Question #07791

1 Answer
Nov 5, 2015

Answer:

#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])#