Question

1. in the reaction below:

fe + cuso₄ → feso₄ + cu
which substance is oxidized?
a. fe
b. cu
c. cuso₄
d. feso₄

1. what is the oxidation number of chromium in cr₂o₇²⁻?

a. +3
b. +4
c. +6
d. +7

1. in the reaction below:

2no + o₂ → 2no₂
what happens to nitrogen?
a. it is reduced
b. it is oxidized
c. it is unchanged
d. it acts as the reducing agent

1. which oxidation number change represents the greatest oxidation?

a. −1 → +1
b. 0 → +2
c. +2 → +5
d. −2 → +2

Explanation:

Question 5

To determine the oxidized substance, we analyze the oxidation states. In \( \text{Fe} + \text{CuSO}_4
ightarrow \text{FeSO}_4 + \text{Cu} \), the oxidation state of Fe changes from 0 (in elemental Fe) to +2 (in \( \text{FeSO}_4 \)). An increase in oxidation state means oxidation. Cu in \( \text{CuSO}_4 \) has an oxidation state of +2, which changes to 0 (in elemental Cu), so it is reduced. \( \text{CuSO}_4 \) is the oxidizing agent, and \( \text{FeSO}_4 \) is a product. So Fe is oxidized.

Step1: Recall oxidation number rules

Oxygen usually has an oxidation number of -2. Let the oxidation number of Cr be \( x \). In \( \text{Cr}_2\text{O}_7^{2-} \), the sum of oxidation numbers is equal to the charge of the ion (-2).

Step2: Set up the equation

There are 2 Cr atoms and 7 O atoms. So \( 2x + 7(-2) = -2 \).

Step3: Solve for \( x \)

\( 2x - 14 = -2 \)
\( 2x = -2 + 14 \)
\( 2x = 12 \)
\( x = 6 \)

In \( 2\text{NO} + \text{O}_2
ightarrow 2\text{NO}_2 \), find the oxidation state of N in NO and \( \text{NO}_2 \). In NO, O is -2, so N is +2. In \( \text{NO}_2 \), O is -2, so for two O atoms (-4 total), N must be +4 (since \( \text{NO}_2 \) is neutral). The oxidation state of N increases from +2 to +4, which means it loses electrons (oxidation). Also, a substance that is oxidized acts as a reducing agent, but the question is about what happens to nitrogen (oxidized or reduced). So N is oxidized.

Answer:

A. Fe

Question 6