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In the reaction 2Cr2O7^2- + 14H+ + 6e^- → 4Cr^3+ + 7H2O, what is the role of Cr2
In the reaction 2Cr2O7^2- + 14H+ + 6e^- → 4Cr^3+ + 7H2O, what is the role of Cr2O7^2-?
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Practice Questions
1 question
Q1
In the reaction 2Cr2O7^2- + 14H+ + 6e^- → 4Cr^3+ + 7H2O, what is the role of Cr2O7^2-?
Oxidizing agent
Reducing agent
Neutral agent
None of the above
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Cr2O7^2- is reduced to Cr^3+, making it the oxidizing agent.
Questions & Step-by-step Solutions
1 item
Q
Q: In the reaction 2Cr2O7^2- + 14H+ + 6e^- → 4Cr^3+ + 7H2O, what is the role of Cr2O7^2-?
Solution:
Cr2O7^2- is reduced to Cr^3+, making it the oxidizing agent.
Steps: 7
Show Steps
Step 1: Identify the reactants in the reaction. Here, we have Cr2O7^2-, H+, and e^-.
Step 2: Look at what happens to Cr2O7^2- in the reaction. It changes to Cr^3+.
Step 3: Understand the meaning of 'reduced'. When a substance gains electrons, it is reduced.
Step 4: Notice that Cr2O7^2- gains electrons (6e^-) in the reaction.
Step 5: Since Cr2O7^2- is gaining electrons, it is being reduced.
Step 6: Identify the role of a substance that is reduced in a reaction. It is called an oxidizing agent.
Step 7: Conclude that Cr2O7^2- is the oxidizing agent because it is reduced to Cr^3+.
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