Question

ni^{2 + }+so_{4}^{2 - }\to nio_{2}+so_{2}
in the above redox reaction, use oxidation numbers to identify the element oxidized, the element reduced, the oxidizing agent and the reducing agent.
name of the element oxidized:
formula of the oxidizing agent:
name of the element reduced:
formula of the reducing agent:

Explanation:

Step1: Determine oxidation numbers of reactants

In $\text{Ni}^{2 + }$, oxidation number of Ni is +2. In $\text{SO}_{4}^{2 - }$, oxygen has an oxidation number of -2. Let the oxidation number of S be $x$. Then $x+4\times(- 2)=-2$, so $x = +6$.

Step2: Determine oxidation numbers of products

In $\text{NiO}_{2}$, oxygen has an oxidation number of -2. Let the oxidation number of Ni be $y$. Then $y + 2\times(-2)=0$, so $y=+4$. In $\text{SO}_{2}$, oxygen has an oxidation number of -2. Let the oxidation number of S be $z$. Then $z+2\times(-2)=0$, so $z = +4$.

Step3: Identify oxidized and reduced elements

Ni changes from +2 to +4, so Ni is oxidized. S changes from +6 to +4, so S is reduced.

Step4: Identify oxidizing and reducing agents

The substance that contains the element that is reduced is the oxidizing agent. So $\text{SO}_{4}^{2 - }$ is the oxidizing agent. The substance that contains the element that is oxidized is the reducing agent. So $\text{Ni}^{2 + }$ is the reducing agent.

Answer:

Name of the element oxidized: Nickel
Name of the element reduced: Sulfur
Formula of the oxidizing agent: $\text{SO}_{4}^{2 - }$
Formula of the reducing agent: $\text{Ni}^{2 + }$