Use the chemical equation to answer the question.
Br2(l)+2NaI(aq)→I2(s)+2NaBr(aq)
Which elements are oxidized and reduced in the reaction?
(1 point)
Responses
Bromine (Br) is oxidized, and sodium (Na) is reduced.
Bromine , left parenthesis cap b r right parenthesis, is oxidized, and sodium , left parenthesis cap n A right parenthesis, is reduced.
Iodine (I) is oxidized, and bromine (Br) is reduced.
Iodine , left parenthesis cap I right parenthesis, is oxidized, and bromine , left parenthesis cap b r right parenthesis, is reduced.
Sodium (Na) is oxidized, and bromine (Br) is reduced.
Sodium , left parenthesis cap n A right parenthesis, is oxidized, and bromine , left parenthesis cap b r right parenthesis, is reduced.
Bromine (Br) is oxidized, and iodine (I)is reduced.
Bromine , left parenthesis cap b r right parenthesis, is oxidized, and iodine , left parenthesis cap i right parenthesis, is reduced.
To determine which elements are oxidized and reduced in a chemical equation, you need to look at the changes in oxidation states (or oxidation numbers) of the elements. Oxidation state refers to the charge an atom would have if the shared electrons in a compound belonged entirely to the more electronegative element.
In the given chemical equation:
Br2(l) + 2NaI(aq) → I2(s) + 2NaBr(aq)
Bromine (Br) changes from an oxidation state of 0 in Br2(l) to -1 in NaBr(aq). It gains an electron, so it is reduced.
Iodine (I) changes from an oxidation state of -1 in NaI(aq) to 0 in I2(s). It loses an electron, so it is oxidized.
Sodium (Na) changes from an oxidation state of +1 in NaI(aq) to +1 in NaBr(aq). Its oxidation state remains the same, so it is neither oxidized nor reduced.
Therefore, the correct answer is: Iodine (I) is oxidized, and bromine (Br) is reduced.