Balance the following oxidation-reduction reaction: fe(s)+na+(aq)→fe2+(aq)+na(s) express the coefficients as integers separated by commas (e.g., 4,1,3,2, where 1 indicates the absence of a coefficient).

The answer to your question is: 1, 2, 1, 2

Explanation:

                       1 Fe(s)  + 2 Na⁺(aq)  → 1 Fe²⁺(aq)  + 2 Na(s)

                             Fe⁰   -   2e⁻       ⇒           Fe⁺²        Oxidases

                             Na⁺   +  1 e⁻       ⇒           Na⁰         Reduces

                      1 x ( 1 Fe⁰      ⇒         1 Fe⁺²)      Interchange number of

                      2 x ( 2Na⁺       ⇒       2 Na⁰ )      electrons

give me them points