The balanced redox reaction in the chemical reaction is given below:
40H2S + 48H+ + 16MnO4¯ ---> 5S8 + 16Mn2+ + 64H2O
Balancing the redox reaction:
Solution:
1) Half-reactions:
H2S ---> S8
MnO4¯ ---> Mn2+
2) Balance:
8H2S ---> S8 + 16H+ + 16e¯
5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O
3) Make the number of electrons equal (note that there are no common factors between 5 and 16 except 1):
40H2S ---> 5S8 + 80H+ + 80e¯ <--- factor of 5
80e¯ + 128H+ + 16MnO4¯ ---> 16Mn2+ + 64H2O <---
factor of 16
4) Thus, the final answer is given below;
40H2S + 48H+ + 16MnO4¯ ---> 5S8 + 16Mn2+ + 64H2O
Oxidation-reduction can simply be defined as a special type of chemical reaction in which the oxidation states of the substrate change.
So therefore, the balanced redox reaction in the chemical reaction is given below:
40H2S + 48H+ + 16MnO4¯ ---> 5S8 + 16Mn2+ + 64H2O
Complete question:
Balance the following redox reaction:
MnO4¯ + H2S ---> Mn2+ + S8
Learn more about oxidation-reduction:
https://brainly.com/question/21851295
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