Question #07791

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Question #07791

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Answer 1 · 2015-11-05T01:38:50

$14 H_{(aq]}^{+} + 6 I_{(aq]}^{-} + {Cr}_{2} O_{7(aq]}^{2 -} \to 3 I_{2(l]} + 2 {Cr}_{(aq]}^{3 +} + 7 H_{2} O_{(l]}$ $14"H"_text((aq])^(+) + 6"I"_text((aq])^(-) + "Cr"_2"O"_text(7(aq])^(2-) -> 3"I"_text(2(l]) + 2"Cr"_text((aq])^(3+) + 7"H"_2"O"_text((l])$

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.

$I_{(aq]}^{-} + {Cr}_{2} O_{7(aq]}^{2 -} \to {Cr}_{(aq]}^{3 +} + I_{2(l]}$ $"I"_text((aq])^(-) + "Cr"_2"O"_text(7(aq])^(2-) -> "Cr"_text((aq])^(3+) + "I"_text(2(l])$

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

$- 1 I^{-} + {+ 6 Cr}_{2} - 2 O_{7}^{2 -} \to + 3 Cr^{3 +} + {0 I}_{2}$ $stackrel(color(blue)(-1))("I")""^(-) + stackrel(color(blue)(+6))("Cr")_2 stackrel(color(blue)(-2))("O")""_7^(2-) -> stackrel(color(blue)(+3))("Cr")""^(3+) + stackrel(color(blue)(0))("I")_2$

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 $- 1$ $color(blue)(-1)$ to $0$ $color(blue)(0)$ .

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

This means that your half-reactions will look like this

oxidation half-reaction

$- 1 I^{-} \to {0 I}_{2}$ $stackrel(color(blue)(-1))("I")""^(-) -> stackrel(color(blue)(0))("I")_2$

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

$2 - 1 I^{-} \to {0 I}_{2}$ $2stackrel(color(blue)(-1))("I")""^(-) -> stackrel(color(blue)(0))("I")_2$

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

$2 - 1 I^{-} \to {0 I}_{2} + 2 e^{-}$ $2stackrel(color(blue)(-1))("I")""^(-) -> stackrel(color(blue)(0))("I")_2 + 2"e"^(-)$

reduction half-reaction

${+ 6 Cr}_{2} O_{7}^{2 -} \to + 3 Cr^{3 +}$ $stackrel(color(blue)(+6))("Cr")_2"O"_7^(2-) -> stackrel(color(blue)(+3))("Cr")""^(3+)$

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

${+ 6 Cr}_{2} O_{7}^{2 -} \to 2 + 3 Cr^{3 +}$ $stackrel(color(blue)(+6))("Cr")_2"O"_7^(2-) -> 2stackrel(color(blue)(+3))("Cr")""^(3+)$

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

${+ 6 Cr}_{2} O_{7}^{2 -} + 6 e^{-} \to 2 + 3 Cr^{3 +}$ $stackrel(color(blue)(+6))("Cr")_2"O"_7^(2-) + 6"e"^(-) -> 2stackrel(color(blue)(+3))("Cr")""^(3+)$

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

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

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

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

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

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 $3$ $3$ to get a total of $6$ $6$ 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

$14"H"_text((aq])^(+) + 6"I"_text((aq])^(-) + "Cr"_2"O"_text(7(aq])^(2-) -> 3"I"_text(2(l]) + 2"Cr"_text((aq])^(3+) + 7"H"_2"O"_text((l])$

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Question #07791

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