Positively charged particles in an
atom are called
electrons
O protons
neurtons

Oxidation-reduction reactions (often called "redox" for short) are reactions that involve the transfer of electrons from one species to another. oxidation states, or oxidation numbers, allow chemists to keep track of these electron transfers. in general, one element will lose electrons (oxidation), with the result that it will increase in oxidation number, and another element will gain electrons (reduction), thereby decreasing in oxidation number. the species that is oxidized is called the reducing agent or reductant. the species that is reduced is called the oxidizing agent or oxidant. to sum up: oxidation = increase in oxidation state = loss of electrons = reducing agent reduction = decrease in oxidation state = gain of electrons = oxidizing agent part a which element is oxidized in this reaction? fe2o3+3co→2fe+3co2 enter the elemental symbol. view available hint(s) is oxidized part b which element is reduced in this reaction? 2hcl+2kmno4+3h2c2o4→6co2+2mno2+2kcl+4h2o enter the elemental symbol. view available hint(s) is reduced

Answer the questions about this reaction: nai(aq) + cl2(g) → nacl(aq) + i2(g) write the oxidation and reduction half-reactions: oxidation half-reaction: reduction half-reaction: based on the table of relative strengths of oxidizing and reducing agents (b-18), would these reactants form these products? write the balanced equation: answer options: a. 0/na -> +1/na+1e- b. nai(aq) + cl2(g) → nacl(aq) + i2(g) c. +1/na+1e- -> 0 /na d. -1/2i -> 0/i2+2e- e. no f. 4nai(aq) + cl2(g) → 4nacl(aq) + i2(g) g. 2nai(aq) + cl2(g) → 2nacl(aq) + i2(g) h. 4nai(aq) + 2cl2(g) → 4nacl(aq) + 2i2(g) i. nai(aq) + cl2(g) → nacl(aq) + i2(g) j. 0/cl2+2e -> -1/2cl- k. yes

In as 1°, 2°, 3°, or 4°. be to . : °b: °c: °d: ° : °b: °c: °d: ° : °b: °c: °d: °e: °f: °g: °h: ° : °b: °c: °d: °e: °f: °g: °h: °i: °