Mno4 + S2 Equation

Examples and Solutions

Example 1: Given Equation: MnO4- + S2-  Mn2+ + S0 Balance the equation by using the ion-electron method.
Step 1 Oxidation: S2-  S0
Reduction: MnO4-  Mn2+ Divide into two half reactions. Identify the element that is being oxidized and the other being reduced. (Use the entire ion or molecule.)
Step
2 Oxidation: S2-  S0
Reduction: MnO4-  Mn2+ Balance all the elements (other than oxygen and hydrogen).
Step
3 Oxidation: S2-  S0
Reduction: 8H+ + MnO4-  Mn2+ + 4H2O Balance O and H. The oxidation requires neither O nor H, but the reduction equation needs 4H2O on the right and 8H+ on the left.
Step
4 Oxidation: S2-  S0 + 2e-
Reduction: 5e- + 8H+ + MnO4-  Mn2+ + 4H2O Balance with electrons critically.
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What do you call the molecules or ions that receive electrons?
a.) Reducing Agents
b.) Combusting Agents
c.) Oxidizing Agents
d.) Redox Agents

5. Which of the following statements is true?
a.) The Oxidation State of the Oxidizing Agent increases as its substance is reduced.
b.) The number of Electrons of the Oxidizing Agent gained as its Oxidation State increased.
c.) The Substance of the Reducing Agent is oxidized as its number of electrons lost.
d.) The Oxidation State of the Reducing Agent decreases as its substance is reduced.

6. It corresponds to the quantity of electrons (e-) that a specific atom gains or loses.
a.) Oxidation State
b.) Oxidizing Agent
c.) Oxidation Reaction
d.) None of the above

7. Which of the following is not true:
a.) Fluorine has always an oxidation number of -1 in compounds.
b.) Elements in Group 17 have an oxidation number of -1 in binary metal compounds with metals or hydrogen.
c.) Oxidization State of Hydrogen generally is +1 in compounds.
d.) Elements in Group 15 have an oxidation number of +3 in binary metal compounds with metals or hydrogen.
8. What is the Oxidation State of an individual

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