Q22.37P - Chemistry: Molecular Nature Of Matter And Change

Question

Selenium is prepared by the reaction of ${\mathbf{\text{H}}}_{\mathbf{\text{2}}}{\mathbf{\text{SeO}}}_{\mathbf{\text{3}}}$ with gaseous ${\mathbf{\text{SO}}}_{\mathbf{\text{2}}}$.

(a) What redox process does the sulfur dioxide undergo? What is the oxidation state of sulfur in the product? 

(b) Given that the reaction occurs in acidic aqueous solution, what is the formula of the sulfur-containing species? 

(c) Write the balanced redox equation for the process.

Step-by-Step Solution

Answer

a) While oxidizing the oxidation number of sulphur changes from ${\text{ + 4 in SO}}_{\text{2}}\text{ to + 6 }$ in ${\text{ - SO}}_{\text{4}}^{\text{2 - }}\text{.}$

b) If the reaction undergoes an acidic aqueous solution, the sulfate ions are hydrated forming ${\text{HSO}}_{\text{4}}^{\text{ - }}\text{(aq)}$ ions.

c) The balanced equation for the reaction would be:

${\text{H}}_{\text{2}}{\text{SeO}}_{\text{3}}{\text{(aq) + 2SO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{O(l) }\to {\text{ Se(s) + 2HSO}}_{\text{4}}^{\text{ - }}{\text{(aq) + 2H}}^{\text{ + }}\text{(aq)}$

Step 1: Selenium

Selenium has chemical characteristics that are similar to sulfur. Selenium reacts directly or in an aqueous solution with metals and numerous non-metals. In appearance, composition, and properties, selenides are similar to sulfides.

Step 2: Obtaining the oxidation state of sulfur

a) At first, the reaction (unbalanced) should be written:

 ${\mathrm{H}}_2{\mathrm{SeO}}_3 +2{\mathrm{SO}}_2 +{\mathrm{H}}_2\mathrm{O}\to \mathrm{Se}+2{\mathrm{H}}_2{\mathrm{SO}}_4$

The selenium, $\text{Se}$, is being reduced, changing its oxidation number from $\text{ + 4 to 0,}$ acting as an oxidizing agent.

Whereas $\text{S}$ is being oxidized, changing its oxidation number from ${\text{ + 4 in SO}}_{\text{2}}\text{ to + 6 }$ in sulfate ion  ${\text{SO}}_{\text{4}}^{\text{2 - }}$.

Step 3: Subpart (b)

If the reaction undergoes in acidic aqueous solution, there are ${\text{H}}^{\text{ + }}$ ions present in the solution, and thus, the sulfate ions are hydrated:

${\text{H}}^{\text{ + }}{\text{(aq) + SO}}_{\text{4}}^{\text{2 - }}\text{(aq) }\to {\text{ HSO}}_{\text{4}}^{-}\text{(aq)}$ 

So, in the reaction in an acidic aqueous solution, the sulfate ions are hydrated forming  ${\text{HSO}}_{\text{4}}^{-}\left(aq\right)$.

Step 4: Subpart (c)

Thee unbalanced reaction in acidic solution:

 ${\text{H}}_{\text{2}}{\text{SeO}}_{\text{3}}{\text{(aq) + SO}}_{\text{2}}\text{(g) }\to {\text{ Se(s) + HSO}}_{\text{4}}^{\text{ - }}\text{(aq)}$

The reduction half-reaction:

 ${\text{H}}_{\text{2}}{\text{SeO}}_{\text{3}}{\text{(aq) + 4e}}^{\text{ - }}\to \text{ Se(s)}$

To balance O and H, adding water to the products

${\text{H}}_{\text{2}}{\text{SeO}}_{\text{3}}{\text{(aq) + 4e}}^{\text{ - }}\to {\text{ Se(s) + 3H}}_{\text{2}}\text{O(l)}$ 

Balancing the charges

${\text{H}}_{\text{2}}{\text{SeO}}_{\text{3}}{\text{(aq) + 4H}}^{\text{ + }}{\text{(aq) + 4e}}^{\text{ - }}\to {\text{S(s) + 3H}}_{\text{2}}\text{O(l)}$ 

The oxidation half-reaction:

${\text{SO}}_{\text{2}}\text{(g) }\to {\text{ HSO}}_{\text{4}}^{\text{ - }}{\text{(aq) + 2e}}^{\text{ - }}$ 

To balance $\text{O and H}$, adding water to the products

 ${\text{SO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{O(l)}\to {\text{HSO}}_{\text{4}}^{\text{ - }}{\text{(aq) + H}}^{\text{ + }}{\text{(aq) + 2e}}^{\text{ - }}$

Balancing the charges

${\text{SO}}_{\text{2}}{\text{(g) + 2H}}_{\text{2}}\text{O(l) }\to {\text{ HSO}}_{\text{4}}^{\text{ - }}{\text{(aq) + 3H}}^{\text{ + }}{\text{(aq) + 2e}}^{\text{ - }}$ 

Now, writing the balanced redox equation:

Given the two half - reactions, the oxidation of ${\text{SO}}_{\text{2}}$ should be multiplied to equalize the number of electrons exchanged:

${\text{H}}_{\text{2}}{\text{SeO}}_{\text{3}}{\text{(aq) + 4H}}^{\text{ + }}{\text{(aq) + 4e}}^{\text{ - }}\to {\text{ Se(s) + 3H}}_{\text{2}}\text{O(l)}$

${\text{2SO}}_{\text{2}}{\text{(g) + 4H}}_{\text{2}}\text{O(l) }\to {\text{ 2HSO}}_{\text{4}}^{\text{ - }}{\text{(aq) + 6H}}^{\text{ + }}{\text{(aq) + 4e}}^{\text{ - }}$ 

Combining two reactions (and crossing out the repeated units on two sides of the equation):

${\text{H}}_{\text{2}}{\text{SeO}}_{\text{3}}{\text{(aq) + 4H}}^{\text{ + }}{\text{(aq) + 4e}}^{\text{ - }}{\text{ + 2SO}}_{\text{2}}{\text{(g) + 4H}}_{\text{2}}\text{O(l) }\to {\text{ Se(s) + 3H}}_{\text{2}}{\text{O(l) + 2HSO}}_{\text{4}}^{\text{ - }}{\text{(aq) + 6H}}^{\text{ + }}{\text{(aq) + 4e}}^{\text{ - }}$ The balanced equation for the reaction would be:

 ${\text{H}}_{\text{2}}{\text{SeO}}_{\text{3}}{\text{(aq) + 2SO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{O(l) }\to {\text{ Se(s) + 2HSO}}_{\text{4}}^{\text{ - }}{\text{(aq) + 2H}}^{\text{ + }}\text{(aq)}$