To analyze the balanced redox reaction:
\[ \text{Br}_2(l) + 2 \text{NaI}(aq) \rightarrow \text{I}_2(s) + 2 \text{NaBr}(aq) \]
Let’s answer each question step by step:
a. What is the oxidation state of Br\(_2\)(l)?
- In its elemental form, the oxidation state of Br in Br\(_2\) is 0.
b. What is the oxidation state of Na in 2NaI(aq)?
- The oxidation state of Na in NaI is +1. Therefore, in 2NaI, it remains +1.
c. What is the oxidation state of I in 2NaI(aq)?
- In NaI, I has an oxidation state of -1.
d. What is the oxidation state of I\(_2\)(s)?
- In its elemental form, the oxidation state of I in I\(_2\) is 0.
e. What is the oxidation state of Na in 2NaBr(aq)?
- Similar to NaI, the oxidation state of Na in NaBr is +1. Therefore, in 2NaBr, it remains +1.
f. What is the oxidation state of Br in 2NaBr(aq)?
- In NaBr, the oxidation state of Br is -1.
g. Which element is oxidized?
- The element that is oxidized is I. In NaI, it goes from -1 in NaI to 0 in I\(_2\), indicating a loss of electrons.
h. Which element is reduced?
- The element that is reduced is Br. In Br\(_2\), it goes from 0 in Br\(_2\) to -1 in NaBr, indicating a gain of electrons.
i. What is the oxidizing agent?
- The oxidizing agent is Br\(_2\), as it causes the oxidation of I by accepting electrons.
j. What is the reducing agent?
- The reducing agent is NaI (or I in it), as it donates electrons and gets oxidized.
k. How many electrons are transferred in the reaction as it is balanced?
- Since each I in NaI goes from -1 to 0 (a change of +1 per I) and there are 2 I atoms in 2NaI, a total of 2 electrons are transferred in the reaction.
In summary:
a. 0
b. +1
c. -1
d. 0
e. +1
f. -1
g. I
h. Br
i. Br\(_2\)
j. NaI
k. 2 electrons
1 answer