To answer the questions regarding the balanced redox reaction:
\[ \text{Br}_2(l) + 2\text{NaI}(aq) = \text{I}_2(s) + 2\text{NaBr}(aq) \]
Let's go through the questions one by one.
a. What is the oxidation state of Br2(l)?
In elemental form, the oxidation state of bromine (\( \text{Br}_2 \)) is 0.
b. What is the oxidation state of Na in 2Nal(aq)?
In sodium iodide (\( \text{NaI} \)), sodium (Na) has an oxidation state of +1.
c. What is the oxidation state of I in 2Nal(aq)?
In sodium iodide (\( \text{NaI} \)), iodine (I) has an oxidation state of -1.
d. What is the oxidation state of I2(s)?
In elemental form, the oxidation state of iodine (\( \text{I}_2 \)) is 0.
e. What is the oxidation state of Na in 2NaBr(aq)?
In sodium bromide (\( \text{NaBr} \)), sodium (Na) has an oxidation state of +1.
f. What is the oxidation state of Br in 2NaBr(aq)?
In sodium bromide (\( \text{NaBr} \)), bromine (Br) has an oxidation state of -1.
g. Which element is oxidized?
Iodine (I) is oxidized because it goes from -1 in \( \text{NaI} \) to 0 in \( \text{I}_2 \).
h. Which element is reduced?
Bromine (Br) is reduced because it goes from 0 in \( \text{Br}_2 \) to -1 in \( \text{NaBr} \).
i. What is the oxidizing agent?
The oxidizing agent is \( \text{Br}_2 \) because it causes the oxidation of iodine by gaining electrons itself.
j. What is the reducing agent?
The reducing agent is \( \text{NaI} \) (specifically the iodide ion), because it provides electrons that reduce bromine.
k. How many electrons are transferred in the reaction as it is balanced?
In this reaction, 2 electrons are transferred: each \( \text{I}^- \) loses 1 electron to become \( \text{I}_2 \), and since there are 2 iodides, a total of 2 electrons are transferred.
In summary:
- a: 0
- b: +1
- c: -1
- d: 0
- e: +1
- f: -1
- g: Iodine (I)
- h: Bromine (Br)
- i: \( \text{Br}_2 \)
- j: \( \text{NaI} \)
- k: 2 electrons
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