Chemistry · Electronic concepts of oxidation and reduction, redox reactions, oxidation number, rules for assigning oxidation number and balancing of redox reactions
The reaction of and results in
The reaction of \( K M n O_{4} \) and \( \mathrm{HCl} \) results in
- A. Oxidation of \( \mathrm{Mn} \) in \( \mathrm{KMnO}_{4} \) and production of \( \mathrm{Cl}_{2} \)
- B. Reduction of Mn in \( K M n O_{4} \) and production of \( H_{2} \)
- C. Oxidation of \( \mathrm{Mn} \) in \( \mathrm{KMnO}_{4} \) and production of \( \mathrm{H}_{2} \)
- D. Reduction of Mn in \( K M n O_{4} \) and production of \( C l_{2} \)
Step-by-step solution
In the reaction 2KMnO4 + 16HCl → 2KCl + 2MnCl2 + 5Cl2 + 8H2O, the oxidation state of Mn decreases from +7 in KMnO4 to +2 in MnCl2, indicating reduction. Chlorine is oxidized from -1 in HCl to 0 in Cl2, producing chlorine gas. Thus, option D is correct.
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