Question #07791
1 Answer
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
Likewise, chromium is being reduced, since its oxidation number goes from
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
#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
#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,
Add
#stackrel(color(blue)(+6))("Cr")_2"O"_7^(2-) + 6"e"^(-) -> 2stackrel(color(blue)(+3))("Cr")""^(3+) + 7"H"_2"O"#
Add
#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
#{(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])#