Question

Question #07791

Answer 1

`14"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"_text((l])`

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.

`"I"_text((aq])^(-) + "Cr"_2"O"_text(7(aq])^(2-) -> "Cr"_text((aq])^(3+) + "I"_text(2(l])`

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

`stackrel(color(blue)(-1))("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 `color(blue)(-1)` to `color(blue)(0)`.

Likewise, chromium is being reduced, since its oxidation number goes from `color(blue)(+6)` to `color(blue)(+3)`.

This means that your half-reactions will look like this

• oxidation half-reaction

`stackrel(color(blue)(-1))("I")""^(-) -> stackrel(color(blue)(0))("I")_2`

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

`2stackrel(color(blue)(-1))("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, `"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"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

`{(2"I"^(-) -> "I"_2 + 2"e"^(-) | xx 3), (14"H"^(+) + "Cr"_2"O"_7^(2-) + 6"e"^(-) -> 2"Cr"^(3+) + 7"H"_2"O") :}`

`{(6"I"^(-) -> 3"I"_2 + 6"e"^(-)), (14"H"^(+) + "Cr"_2"O"_7^(2-) + 6"e"^(-) -> 2"Cr"^(3+) + 7"H"_2"O") :}`

Now add the two half-reactions to get

`6"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"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"_text((l])`