Identify the element oxidized and the element reduced. Give the number of electrons lost or gained by each atom. Li + H2O → LiOH + H2

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kony345p

14.04.2021

Chemistry

A) ${MnO_4}^- + 3 H^+ + \frac{5}{2} H_2O_2 \rightarrow Mn^{2+} + 4 H_2O + \frac{5}{2} O_2$

B) 7.5 molar

Explanation:

A) Reduction

${MnO_4}^- + 8 H^+ + 5e^- \rightarrow Mn^{2+} + 4 H_2O$

Oxidation

$H_2O_2 \rightarrow O_2 + 2 H^+ + 2e^-$

Multiplying the oxidation reaction by 5/2 and adding it to the reduction equation:

${MnO_4}^- + 8 H^+ + 5e^- \rightarrow Mn^{2+} + 4 H_2O$

$\frac{5}{2} H_2O_2 \rightarrow \frac{5}{2} O_2 + 5 H^+ + 5e^-$

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${MnO_4}^- + 3 H^+ + \frac{5}{2} H_2O_2 \rightarrow Mn^{2+} + 4 H_2O + \frac{5}{2} O_2$

B) 10 ml = 0.01 L

20 ml = 0.02 L

mol of MnO4− = molarity*volume = 1.5*0.02 = 0.03

1 mol of MnO4− reacts with 5/2 mol of H2O2, then:

mol of H2O2 = 0.03*5/2 = 0.075

molarity = mol/volume = 0.075/0.01 = 7.5 molar

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