the power of minus sign Oksidasi colon space Cr to the power of 3 plus sign yields Cr O subscript 4 to the power of 2 minus sign end exponent Oksidasi colon space 3 I to the power of minus sign yields 3 I O subscript 4 to the power of minus sign  
 

5. Menambahkan H subscript bold 2 O pada sisi yang kelebihan atom O

Reduksi space colon Cl subscript 2 yields 2 Cl to the power of minus sign Oksidasi colon space Cr to the power of 3 plus sign yields Cr O subscript 4 to the power of 2 minus sign end exponent plus 4 H subscript 2 O Oksidasi colon space 3 I to the power of minus sign yields 3 I O subscript 4 to the power of minus sign plus 12 H subscript 2 O 
 

6. Setarakan unsur H dengan menambahkan O H to the power of bold minus sign pada suasana basa 

Reduksi space colon Cl subscript 2 yields 2 Cl to the power of minus sign Oksidasi colon space Cr to the power of 3 plus sign and 8 O H to the power of minus sign yields Cr O subscript 4 to the power of 2 minus sign end exponent plus 4 H subscript 2 O Oksidasi colon space 3 I to the power of minus sign and 24 O H to the power of minus sign yields 3 I O subscript 4 to the power of minus sign plus 12 H subscript 2 O 
 

7. Menyetarakan jumlah muatan dengan penambahan elektron

Reduksi space colon Cl subscript 2 and 2 e to the power of minus sign yields 2 Cl to the power of minus sign Oksidasi colon space Cr to the power of 3 plus sign and 8 O H to the power of minus sign yields Cr O subscript 4 to the power of 2 minus sign end exponent plus 4 H subscript 2 O and 3 e to the power of minus sign Oksidasi colon space 3 I to the power of minus sign and 24 O H to the power of minus sign yields 3 I O subscript 4 to the power of minus sign plus 12 H subscript 2 O and 24 e to the power of minus sign 
 

8. Menyetarakan jumlah elektron dikedua reaksi

Reduksi space colon Cl subscript 2 and 2 e to the power of minus sign yields 2 Cl to the power of minus sign space vertical line cross times 27 Oksidasi colon space Cr to the power of 3 plus sign and 8 O H to the power of minus sign yields Cr O subscript 4 to the power of 2 minus sign end exponent plus 4 H subscript 2 O and 3 e to the power of minus sign space vertical line space cross times 2 Oksidasi colon space 3 I to the power of minus sign and 24 O H to the power of minus sign yields 3 I O subscript 4 to the power of minus sign plus 12 H subscript 2 O and 24 e to the power of minus sign space vertical line cross times 2  Reduksi space colon 27 Cl subscript 2 and 54 e to the power of minus sign yields 54 Cl to the power of minus sign space Oksidasi colon space 2 Cr to the power of 3 plus sign and 16 O H to the power of minus sign yields 2 Cr O subscript 4 to the power of 2 minus sign end exponent plus 8 H subscript 2 O and 6 e to the power of minus sign Oksidasi colon space 6 I to the power of minus sign and 48 O H to the power of minus sign yields 6 I O subscript 4 to the power of minus sign plus 24 H subscript 2 O and 48 e to the power of minus sign 
 

9. Menjumlahkan kedua reaksi

27 Cl subscript 2 plus bold 54 italic e to the power of bold minus sign yields 54 Cl to the power of minus sign space 2 Cr to the power of 3 plus sign and bold 16 O H to the power of bold minus sign yields 2 Cr O subscript 4 to the power of 2 minus sign end exponent plus bold 8 H subscript bold 2 O plus bold 6 italic e to the power of bold minus sign 6 I to the power of minus sign and bold 48 O H to the power of minus sign yields 6 I O subscript 4 to the power of minus sign plus bold 24 H subscript bold 2 O plus bold 48 italic e to the power of bold minus sign  bold 27 Cl subscript bold 2 bold and bold 2 Cr to the power of bold 3 bold plus sign bold and bold 6 I to the power of bold minus sign bold and bold 64 O H to the power of bold minus sign bold yields bold 54 Cl to the power of bold minus sign bold and bold 2 Cr O subscript bold 4 to the power of bold 2 bold minus sign end exponent bold plus bold 6 I O subscript bold 4 to the power of bold minus sign bold plus bold 32 H subscript bold 2 O 
 

10. Mengembalikan ke reaksi semula

bold 2 Cr I subscript bold 3 bold and bold 64 K O H bold and bold 27 Cl subscript bold 2 bold yields bold 2 K subscript bold 2 Cr O subscript bold 4 bold plus bold 6 K I O subscript bold 4 bold and bold 54 K Cl bold and bold 32 H subscript bold 2 O 

Jadi reaksi setaranya dalam suasana basa adalah:

bold 2 Cr I subscript bold 3 bold and bold 64 K O H bold and bold 27 Cl subscript bold 2 bold yields bold 2 K subscript bold 2 Cr O subscript bold 4 bold plus bold 6 K I O subscript bold 4 bold and bold 54 K Cl bold and bold 32 H subscript bold 2 O

Latihan Bab

Konsep Redoks dan Biloks

Reaksi Redoks Spesial

Metode Setengah Reaksi

Metode Biloks

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