Question

Write the oxidation and reduction half reactions;

a. Fe2+, Fe3+ ? Br2, Br-

b. Mg , Mg2+ ? Cr3+, Cr2+

Answer 1

a.

Fe²⁺ → Fe³⁺ + 1 e⁻

Br₂ + 2 e⁻ → 2 Br⁻

b.

Mg → Mg²⁺ + 2 e⁻

Cr³⁺ + 1 e⁻ → Cr²⁺

Step-by-step explanation:

a.

Fe is oxidized and its oxidation number increases from 2+ to 3+, according to the following oxidation half-reaction.

Fe²⁺ → Fe³⁺ + 1 e⁻

Br is reduced and its oxidation number decreases from 0 to 1-, according to the following reduction half-reaction.

Br₂ + 2 e⁻ → 2 Br⁻

b.

Mg is oxidized and its oxidation number increases from 0 to 2+, according to the following oxidation half-reaction.

Mg → Mg²⁺ + 2 e⁻

Cr is reduced and its oxidation number decreases from 3+ to 2+, according to the following reduction half-reaction.

Cr³⁺ + 1 e⁻ → Cr²⁺

Answer 2

a)

`Fe^(2+)`⇒
`Fe^(3+)+e^-`

`Br_2+2e^-`⇒
`2Br^-`

b)

`Mg`⇒
`Mg^(2+)+2e^-`

`Cr^(3+)+e^-`⇒
`Cr^(3+)`

Step-by-step explanation:

A)

Remember that positive number superscripts mean electrons lack and negative numbers mean electrons 'excess' (if we compare it with the neutral element). So, for the case of Fe2+ which is converted to Fe3+, we know that in Fe2+ there is a two electrons lack, while in Fe3+ there is a 3 electrons lack; it means that Fe2+ was converted to Fe3+ but releasing one electron:

`Fe^(2+)`⇒
`Fe^(3+)+e^-`

The same analysis is applied to Br2; Br2 is a molecule which is said to have a zero superscript because it is an apolar covalent bond; and it is converted to Br-, which, according to what I wrote above, means that there is a one electron excess. So, Br2 must have received an electron in order to change to Br-; but Br2 can't change to Br- as simple as that because Br2 is a molecule, not an atom; it is a molecule that has two Br atoms, so, Br2 must give two Br- ions as products, but receiving one electron for each one:

`Br_2+2e^-`⇒
`2Br^-`

b)

Applying the same, in Mg2+ there is a 2 electrons lack, and in Mg is not electron lack (its superscript is zero), so Mg must have released two electrons in order to change to Mg2+:

`Mg`⇒
`Mg^(2+)+2e^-`

Cr3+ has a 3 electrons lack, and Cr2+ a two electrons one, so, Cr3+ must receive an electron to convert to Cr2+:

`Cr^(3+)+e^-`⇒
`Cr^(3+)`