Question

aleks g For each reaction, write the chemical formulae of the oxidized reactants in the space provided. Write the chemical formulae of the reduced reactants in the space provided. Fe

Answer 1

(a) Iron metal is oxidized, lead(II) cation is reduced;

(b) Silver cation is reduced, magnesium metal is oxidized;

(c) Copper(II) cation is reduced, magnesium metal is oxidized.

Step-by-step explanation:

The original question from ALEKS contains 3 reactions, we'll use them to illustrate the concept as well:

(a)
`2 Fe (s) + 3 Pb(NO_3)_2 (aq)\rightarrow 3 Pb (s) + 2 Fe(NO_3)_3 (aq)`;

(b)
`2 AgNO_3 (aq) + Mg (s)\rightarrow 2 Ag (s) + Mg(NO_3)_2 (aq)`;

(c)
`CuCl_2 (aq) + Mg (s)\rightarrow MgCl_2 (aq) + Cu (s)`

Using the abbreviation OILRIG, we know that oxidation is loss, reduction is gain (of electrons).

(a) We start with Fe at oxidation state of 0, as it's not bonded to anything and end up with iron(III) nitrate, iron has a charge of 3+ in this compound, as nitrate has a charge of 1- and 3 nitrate anions should be balanced by 3+:

`Fe (s)\rightarrow Fe^(3+) (aq) + 3e^-`

Iron metal loses 3 electrons, so it's oxidized.

Similarly, we start with lead with a charge of 2+ and end up with solid lead which is neutral:

`Pb^(2+) (aq) + 2e^- \rightarrow Pb (s)`

Lead(II) cation gains electrons, so it's reduced.

(b) Similarly, in this reaction we notice:

`Ag^+ (aq) + e^- \rightarrow Ag (s)`

Silver cation gains an electron, it's reduced.

`Mg (s)\rightarrow Mg^(2+) (aq) + 2e^-`

Magnesium metal is oxidized, as it loses electrons.

(c) Similarly:

`Cu^(2+) (aq) + 2e^- \rightarrow Cu (s)`

Copper(II) cation gains 2 electrons, so it's reduced.

`Mg (s)\rightarrow Mg^(2+) (aq) + 2e^-`

Magnesium metal loses 2 electrons, so it's oxidized.