Cho 8,36 gam hỗn hợp nhôm và đồng tác dụng vừa đủ với dung dịch HCl 20%. Sau phản ứng thu được 4,4622 lít khí hydrogen ở đkc và
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Answer:
Mn is the oxidizing agent.
N is the reducing agent.
Explanation:
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In this case, according to the undergoing chemical reaction, it is seen that the manganese in KMnO4 has an oxidation state of 7+, in MnSO4 of 2+ and nitrogen in KNO2 is 3+ and in KNO3 is 5+; thus we have the following half-reactions:
Thus, since manganese is undergoing a decrease in the oxidation state, we infer it is the oxidizing agent whereas nitrogen, undergoing an increase in the oxidation state is the reducing agent.
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