To analyze the balanced redox reaction:
\[ \text{Br}_2(l) + 2\text{NaI}(aq) \rightarrow \text{I}_2(s) + 2\text{NaBr}(aq), \]
we need to determine the oxidation states of the relevant elements and identify the components of the redox reaction.
a. What is the oxidation state of Br in Br2(l)?
- In \( \text{Br}_2 \) (bromine in its elemental form), the oxidation state is \( 0 \).
b. What is the oxidation state of Na in 2NaI(aq)?
- In \( \text{NaI} \), sodium (Na) has an oxidation state of \( +1 \).
c. What is the oxidation state of I in 2NaI(aq)?
- In \( \text{NaI} \), iodine (I) has an oxidation state of \( -1 \).
d. What is the oxidation state of I in I2(s)?
- In \( \text{I}_2 \) (iodine in its elemental form), the oxidation state is \( 0 \).
e. What is the oxidation state of Na in 2NaBr(aq)?
- In \( \text{NaBr} \), sodium (Na) has an oxidation state of \( +1 \).
f. What is the oxidation state of Br in 2NaBr(aq)?
- In \( \text{NaBr} \), bromine (Br) has an oxidation state of \( -1 \).
g. Which element is oxidized?
- Iodine (I) is oxidized since it goes from an oxidation state of \( -1 \) in \( \text{NaI} \) to \( 0 \) in \( \text{I}_2 \).
h. Which element is reduced?
- Bromine (Br) is reduced since it goes from an oxidation state of \( 0 \) in \( \text{Br}_2 \) to \( -1 \) in \( \text{NaBr} \).
i. What is the oxidizing agent?
- The oxidizing agent is \( \text{Br}_2 \) because it gains electrons (is reduced).
j. What is the reducing agent?
- The reducing agent is \( \text{NaI} \) because it loses electrons (is oxidized).
k. How many electrons are transferred in the reaction as it is balanced?
- In this reaction, 2 electrons are transferred: each iodine atom in \( 2NaI \) is oxidized (2 iodines total, each losing 1 electron).
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 are transferred.
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