Question

7. Determine whether each of the following half-reactions represents an oxidation or a reduction. Supply the correct number ofelectrons to the appropriate side to balance the equation.(a) Na > Na (+)(b) C2O4 (2-) > 2 CO2(c) 2 I (-) -› I2(d) Cr2O7 (2-) + 14 H (+) - › 2 Cr (3+) + 7 H2O

Answer 1

1) What are oxidation and reduction?

Oxidation: This is the loss of electrons. The oxidation number goes up.

Reduction: This is the gain of electrons. The oxidation number goes down.

$a)\text{ }Na\rightarrow Na^+$

The oxidation number of Sodium changes from 0 to +1.

This is an oxidation.

$Na\rightarrow Na^++e^-$$b)\text{ }C_2O_4^{2-}\rightarrow2CO_2$

Let's analyze C2O4 (2-). The oxidation number of Oxigen is usually -2 and the charge of the molecule is 2-. The total negative charge is 8-. So, there must be a +6 charge in total. Since there are two carbons, each carbon has an oxidation number of +3.

CO2. The oxidation number of Oxigen is usually -2 and the charge of the molecule is 2-. The total negative charge is 4-. So, there must be a +4 charge in total. Since there are two carbons, each carbon has an oxidation number of +2.

This is a reduction.

$C_2O_4^{2-}+2e^-\rightarrow2CO_2$

----

$c)\text{ }2\text{ }I^-\rightarrow I_2$

The oxidation number of an ion is the ion charge. So, the oxidation number of I (-) is -1. The oxidation number of an element is 0.

This is an oxidation.

$2\text{ }I^-\rightarrow I_2+2e^-$

----

$d)\text{ }Cr_2O_7^{2-}+14H^+\rightarrow2Cr^{3+}+7H_2O$

The oxidation number of Oxigen is usually -2 and the charge of the molecule is 2-. The total negative charge is -14. So, there must be a +12 charge in total. Since we have two Cr, each of those has a +6 charge. The oxidation number of monoatomic ions is the ion charge. So, in the products, the oxidation number of Cr is +3.

This is a reduction.

$6e^-+Cr_2O_7^{2-}+14H^+\rightarrow2Cr^{3+}+7H_2O$

.