Redox Reactions and Oxidation States

This video tutorial explains how to balance the chemical equation for the reaction between lead(II) sulfide (PbS) and iodine gas (I2). It describes the process of balancing the equation, identifying it as a single displacement reaction, and highlights the redox nature of the reaction by discussing the changes in oxidation states of sulfur and iodine. The video concludes with a mention of additional resources for further help.

What type of reaction occurs when iodine displaces sulfur in lead(II) sulfide?

• Double displacement reaction

• Single displacement reaction

• Synthesis reaction

• Decomposition reaction

How many iodine atoms are present on the product side of the balanced equation?

• Four

• Three

• Two

• One

What is the coefficient for lead(II) sulfide in the balanced equation?

• Three

• One

• Zero

• Two

In the redox reaction, what happens to the oxidation state of sulfur?

• It decreases from 0 to -2

• It increases from 0 to +2

• It remains the same

• It increases from -2 to 0

What is the initial oxidation state of iodine in the reaction?

• -1

• 0

• +1

• +2

Which element is reduced in the redox reaction?

• Lead

• Sulfur

• Oxygen

• Iodine

Where can you find additional help for balancing oxidation-reduction reactions?

• In the textbook

• In the glossary

• In the video description

• In the appendix