Answer:
a)
[tex]Fe^{2+}[/tex]⇒[tex]Fe^{3+}+e^-[/tex]
[tex]Br_2+2e^-[/tex]⇒[tex]2Br^-[/tex]
b)
[tex]Mg[/tex]⇒[tex]Mg^{2+}+2e^-[/tex]
[tex]Cr^{3+}+e^-[/tex]⇒[tex]Cr^{3+}[/tex]
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:
[tex]Fe^{2+}[/tex]⇒[tex]Fe^{3+}+e^-[/tex]
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:
[tex]Br_2+2e^-[/tex]⇒[tex]2Br^-[/tex]
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+:
[tex]Mg[/tex]⇒[tex]Mg^{2+}+2e^-[/tex]
Cr3+ has a 3 electrons lack, and Cr2+ a two electrons one, so, Cr3+ must receive an electron to convert to Cr2+:
[tex]Cr^{3+}+e^-[/tex]⇒[tex]Cr^{3+}[/tex]
Answer:
a.
Fe²⁺ → Fe³⁺ + 1 e⁻
Br₂ + 2 e⁻ → 2 Br⁻
b.
Mg → Mg²⁺ + 2 e⁻
Cr³⁺ + 1 e⁻ → Cr²⁺
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⁺
