Question

CrO42- + NO+ 4H+NO3- + Cr3++ 2H2O 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: name of the element reduced: formula of the oxidizing agent: formula of the reducing agent:

Answer 1

Answer:

- Element oxidized: N.

- The element reduced: Cr.

- The oxidizing agent: Chromium.

- The reducing agent: Nitrogen.

- Formula of the oxidizing agent: (CrO₄)²⁻

- Formula of the reducing agent: NO.

- name of the element oxidized: nitrogen.

- name of the element reduced: chromium.

Explanation:

Hello,

In this case, the redox reaction is:

$(CrO_4)^{2-} + NO + 4H^+\rightarrow (NO_3)^- + Cr^{3+} + 2H_2O$

In such a way, each element has the following oxidation state distribution:

$(Cr^{6+}O_4)^{2-} + N^{2+}O^{2-} + 4H^+\rightarrow (N^{5+}O^{2-}_3)^- + Cr^{3+} + 2H_2^{+}O^{2-}$

Thus, it seen that:

- The oxidized element is nitrogen as its oxidation state changes from 2+ to 5+. In addition, it is the reducing agent since it undergoes oxidation.

- The reduced element is chromium as its oxidation state changes from 6+ to 3+. In addition, it is the oxidizing agent since it undergoes reduction.

Hence, in order to answer:

- Element oxidized: N.

- The element reduced: Cr.

- The oxidizing agent: Chromium.

- The reducing agent: Nitrogen.

- Formula of the oxidizing agent: (CrO₄)²⁻

- Formula of the reducing agent: NO.

- name of the element oxidized: nitrogen.

- name of the element reduced: chromium.

Best regards.