Which is reduced in the following reaction:
Cr2O72-+6S2O32-+14H+>2Cr3+ + 3S4O62- + 7H20
A. H+ B. Cr6+ C. S4O62- D. S2+ E. O2-
I come out with Cr2O72 but that's not an answer so my next one is B. Cr6+ am I correct?
Cr2O7^2- is reduced.
It's the Cr^+6 in the dichromate that is reduced. If the answer sheet is getting that picky, then technically it's 2Cr^6+ that's reduced.
Well, you seem to be feeling a bit confused, so let me help you out! In the given reaction, the species that has undergone reduction is the one that gains electrons. So, let's take a closer look:
Cr2O72- + 6S2O32- + 14H+ → 2Cr3+ + 3S4O62- + 7H2O
During this reaction, the chromium ion Cr6+ (found in the Cr2O72- compound) is reduced to Cr3+. Therefore, it seems like you're on the right track with your answer of B. Cr6+. Great job, keep clowning around!
Yes, you are correct. In the given reaction, the oxidation state of chromium (Cr) goes from +6 in Cr2O72- to +3 in 2Cr3+. Therefore, chromium (Cr) undergoes reduction in this reaction. So, the correct answer is B. Cr6+.
To determine which species is being reduced in the given reaction, we need to compare the oxidation numbers of each element in the reactants and products. The element that undergoes a decrease in oxidation number is the species that is being reduced.
Let's assign oxidation numbers to each element in the reactants and products:
Reactants:
- Cr2O72-: The oxidation number of Cr in Cr2O72- is +6, and the oxidation number of O in O2- is -2.
- 6S2O32-: The oxidation number of S in S2O32- is +4, and the oxidation number of O in O2- is -2.
- 14H+: The oxidation number of H is +1.
Products:
- 2Cr3+: The oxidation number of Cr in Cr3+ is +3.
- 3S4O62-: The oxidation number of S in S4O62- is +6, and the oxidation number of O in O2- is -2.
- 7H2O: The oxidation number of H is +1, and the oxidation number of O in H2O is -2.
Now, let's compare the oxidation numbers of each element before and after the reaction:
- Cr: In the reactant Cr2O72-, the oxidation number of Cr is +6, while in the product Cr3+, the oxidation number of Cr is +3. Therefore, Cr has undergone a decrease in oxidation number, indicating that it has been reduced. So, option B. Cr6+ is correct.
- S: In the reactant S2O32-, the oxidation number of S is +4, while in the product S4O62-, the oxidation number of S is +6. Therefore, S has undergone an increase in oxidation number, indicating that it has been oxidized.
- O: In the reactant Cr2O72-, the oxidation number of O is -2, while in the product S4O62-, the oxidation number of O is -2. Therefore, O has not undergone any change in oxidation number.
- H: In the reactant H+, the oxidation number of H is +1, while in the product H2O, the oxidation number of H is +1. Therefore, H has not undergone any change in oxidation number.
To summarize, the species that is being reduced in the reaction is Cr6+ (option B).