Chemistry, 21.10.2019 20:20 elrancho

For each chemical reaction listed in the table below, decide whether the highlighted atom is being oxidized or reduced.

For each chemical reaction listed in the table below, decide whether the highlighted atom is being o

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Answer from: boonkgang6821

1 = copper get reduced.

2 = carbon get oxidized.

3 = carbon neither oxidized nor reduced.

4 = sulfur neither oxidized nor reduced.

Explanation:

The oxidation reduction reactions are called redox reaction. These reactions are take place by gaining or losing the electrons and oxidation state of elements are changed.

Oxidation:

Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.

Reduction:

Reduction involve the gain of electron and oxidation number is decreased.

Consider the following reactions.

4KI + 2CuCl₂ → 2CuI + I₂ + 4KCl

the oxidation state of copper is changed from +2 to +1 so copper get reduced.

CO + H₂O → CO₂ + H₂

the oxidation state of carbon is +2 on reactant side and on product side it becomes +4 so carbon get oxidized.

Na₂CO₃ + H₃PO₄ → Na₂HPO₄ + CO₂ + H₂O

The oxidation state of carbon on reactant side is +4. while on product side is also +4 so it neither oxidized nor reduced.

H₂S + 2NaOH → Na₂S + 2H₂O

The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.

Answer from: Quest

answeridentity;

Answer from: Quest

Answerthe half-reaction having the greatest tendency to occur at the anode: i₂(aq) + 2e⁻ ⇄ 2i⁻ ; e = +0.53 v====== $\begin{aligned} \text{i}_2\text{\footnotesize (aq)}+2\,\text{e}^{-} & \rightleftharpoons 2\,\text{i}^{-}\text{\footnotesize (aq)}& \text{e} & = +0.53\text{ v} \\ \text{co}^{3+}\text{\footnotesize (aq)} + \text{e}^{-} & \rightleftharpoons \text{co}^{2+}\text{\footnotesize (aq)} & \text{e} & = +1.80\text{ v} \\ \text{o}_2\text{\footnotesize (g)}+4\,\text{h}^{+}\text{\footnotesize (aq)} + 4\,\text{e}^{-} & \rightleftharpoons 2\,\text{h}_2\text{o}\text{\footnotesize(l)}& \text{e} & = +1.23\text{ v} \\ \end{aligned}$ oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be i₂(aq) + 2e⁻ ⇄ 2i⁻ e = +0.53 vbecause its reduction potential is the lowest out of the three.