Roboguru

Setarakan reaksi redoks berikut. (suasana basa)

Setarakan reaksi redoks berikut.


undefined (suasana basa)

  1. ... space  

  2. ... undefined 

Jawaban:

Tentukan atom yang mengalami perubahan bilangan oksidasi dan setarakan jumlah atom yang berubah bilangan oksidasinya.undefined 

undefined 

Kalikan pengali dengan koefisien:undefined 

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

Setarakan jumlah O dengan menambahkan undefined:undefined 

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

Setarakan jumlah H dengan menambahkan ion undefined:undefined  

begin mathsize 14px style 2 Mn O subscript 4 to the power of minus sign left parenthesis italic a italic q right parenthesis plus 3 C subscript 2 O subscript 4 to the power of 2 minus sign end exponent left parenthesis italic a italic q right parenthesis plus bold 8 H to the power of bold plus sign bold left parenthesis italic a italic q bold right parenthesis yields 2 Mn O subscript 2 open parentheses italic s close parentheses and 6 C O subscript 2 open parentheses italic g close parentheses and 4 H subscript 2 O open parentheses italic l close parentheses 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:undefined 

begin mathsize 14px style 2 Mn O subscript 4 to the power of minus sign left parenthesis italic a italic q right parenthesis plus 3 C subscript 2 O subscript 4 to the power of 2 minus sign end exponent left parenthesis italic a italic q right parenthesis plus 8 H to the power of plus sign left parenthesis italic a italic q right parenthesis plus bold 8 O H to the power of bold minus sign bold left parenthesis italic a italic q bold right parenthesis yields 2 Mn O subscript 2 open parentheses italic s close parentheses and 6 C O subscript 2 open parentheses italic g close parentheses and 4 H subscript 2 O open parentheses italic l close parentheses and bold 8 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:undefined 

begin mathsize 14px style 2 Mn O subscript 4 to the power of minus sign left parenthesis italic a italic q right parenthesis plus 3 C subscript 2 O subscript 4 to the power of 2 minus sign end exponent left parenthesis italic a italic q right parenthesis plus 4 H subscript 2 O open parentheses italic l close parentheses yields 2 Mn O subscript 2 open parentheses italic s close parentheses and 6 C O subscript 2 open parentheses italic g close parentheses and 8 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