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## Q. 4.18

Iodine is slightly soluble in water and dissolves to make a yellow-brown solution of $I_{2}(aq)$. When colorless $Sn^{2+}(aq)$ is dissolved in it (Figure 4.28), the solution turns colorless as $I_{2}(aq)$ forms and $Sn^{2+}(aq)$ is converted into $Sn^{4+}(aq)$. (a) Is this a redox reaction? (b) Balance the equation that describes the reaction.

## Verified Solution

Collect and Organize The reactants in this process are $I_{2}(aq)$ and $Sn^{2+}(aq)$; the products are $I_{2}(aq)$ and $Sn^{4+}(aq)$.

Analyze The oxidation number of tin changes from +2 in $Sn^{2+}(aq)$ to +4 in $Sn^{4+}(aq)$. The oxidation number of iodine changes from 0 in $I_{2}(aq)$ to -1 in $I^{-}(aq)$. We write tin and iodine in separate equations when we write the respective half reactions.

Solve
a. Because the oxidation numbers change, this is a redox reaction.

b. The unbalanced equation is

$Sn^{2+}(aq)+I_{2}(aq) → I^{-}(aq)+Sn^{4+}(aq)$

We balance it via our five-step procedure.

1. Separate the half-reactions.
Oxidation:           $Sn^{2+}(aq) → Sn^{4+}(aq)$
Reduction:          $I_{2}(aq) → I^{-}(aq)$
2.  Balance masses.
Oxidation:           $Sn^{2+}(aq) → Sn^{4+}(aq)$
Reduction:          $I_{2}(aq) → 2I^{-}(aq)$
3. Balance the charges by adding electrons.
Oxidation:             $Sn^{2+}(aq) → Sn^{4+}(aq)+2e^{-}$
Reduction:           $2e^{-} + I_{2}(aq) → 2I^{-}(aq)$
4. Balance the numbers of electrons. This step is not needed here because the numbers of electrons are the same in the two half-reactions.
Oxidation:          $Sn^{2+}(aq) → Sn^{4+}(aq)+ \sout{2e^{-}}$
Reduction:          $\sout{2e^{-}} + I_{2}(aq) → 2I^{-}(aq)$
Overall equation: $Sn^{2+}(aq) + I_{2}(aq) →2I^{-}(aq) + Sn^{4+}(aq)$
Check the overall equation for mass balance: $1 Sn + 2 I = 2 I + 1 Sn$. Check the
charge balance: reactant side 2+, product side $2(1-) + (4+) = 2+$.
Think About It We can balance redox reactions by using half-reactions. Adding the half-reactions gives us the net ionic equation and tells us the number of electrons transferred from the reducing agent to the oxidizing agent. Two moles of electrons are transferred to iodine for each mole of $Sn^{2+}$ (aq) that is oxidized. A balanced redox equation must be balanced with regard to electrons transferred as well as numbers of atoms.