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高锰酸钾 (KMnO4)\left(\mathrm{KMnO}_{4}\right) 为重要的化合物，具有众多应用。 （1）高锰酸钾的制备可以通过氯酸钾 $\left(...
Mar 19, 2024
高锰酸钾 (KMnO4)\left(\mathrm{KMnO}_{4}\right) 为重要的化合物，具有众多应用。 （1）高锰酸钾的制备可以通过氯酸钾 (KClO3)\left(\mathrm{KClO}_{3}\right) 、氢氧化钾 (KOH)(\mathrm{KOH}) 和二氧化锰 (MnO2)\left(\mathrm{MnO}_{2}\right) 在加热条件下反应先生成锰酸钾 (K2MnO4)\left(\mathrm{K}_{2} \mathrm{MnO}_{4}\right),再由 K2MnO4\mathrm{K}_{2} \mathrm{MnO}_{4} 通过氧化还原化学反应转化成 KMnO4\mathrm{KMnO}_{4} 。 配平化学反应方程式: MnO2+…KClO3+…KOH≜_K2MnO4+…KCl+…H2O \mathrm{MnO}_{2}+\ldots \mathrm{KClO}_{3}+\ldots \mathrm{KOH} \triangleq \_\mathrm{K}_{2} \mathrm{MnO}_{4}+\ldots \mathrm{KCl}+\ldots \mathrm{H}_{2} \mathrm{O} ( 2 ) 接着向 K2MnO4\mathrm{K}_{2} \mathrm{MnO}_{4} 溶液中通入 CO2\mathrm{CO}_{2} ，可使 K2MnO4\mathrm{K}_{2} \mathrm{MnO}_{4} 发生歧化反应得到 KMnO4\mathrm{KMnO}_{4} 和 MnO2\mathrm{MnO}_{2} 。具体反应为: 3 K2MnO4+2CO2=2KMnO4+MnO2+2 K2CO3 。  3 \mathrm{~K}_{2} \mathrm{MnO}_{4}+2 \mathrm{CO}_{2}=2 \mathrm{KMnO}_{4}+\mathrm{MnO}_{2}+2 \mathrm{~K}_{2} \mathrm{CO}_{3} \text { 。 } (1) 用双线桥法标出反应的电子转移: (2) 岐化反应中氧化剂和还原剂的物质的量之比为 （3）用流程百的方法制备 KMnO4， K2MnO4\mathrm{KMnO}_{4} ， \mathrm{~K}_{2} \mathrm{MnO}_{4} 利用率低，仅 23\frac{2}{3} 转化为 KMnO4\mathrm{KMnO}_{4} 。为使 K2MnO4\mathrm{K}_{2} \mathrm{MnO}_{4} 完全转化为 KMnO4\mathrm{KMnO}_{4} ，可以通过电解法。离子方程式为 MnO42−+⟶H2O→ 通电 \mathrm{MnO}_{4}^{2-}+\longrightarrow \mathrm{H}_{2} \mathrm{O} \xrightarrow{\text { 通电 }} 。 通过化合价升降分析，确定该反应的还原产物为 ; 补全并配平电解反应的离子反应方程式
Solution
1
Balance the reaction equation: To balance the chemical equation for the preparation of potassium manganate (K2MnO4\mathrm{K_2MnO_4}), we need to ensure that the number of atoms of each element is the same on both sides of the equation
2
Balancing the atoms other than oxygen and hydrogen: Start by balancing the manganese atoms. There is one MnO2\mathrm{MnO_2} on the reactant side and one K2MnO4\mathrm{K_2MnO_4} on the product side, so manganese is already balanced. Next, balance the potassium atoms by adding three KOH\mathrm{KOH} for every K2MnO4\mathrm{K_2MnO_4} produced
3
Balancing oxygen atoms: There are four oxygen atoms in K2MnO4\mathrm{K_2MnO_4}, two in MnO2\mathrm{MnO_2}, and three in each KClO3\mathrm{KClO_3}. To balance the oxygen atoms, we need to add three KOH\mathrm{KOH} and three KClO3\mathrm{KClO_3} to the reactant side
4
Balancing hydrogen atoms: Since we have three KOH\mathrm{KOH} on the reactant side, we will produce three H2O\mathrm{H_2O} on the product side to balance the hydrogen atoms
5
Balancing chlorine atoms: With three KClO3\mathrm{KClO_3} on the reactant side, we will have three KCl\mathrm{KCl} on the product side to balance the chlorine atoms
6
Final balanced equation: The balanced chemical equation is: MnO2+3KClO3+3KOH→△K2MnO4+3KCl+3H2O \mathrm{MnO_2} + 3\mathrm{KClO_3} + 3\mathrm{KOH} \xrightarrow{\triangle} \mathrm{K_2MnO_4} + 3\mathrm{KCl} + 3\mathrm{H_2O}
Answer
MnO2+3KClO3+3KOH→△K2MnO4+3KCl+3H2O\mathrm{MnO_2} + 3\mathrm{KClO_3} + 3\mathrm{KOH} \xrightarrow{\triangle} \mathrm{K_2MnO_4} + 3\mathrm{KCl} + 3\mathrm{H_2O}
Key Concept
Balancing chemical equations involves ensuring that the number of atoms for each element is the same on both sides of the equation.
Explanation
The balanced equation is achieved by adjusting the coefficients of the reactants and products to ensure that the law of conservation of mass is satisfied.
Solution
1
Identify the oxidizing and reducing agents: In the disproportionation reaction of K2MnO4\mathrm{K_2MnO_4}, the same substance is both oxidized and reduced
2
Determine the ratio of oxidizing to reducing agents: The balanced equation for the disproportionation reaction is: 3K2MnO4+2CO2→2KMnO4+MnO2+2K2CO3 3\mathrm{K_2MnO_4} + 2\mathrm{CO_2} \rightarrow 2\mathrm{KMnO_4} + \mathrm{MnO_2} + 2\mathrm{K_2CO_3} From this equation, we can see that two moles of KMnO4\mathrm{KMnO_4} are produced for every one mole of MnO2\mathrm{MnO_2} produced
Answer
The ratio of oxidizing agent (KMnO4\mathrm{KMnO_4}) to reducing agent (MnO2\mathrm{MnO_2}) is 2:1.
Key Concept
The ratio of oxidizing to reducing agents in a disproportionation reaction can be determined from the balanced chemical equation.
Explanation
In the disproportionation reaction, K2MnO4\mathrm{K_2MnO_4} acts as both the oxidizing and reducing agent, and the ratio of the produced KMnO4\mathrm{KMnO_4} to MnO2\mathrm{MnO_2} is 2:1.
Solution
1
Analyze the electrolysis reaction: During the electrolysis of K2MnO4\mathrm{K_2MnO_4}, MnO42−\mathrm{MnO_4}^{2-} is oxidized to MnO4−\mathrm{MnO_4}^{-}
2
Determine the reduction product: Water (H2O\mathrm{H_2O}) is reduced during the reaction, producing OH−\mathrm{OH}^{-} and H2\mathrm{H_2} gas
3
Balance the half-reactions: The oxidation half-reaction is: MnO42−→MnO4−+e− \mathrm{MnO_4}^{2-} \rightarrow \mathrm{MnO_4}^{-} + e^- The reduction half-reaction for water is: 2H2O+2e−→H2+2OH− 2\mathrm{H_2O} + 2e^- \rightarrow \mathrm{H_2} + 2\mathrm{OH}^-
4
Combine and balance the overall reaction: To balance the electrons, we need to multiply the oxidation half-reaction by 2 and the reduction half-reaction by 1. The balanced overall reaction is: 2MnO42−+2H2O→electrolysis2MnO4−+H2+2OH− 2\mathrm{MnO_4}^{2-} + 2\mathrm{H_2O} \xrightarrow{\text{electrolysis}} 2\mathrm{MnO_4}^{-} + \mathrm{H_2} + 2\mathrm{OH}^-
Answer
2MnO42−+2H2O→electrolysis2MnO4−+H2+2OH−2\mathrm{MnO_4}^{2-} + 2\mathrm{H_2O} \xrightarrow{\text{electrolysis}} 2\mathrm{MnO_4}^{-} + \mathrm{H_2} + 2\mathrm{OH}^-
Key Concept
Electrolysis involves the oxidation and reduction of species at the electrodes, and the overall reaction is balanced by ensuring the number of electrons lost equals the number gained.
Explanation
The reduction product of water during electrolysis is OH−\mathrm{OH}^{-} and H2\mathrm{H_2} gas, and the balanced electrolysis reaction shows the conversion of MnO42−\mathrm{MnO_4}^{2-} to MnO4−\mathrm{MnO_4}^{-}.
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