# Question 07791

Nov 5, 2015

$14 {\text{H"_text((aq])^(+) + 6"I"_text((aq])^(-) + "Cr"_2"O"_text(7(aq])^(2-) -> 3"I"_text(2(l]) + 2"Cr"_text((aq])^(3+) + 7"H"_2"O}}_{\textrm{\left(l\right]}}$

#### Explanation:

You're dealing with a redox reaction in which the iodide anions are being oxidized to iodine and the dichromate anions are being reduced to chromium(III) cations.

${\text{I"_text((aq])^(-) + "Cr"_2"O"_text(7(aq])^(2-) -> "Cr"_text((aq])^(3+) + "I}}_{\textrm{2 \left(l\right]}}$

Before doing anything else, assign oxidation numbers to the atoms that take part in the reaction.

${\stackrel{\textcolor{b l u e}{- 1}}{\text{I")""^(-) + stackrel(color(blue)(+6))("Cr")_2 stackrel(color(blue)(-2))("O")""_7^(2-) -> stackrel(color(blue)(+3))("Cr")""^(3+) + stackrel(color(blue)(0))("I}}}_{2}$

The oxidation numbers of the atoms on the reactants' side and on the products' side confirm that iodide is indeed being oxidized to iodine, since its oxidation number goes from $\textcolor{b l u e}{- 1}$ to $\textcolor{b l u e}{0}$.

Likewise, chromium is being reduced, since its oxidation number goes from $\textcolor{b l u e}{+ 6}$ to $\textcolor{b l u e}{+ 3}$.

This means that your half-reactions will look like this

• oxidation half-reaction

${\stackrel{\textcolor{b l u e}{- 1}}{\text{I")""^(-) -> stackrel(color(blue)(0))("I}}}_{2}$

Balance the iodine atoms by multiplying the iodide anions by $2$

$2 {\stackrel{\textcolor{b l u e}{- 1}}{\text{I")""^(-) -> stackrel(color(blue)(0))("I}}}_{2}$

Each iodie anion will lose one electron, which means that wo iodide anions will lose a total of two electrons

2stackrel(color(blue)(-1))("I")""^(-) -> stackrel(color(blue)(0))("I")_2 + 2"e"^(-)

• reduction half-reaction

stackrel(color(blue)(+6))("Cr")_2"O"_7^(2-) -> stackrel(color(blue)(+3))("Cr")""^(3+)

Multiply the chromium(III) cations by $2$ to get

stackrel(color(blue)(+6))("Cr")_2"O"_7^(2-) -> 2stackrel(color(blue)(+3))("Cr")""^(3+)

Each chromium atom will gain three electrons, which means that two chromium atoms will gain a total of six electrons

stackrel(color(blue)(+6))("Cr")_2"O"_7^(2-) + 6"e"^(-) -> 2stackrel(color(blue)(+3))("Cr")""^(3+)

You're in acidic solution, so you can balance the oxygen atoms by adding wter molecules and the hydrogen atoms by adding protons, ${\text{H}}^{+}$.

Add $7$ water molecules to the products' side to balance the $7$ oxygen atoms present on the reactants' side

stackrel(color(blue)(+6))("Cr")_2"O"_7^(2-) + 6"e"^(-) -> 2stackrel(color(blue)(+3))("Cr")""^(3+) + 7"H"_2"O"#

Add $14$ protons on the reactants' side to balance the hydrogen atoms

$14 \text{H"^(+) + stackrel(color(blue)(+6))("Cr")_2"O"_7^(2-) + 6"e"^(-) -> 2stackrel(color(blue)(+3))("Cr")""^(3+) + 7"H"_2"O}$

Now, in any redox reaction, the number of electrons gained in the reduction half-raection must be equal to the number of electrons lost in the oxidation half-reaction.

This means that you must multiply the oxidation half-reaction by $3$ to get a total of $6$ electrons transferred in the redox reaction

$\left\{\begin{matrix}2 \text{I"^(-) -> "I"_2 + 2"e"^(-) | xx 3 \\ 14"H"^(+) + "Cr"_2"O"_7^(2-) + 6"e"^(-) -> 2"Cr"^(3+) + 7"H"_2"O}\end{matrix}\right.$

$\left\{\begin{matrix}6 \text{I"^(-) -> 3"I"_2 + 6"e"^(-) \\ 14"H"^(+) + "Cr"_2"O"_7^(2-) + 6"e"^(-) -> 2"Cr"^(3+) + 7"H"_2"O}\end{matrix}\right.$

Now add the two half-reactions to get

$6 \text{I"^(-) + 14"H"^(+) + "Cr"_2"O"_7^(2-) + color(red)(cancel(color(black)(6"e"^(-)))) -> 3"I"_2 + color(red)(cancel(color(black)(6"e"^(-)))) + 2"Cr"^(3+) + 7"H"_2"O}$

Finally, the balanced chemical equation for this redox reaction is

$14 {\text{H"_text((aq])^(+) + 6"I"_text((aq])^(-) + "Cr"_2"O"_text(7(aq])^(2-) -> 3"I"_text(2(l]) + 2"Cr"_text((aq])^(3+) + 7"H"_2"O}}_{\textrm{\left(l\right]}}$