Chapter 22, Problem 22.78QP

Interpretation Introduction

Interpretation:

The balanced chemical equation for the reaction of concentrated sulfuric acid and sodium iodide has to be written.

Concept Introduction:

Balanced equation:

It is a chemical equation in which the number of each type of atoms will be equal in the stoichiometric ratios on both sides of the equation.  A balanced chemical equation obeys the law of conservation of mass.

Law of conservation of mass:

This law indicates that in the chemical reaction, mass is neither created nor destroyed.

Balancing redox reaction:

• Balance the atoms other than Oxygen and Hydrogen atom.
• Balance the Oxygen atoms by adding water on the appropriate side.
• Balance the hydrogen atoms by adding equal number of ${\text{H}}^{+}$ on the opposite side.
• Balance the charge by adding appropriate number of electrons at the appropriate side.
• Multiply the reduction and oxidation half, by the whole number such that the number of electrons released in the reduction half should be equal to the number of electrons gained in the oxidation half.
• Add both the oxidation and reduction half-reaction and cancel out the electrons on both the sides of the reaction.
• Write the final equation and count the number of atoms and check the charge to verify the balanced reaction.

Explanation of Solution

Concentrated sulfuric acid-${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ reacts with sodium iodide-$\text{NaI}$ to produce molecular iodine-${\text{I}}_{\text{2}}$, hydrogen sulphide-${\text{H}}_2\text{S}$ and sodium hydrogen sulfate-${\text{NaHSO}}_{\text{4}}$.

So, the chemical formula can be written as:

${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)+\text{NaI}\left(\text{aq}\right)\stackrel{}{\to }{\text{I}}_{\text{2}}\left(\text{s}\right)\text{+}{\text{H}}_2\text{S}\left(\text{g}\right)\text{+}{\text{NaHSO}}_{\text{4}}\left(\text{aq}\right)$

The half-reactions that are involved in the above reaction are:

Oxidation half- reaction:

${\text{2I}}^{-}\left(\text{aq}\right)\stackrel{}{\to }{\text{I}}_{\text{2}}\left(\text{s}\right)+{\text{2e}}^{\text{-}}$

Reduction half-reaction:

${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)\stackrel{}{\to }{\text{H}}_{\text{2}}\text{S}\left(\text{g}\right)$

Balancing oxygen by the addition of water molecules:

${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)\stackrel{}{\to }{\text{H}}_{\text{2}}\text{S}\left(\text{g}\right)+{\text{4H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Balancing hydrogen by the addition of protons:

${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)+{\text{8H}}^{\text{+}}\left(\text{aq}\right)\stackrel{}{\to }{\text{H}}_{\text{2}}\text{S}\left(\text{g}\right)+{\text{4H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Balancing the charge on both sides of the equation:

${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)+{\text{8H}}^{\text{+}}\left(\text{aq}\right){\text{+8e}}^{\text{-}}\stackrel{}{\to }{\text{H}}_{\text{2}}\text{S}\left(\text{g}\right)+{\text{4H}}_{\text{2}}\text{O}\left(\text{l}\right)$

The reduction and oxidation half reactions have to be multiplied by the whole number such that the number of electrons gained in the reduction half should be equal to the number of electrons released in the oxidation half.

Then the half reactions are added to cancel out the electrons.

These math operations result in the final equation.

Determining the overall balanced equation:

The balanced reduction half-reaction is

${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)+{\text{8H}}^{\text{+}}\left(\text{aq}\right){\text{+8e}}^{\text{-}}\stackrel{}{\to }{\text{H}}_{\text{2}}\text{S}\left(\text{g}\right)+{\text{4H}}_{\text{2}}\text{O}\left(\text{l}\right)$

The balanced oxidation half-reaction is

${\text{2I}}^{-}\left(\text{aq}\right)\stackrel{}{\to }{\text{I}}_{\text{2}}\left(\text{s}\right)+{\text{2e}}^{\text{-}}$

$\begin{array}{l}\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)+{\text{8H}}^{\text{+}}\left(\text{aq}\right){\text{+8e}}^{\text{-}}\stackrel{}{\to }{\text{H}}_{\text{2}}\text{S}\left(\text{g}\right)+{\text{4H}}_{\text{2}}\text{O}\left(\text{l}\right)\right)\\ 4\left({\text{2I}}^{-}\left(\text{aq}\right)\stackrel{}{\to }{\text{I}}_{\text{2}}\left(\text{s}\right)+{\text{2e}}^{\text{-}}\right)\end{array}$

$\begin{array}{l}--------------------------\\ {\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)\text{+}8{\text{I}}^{-}\left(\text{aq}\right)\text{+}{\text{8H}}^{\text{+}}\left(\text{aq}\right)\text{+}{\bcancel{\text{8}}\bcancel{\text{e}}}^{\text{-}}\stackrel{}{\to }4{\text{I}}_{\text{2}}\left(\text{s}\right){\text{+H}}_{\text{2}}\text{S}\left(\text{g}\right)+{\text{4H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+}\bcancel{\text{8}}\bcancel{\text{e}}}^{\text{-}}\\ ---------------------------\end{array}$

${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)\text{+}8{\text{I}}^{-}\left(\text{aq}\right)\text{+}{\text{8H}}^{\text{+}}\left(\text{aq}\right)\stackrel{}{\to }4{\text{I}}_{\text{2}}\left(\text{s}\right){\text{+H}}_{\text{2}}\text{S}\left(\text{g}\right)+{\text{4H}}_{\text{2}}\text{O}\left(\text{l}\right)$

One sodium ion has to be added for each iodine ion to obtain sodium on the product side:

${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)\text{+}8{\text{Na}}^{\text{+}}{\text{I}}^{-}\left(\text{aq}\right)\text{+}{\text{8H}}^{\text{+}}\left(\text{aq}\right)\stackrel{}{\to }4{\text{I}}_{\text{2}}\left(\text{s}\right){\text{+H}}_{\text{2}}\text{S}\left(\text{g}\right)+{\text{4H}}_{\text{2}}\text{O}\left(\text{l}\right)+8{\text{Na}}^{\text{+}}\left(\text{aq}\right)$

Extra moles of sulfuric acid needed to obtain the sulfate salt of sodium on the product side.

${\text{9H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)\text{+}8{\text{Na}}^{\text{+}}{\text{I}}^{-}\left(\text{aq}\right)\text{+}{\text{8H}}^{\text{+}}\left(\text{aq}\right)\stackrel{}{\to }4{\text{I}}_{\text{2}}\left(\text{s}\right){\text{+H}}_{\text{2}}\text{S}\left(\text{g}\right)+{\text{4H}}_{\text{2}}\text{O}\left(\text{l}\right)+8{\text{NaHSO}}_4\left(\text{aq}\right)$

This is the overall balanced redox reaction.