Chemistry, 25.07.2019 20:00 hrcatena

Onsider a galvanic cell in which al3 is reduced to elemental aluminum, and magnesium metal is oxidized to mg2 . write the balanced half-cell reactions that take place at the cathode and at the anode.

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Two Half Cell reactions are as follow,

3e⁻ + Al⁺³ → Al (Reduction Reaction)

Mg → Mg⁺² + 2e⁻ (Oxidation Reaction)

In order to balance the number of electrons, multiply reduction reaction by 2 and oxidation reaction by 3.
So,
6e⁻ + 2 Al⁺³ → 2 Al

3 Mg → 3 Mg⁺² + 6e⁻ (e⁻ cancelled)

2 Al⁺³ + 3 Mg → 2 Al + 3 Mg⁺²

Result:
At Anode Magnesium metal is Oxidized and At Cathode Aluminium Ions are reduced.

Answer from: Quest

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form( $n_2$ ) .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

Answer from: Quest

the answer to this question is 3.

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Onsider a galvanic cell in which al3 is reduced to elemental aluminum, and magnesium metal is oxidiz...