Roboguru

Setarakan reaksi redoks berikut. (suasana basa)

Setarakan reaksi redoks berikut.


begin mathsize 14px style Cr O subscript 4 to the power of 2 minus sign open parentheses italic a italic q close parentheses plus Fe open parentheses O H close parentheses subscript 2 open parentheses italic s close parentheses rightwards arrow Cr subscript 2 O subscript 3 open parentheses italic s close parentheses plus Fe open parentheses O H close parentheses subscript 3 left parenthesis italic s right parenthesis end style (suasana basa)

  1. ... space  

  2. ... undefined 

Jawaban:

Tentukan atom yang mengalami perubahan bilangan oksidasi dan setarakan jumlah atom yang berubah bilangan oksidasinya.begin mathsize 14px style space end style 

undefined 

Kalikan pengali dengan koefisien:undefined 

begin mathsize 14px style bold 2 Cr O subscript 4 to the power of 2 minus sign end exponent left parenthesis italic a italic q right parenthesis plus bold 6 Fe open parentheses O H close parentheses subscript 2 left parenthesis italic a italic q right parenthesis yields Cr subscript 2 O subscript 3 open parentheses italic s close parentheses and bold 6 Fe open parentheses O H close parentheses subscript 3 left parenthesis italic s right parenthesis end style 

Setarakan jumlah O dengan menambahkan begin mathsize 14px style H subscript 2 O end style:undefined 

begin mathsize 14px style 2 Cr O subscript 4 to the power of 2 minus sign end exponent left parenthesis italic a italic q right parenthesis plus 6 Fe open parentheses O H close parentheses subscript 2 left parenthesis italic a italic q right parenthesis plus H subscript bold 2 O bold open parentheses italic l bold close parentheses yields Cr subscript 2 O subscript 3 open parentheses italic s close parentheses and 6 Fe open parentheses O H close parentheses subscript 3 left parenthesis italic s right parenthesis end style 

Setarakan jumlah H dengan menambahkan ion begin mathsize 14px style H to the power of plus sign end style:undefined 

begin mathsize 14px style 2 Cr O subscript 4 to the power of 2 minus sign end exponent left parenthesis italic a italic q right parenthesis plus 6 Fe open parentheses O H close parentheses subscript 2 left parenthesis italic a italic q right parenthesis plus H subscript 2 O open parentheses italic l close parentheses and bold 4 H to the power of bold plus sign bold left parenthesis italic a italic q bold right parenthesis yields Cr subscript 2 O subscript 3 open parentheses italic s close parentheses and 6 Fe open parentheses O H close parentheses subscript 3 left parenthesis italic s right parenthesis end style 

Tambahkan ion begin mathsize 14px style O H to the power of minus sign end style di kedua ruas sebanyak jumlah ion undefined:  

begin mathsize 14px style 2 Cr O subscript 4 to the power of 2 minus sign end exponent left parenthesis italic a italic q right parenthesis plus 6 Fe open parentheses O H close parentheses subscript 2 left parenthesis italic a italic q right parenthesis plus H subscript 2 O open parentheses italic l close parentheses and 4 H to the power of plus sign left parenthesis italic a italic q right parenthesis plus bold 4 O H to the power of bold minus sign bold left parenthesis italic a italic q bold right parenthesis yields Cr subscript 2 O subscript 3 open parentheses italic s close parentheses and 6 Fe open parentheses O H close parentheses subscript 3 left parenthesis italic s right parenthesis plus bold 4 O H to the power of bold minus sign bold left parenthesis italic a italic q bold right parenthesis end style 

Ion undefined dan undefined akan membentuk undefined. Eliminasi undefined pada kedua ruas sehingga diperoleh reaksi setara:

begin mathsize 14px style 2 Cr O subscript 4 to the power of 2 minus sign end exponent left parenthesis italic a italic q right parenthesis plus 6 Fe open parentheses O H close parentheses subscript 2 left parenthesis italic a italic q right parenthesis plus 5 H subscript 2 O open parentheses italic l close parentheses yields Cr subscript 2 O subscript 3 open parentheses italic s close parentheses and 6 Fe open parentheses O H close parentheses subscript 3 left parenthesis italic s right parenthesis plus 4 O H to the power of minus sign left parenthesis italic a italic q right parenthesis end style 

0

Ruangguru

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved