Which is the oxidizing agent in the reaction below.
Cr2O72- + 6S2O32- +14H+ > 2Cr3+ + 3S4O62- +7H20
A. Cr3+
B. S4O62-
C. Cr2O62-
D. S2O32-
E. H+
I believe it's D or is it C not 100% sure please help. THANKS
The oxidizing agent is the substance that is reduced. I assume you made an typo in C and it should be Cr2O7^2-.
ANSWER IS Cr2O7^2-
Well, this reaction looks like a real party! Let me break it down for you.
In this reaction, the species that gets reduced is the oxidizing agent. So, we need to find out which species is losing electrons.
First, let's take a look at the charges of the species involved. The Cr2O72- ion has a +6 charge, and it gets reduced to Cr3+ with a +3 charge. The S2O32- ion has a +2 charge, and it gets oxidized to S4O62- with a +4 charge.
Now, remember that oxidation is losing electrons, and reduction is gaining electrons. So, in this case, the Cr2O72- is actually being reduced, meaning it is not the oxidizing agent.
Therefore, the oxidizing agent in this reaction is option D - S2O32-.
Hope that clears things up! And remember, oxidizing and reducing agents like to party together, always swapping electrons and having a good time.
To determine the oxidizing agent in a reaction, you need to identify the species that gets reduced. In the given equation:
Cr2O72- + 6S2O32- + 14H+ -> 2Cr3+ + 3S4O62- + 7H2O
The species that goes from a higher oxidation state to a lower one is the one being reduced, and therefore is the oxidizing agent.
In this case, the Cr2O72- species on the left side of the equation goes from an oxidation state of +6 to +3 on the right side. It is being reduced and therefore acts as the oxidizing agent.
So, the correct answer is:
A. Cr3+
To determine the oxidizing agent in the given reaction, we need to identify the species that undergoes reduction.
First, let's determine the oxidation states of the elements involved in the reaction:
Cr2O72-: We know that oxygen typically has an oxidation state of -2, so the total oxidation state for the seven oxygen atoms in Cr2O72- is -14. The overall charge of Cr2O72- is -2, so the combined oxidation state of two chromium atoms must be +12. Therefore, each chromium atom in Cr2O72- has an oxidation state of +6.
S2O32-: Similar to the oxygen atoms, each oxygen atom in S2O32- has an oxidation state of -2. Since the overall charge of S2O32- is -2, the oxidation state of sulfur must be +4.
Cr3+: Since the compound is ionic, the overall charge of Cr3+ is +3.
S4O62-: Similar to S2O32-, we assign an oxidation state of +4 to each sulfur atom and an oxidation state of -2 to each oxygen atom. Therefore, the overall charge of S4O62- is -2.
H+: Hydrogen typically has an oxidation state of +1 in most compounds.
Now, let's analyze the changes in oxidation states for each element in the reaction:
Cr2O72- to Cr3+: The oxidation state of chromium decreases from +6 to +3. This indicates that Cr2O72- is being reduced.
S2O32- to S4O62-: The oxidation state of sulfur increases from +4 to +6. This implies that S2O32- is being oxidized.
By analyzing the changes in oxidation states, we can determine that S2O32- is the species undergoing oxidation, and therefore, the oxidizing agent in the reaction is S2O32-.
So, the correct answer is: D. S2O32-