Please check. Thank you.
2.
What is the reduction half-reaction for the following unbalanced redox equation?
Cr2O72– + NH4+ Cr2O3 + N2
A.) Cr2O3 -> Cr2O72–
B.) Cr2O72– -> Cr2O3
C.) NH4+ -> N2
D.) N2 -> NH4+
3.
Which oxidation-reduction reactions are best balanced by the half-reaction method?
A.) covalent reactions
B.) acid-base reactions
C.) ionic reactions
D.) intermolecular reactions
Thank You.
what are your answers?
2. B 3. C
wrong, and wrong.
I'm really trying to figure this out. Is it D for number 2 and B for number 3? I know you already explained it once and if i'm wrong, any additional help would be greatly appreciated.
To solve this problem, we need to analyze the given redox equation and identify the half-reactions for the oxidation and reduction processes that occur.
For the given redox equation:
Cr2O72– + NH4+ Cr2O3 + N2
We can see that Cr2O72– is being reduced to Cr2O3, and NH4+ is being oxidized to N2.
The reduction half-reaction involves the gain of electrons, so the reduction half-reaction can be written as:
Cr2O72– + 14H+ + 6e– Cr2O3 + 7H2O
The oxidation half-reaction involves the loss of electrons, so the oxidation half-reaction can be written as:
2NH4+ – 8e– N2 + 8H+
Now, to find the correct option for the reduction half-reaction, we need to compare the given options with the reduction half-reaction we derived.
Looking at the options, none of them perfectly match the reduction half-reaction we derived. However, option B.) Cr2O72– -> Cr2O3 is the closest match, as it involves the Cr2O72– ion being reduced to Cr2O3.
Therefore, the reduction half-reaction for the given redox equation is:
B.) Cr2O72– -> Cr2O3
Moving on to the second question. The half-reaction method is best suited for balancing ionic reactions.
Therefore, the correct answer to the second question is:
C.) ionic reactions
I hope this helps! Let me know if you have any further questions.