Chapter 14 Question 3

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emilyharland
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Chapter 14 Question 3

Postby emilyharland » Mon Feb 06, 2017 10:53 am

This problem states, balance the following skeletal equations by using oxidation and reduction half reactions. All the reactions take place in acidic solution. The reaction is Cl2(g) + S2O3 2-(aq) --> Cl- (aq)+ SO4 2- (aq). I understand how to get which agent is oxidizing and which one is reducing but am having trouble writing out the half reactions specifically where the H2O and H+ come from?

Ryan Doyle 3E
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Re: Chapter 14 Question 3

Postby Ryan Doyle 3E » Mon Feb 06, 2017 12:43 pm

Because the reaction is aqueous, it is taking place in water; therefore, the water will actually provide the mechanism to balance the half reactions and electron transfer. To balance the Oxygen in the half reactions, one must add H2O to the opposite side and to balance Hydrogen, one must add H+ to the opposite side.

Julia Nakamura 2D
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Re: Chapter 14 Question 3

Postby Julia Nakamura 2D » Mon Feb 06, 2017 2:15 pm

Yes, so after finding which species are being oxidized and reduced, you can write two unbalanced equations for the half reactions. The next steps are to balance all of the elements, then balance the oxygens by using H2O, and then balance the hydrogens by using H+. There is a clearer outline of these steps on page 563 of the book.

Madeline_Foo_3J
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Re: Chapter 14 Question 3

Postby Madeline_Foo_3J » Mon Feb 06, 2017 3:07 pm

Why does the solution manual give S2O3(2-) 8 electrons when balancing the equation? I thought the Sulfur went from a 2+ to a 6+?

Josue_Marin_3I
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Re: Chapter 14 Question 3

Postby Josue_Marin_3I » Tue Feb 07, 2017 3:31 pm

How did you guys get Cl2 to be the oxidizing agent and (S203)2- to be the reduction agent?

Chem_Mod
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Re: Chapter 14 Question 3

Postby Chem_Mod » Tue Feb 07, 2017 3:54 pm

First, the sulfur in S2O32- does go from oxidation number 2+ to 6+ when it turns into SO42-. However, since when balanced there are two SO42-, the total number of electrons that are oxidized are 8 (4 from each sulfur).


To answer the second question about how to determine whether something is the oxidizing agent or reducing agent, you must look at the oxidation number of the elements before and after the reaction. If something gets a higher oxidation number, like sulfur in this question, it is being oxidized. Since it is oxidized, it acts as the reducing agent as it gives up electrons allowing the other entity to be reduced. Conversely, Cl2 being reduced because it's oxidation number decreases. Thus it is gaining electrons and acts as the oxidizing agent. By taking the electrons in, it is allowing the other entity to be oxidized.

Josue_Marin_3I
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Re: Chapter 14 Question 3

Postby Josue_Marin_3I » Tue Feb 07, 2017 4:04 pm

Oh okay! So to be clear, the oxidizing as well as the reducing agents are dependent on the other entities?

Chem_Mod
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Re: Chapter 14 Question 3

Postby Chem_Mod » Tue Feb 07, 2017 11:18 pm

Yes.

What is oxidized is also the reducing agent and what is reduced is also the oxidizing agent.


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