Reaction between copper(II) iodate and iodideWhich ion of iron is produced in a reaction between iron and copper(II) sulfate?Why should the redox reaction happen between the copper ions and iron atoms?Can the thermodynamic predictions of redox reactions based on E and dG contradict each other?Equation for reaction of tin(IV) and iodine to produce tin(IV) iodideWhat is the balanced equation for the reaction of potassium permanganate, iron sulfate, and sulfuric acid?Spontaneous reaction between chromium(III) and aluminumWhy does KMnO4 oxidize iodide to iodine in acidic medium, whereas iodide to iodate in alkaline medium?Stoichiometric reaction of hydrochloric acid, potassium nitrate, and copperCopper Mirror ReactionNo sign flipping while figuring out the emf of voltaic cell?
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Reaction between copper(II) iodate and iodide
Which ion of iron is produced in a reaction between iron and copper(II) sulfate?Why should the redox reaction happen between the copper ions and iron atoms?Can the thermodynamic predictions of redox reactions based on E and dG contradict each other?Equation for reaction of tin(IV) and iodine to produce tin(IV) iodideWhat is the balanced equation for the reaction of potassium permanganate, iron sulfate, and sulfuric acid?Spontaneous reaction between chromium(III) and aluminumWhy does KMnO4 oxidize iodide to iodine in acidic medium, whereas iodide to iodate in alkaline medium?Stoichiometric reaction of hydrochloric acid, potassium nitrate, and copperCopper Mirror ReactionNo sign flipping while figuring out the emf of voltaic cell?
$begingroup$
I was attempting the following question from IChO 1989, which involved balancing of redox equations. One may just expect this to be a typical iodate-iodide redox reaction, usually employed in iodometry. However, the presence of $ceCu(II)$ makes things a bit more complicated since it can also be reduced to give $ceCuI$.
To determine the solubility product of copper(II) iodate, $ceCu(IO3)2$, by iodometric titration in an acidic solution ($pu25 ^circ C$), $pu30.00 cm3$ of a $0.100$ molar sodium thiosulfate solution are needed to titrate $pu20.00 cm3$ of a saturated aqueous solution of $ceCu(IO3)2$.
1.1 Write the sequence of balanced equations for the above described reactions. [...]
After realising that $ce Cu^2+$ is reduced as well, I wrote the following overall equation :
$$ce2Cu^2+ + 2IO3^- + 12H^+ + 14I^- -> 7I2 + 2 CuI + 6H2O tag1$$
which is based on
$$beginalign
ce2Cu^2+ + 4I^- &-> 2CuI + I2 tag2 labeleq:cuii \
ce2IO3^- + 12H^+ + 10I^- &-> 6I2 + 6H2O tag3 labeleq:iodate
endalign$$
However, it turns out that the equation given in the answer scheme seems to have multiplied eq. $(refeq:iodate)$ by $2$ and the overall equation given is:
$$ce2Cu^2+ + 4IO3^- + 24H^+ + 24I^- -> 13I2 + 2CuI + 12H2O tag4$$
Usually, this is a trivial thing and does not matter. But in this case, it does since the reaction stiochiometry is completely different based on the equation given. Have I gone wrong somewhere?
redox stoichiometry
edited May 27 at 12:19
orthocresol♦
41.7k7125256
asked May 27 at 0:55
Tan Yong BoonTan Yong Boon
4,45511551
$endgroup$
add a comment |
$begingroup$
I was attempting the following question from IChO 1989, which involved balancing of redox equations. One may just expect this to be a typical iodate-iodide redox reaction, usually employed in iodometry. However, the presence of $ceCu(II)$ makes things a bit more complicated since it can also be reduced to give $ceCuI$.
To determine the solubility product of copper(II) iodate, $ceCu(IO3)2$, by iodometric titration in an acidic solution ($pu25 ^circ C$), $pu30.00 cm3$ of a $0.100$ molar sodium thiosulfate solution are needed to titrate $pu20.00 cm3$ of a saturated aqueous solution of $ceCu(IO3)2$.
1.1 Write the sequence of balanced equations for the above described reactions. [...]
After realising that $ce Cu^2+$ is reduced as well, I wrote the following overall equation :
$$ce2Cu^2+ + 2IO3^- + 12H^+ + 14I^- -> 7I2 + 2 CuI + 6H2O tag1$$
which is based on
$$beginalign
ce2Cu^2+ + 4I^- &-> 2CuI + I2 tag2 labeleq:cuii \
ce2IO3^- + 12H^+ + 10I^- &-> 6I2 + 6H2O tag3 labeleq:iodate
endalign$$
However, it turns out that the equation given in the answer scheme seems to have multiplied eq. $(refeq:iodate)$ by $2$ and the overall equation given is:
$$ce2Cu^2+ + 4IO3^- + 24H^+ + 24I^- -> 13I2 + 2CuI + 12H2O tag4$$
Usually, this is a trivial thing and does not matter. But in this case, it does since the reaction stiochiometry is completely different based on the equation given. Have I gone wrong somewhere?
redox stoichiometry
edited May 27 at 12:19
orthocresol♦
41.7k7125256
asked May 27 at 0:55
Tan Yong BoonTan Yong Boon
4,45511551
$endgroup$
$begingroup$
Well, I get what you got. Iodate to iodide is +6e balanced with protons and water. Cupric to cuprous is + 1e. A total of +7e for reduction. Two iodide to iodine is -2e oxidation. Multiply reduction by 2 and the oxidation by 7 and it works out. Can't speak for the "answer scheme".
$endgroup$
– user55119
May 27 at 1:51
$begingroup$
This question is clearly not closeable as homework because there is effort provided. If you want to close it, please describe your reasoning in a comment.
$endgroup$
– M.A.R. ಠ_ಠ
May 27 at 7:28
add a comment |
$begingroup$
I was attempting the following question from IChO 1989, which involved balancing of redox equations. One may just expect this to be a typical iodate-iodide redox reaction, usually employed in iodometry. However, the presence of $ceCu(II)$ makes things a bit more complicated since it can also be reduced to give $ceCuI$.
To determine the solubility product of copper(II) iodate, $ceCu(IO3)2$, by iodometric titration in an acidic solution ($pu25 ^circ C$), $pu30.00 cm3$ of a $0.100$ molar sodium thiosulfate solution are needed to titrate $pu20.00 cm3$ of a saturated aqueous solution of $ceCu(IO3)2$.
1.1 Write the sequence of balanced equations for the above described reactions. [...]
After realising that $ce Cu^2+$ is reduced as well, I wrote the following overall equation :
$$ce2Cu^2+ + 2IO3^- + 12H^+ + 14I^- -> 7I2 + 2 CuI + 6H2O tag1$$
which is based on
$$beginalign
ce2Cu^2+ + 4I^- &-> 2CuI + I2 tag2 labeleq:cuii \
ce2IO3^- + 12H^+ + 10I^- &-> 6I2 + 6H2O tag3 labeleq:iodate
endalign$$
However, it turns out that the equation given in the answer scheme seems to have multiplied eq. $(refeq:iodate)$ by $2$ and the overall equation given is:
$$ce2Cu^2+ + 4IO3^- + 24H^+ + 24I^- -> 13I2 + 2CuI + 12H2O tag4$$
Usually, this is a trivial thing and does not matter. But in this case, it does since the reaction stiochiometry is completely different based on the equation given. Have I gone wrong somewhere?
redox stoichiometry
edited May 27 at 12:19
orthocresol♦
41.7k7125256
asked May 27 at 0:55
Tan Yong BoonTan Yong Boon
4,45511551
$endgroup$
I was attempting the following question from IChO 1989, which involved balancing of redox equations. One may just expect this to be a typical iodate-iodide redox reaction, usually employed in iodometry. However, the presence of $ceCu(II)$ makes things a bit more complicated since it can also be reduced to give $ceCuI$.
To determine the solubility product of copper(II) iodate, $ceCu(IO3)2$, by iodometric titration in an acidic solution ($pu25 ^circ C$), $pu30.00 cm3$ of a $0.100$ molar sodium thiosulfate solution are needed to titrate $pu20.00 cm3$ of a saturated aqueous solution of $ceCu(IO3)2$.
1.1 Write the sequence of balanced equations for the above described reactions. [...]
After realising that $ce Cu^2+$ is reduced as well, I wrote the following overall equation :
$$ce2Cu^2+ + 2IO3^- + 12H^+ + 14I^- -> 7I2 + 2 CuI + 6H2O tag1$$
which is based on
$$beginalign
ce2Cu^2+ + 4I^- &-> 2CuI + I2 tag2 labeleq:cuii \
ce2IO3^- + 12H^+ + 10I^- &-> 6I2 + 6H2O tag3 labeleq:iodate
endalign$$
However, it turns out that the equation given in the answer scheme seems to have multiplied eq. $(refeq:iodate)$ by $2$ and the overall equation given is:
$$ce2Cu^2+ + 4IO3^- + 24H^+ + 24I^- -> 13I2 + 2CuI + 12H2O tag4$$
Usually, this is a trivial thing and does not matter. But in this case, it does since the reaction stiochiometry is completely different based on the equation given. Have I gone wrong somewhere?
redox stoichiometry
redox stoichiometry
edited May 27 at 12:19
orthocresol♦
41.7k7125256
asked May 27 at 0:55
Tan Yong BoonTan Yong Boon
4,45511551
edited May 27 at 12:19
orthocresol♦
41.7k7125256
asked May 27 at 0:55
Tan Yong BoonTan Yong Boon
4,45511551
edited May 27 at 12:19
orthocresol♦
41.7k7125256
edited May 27 at 12:19
orthocresol♦
41.7k7125256
edited May 27 at 12:19
orthocresol♦
41.7k7125256
41.7k7125256
asked May 27 at 0:55
Tan Yong BoonTan Yong Boon
4,45511551
asked May 27 at 0:55
Tan Yong BoonTan Yong Boon
4,45511551
asked May 27 at 0:55
Tan Yong BoonTan Yong Boon
4,45511551
4,45511551
$begingroup$
Well, I get what you got. Iodate to iodide is +6e balanced with protons and water. Cupric to cuprous is + 1e. A total of +7e for reduction. Two iodide to iodine is -2e oxidation. Multiply reduction by 2 and the oxidation by 7 and it works out. Can't speak for the "answer scheme".
$endgroup$
– user55119
May 27 at 1:51
$begingroup$
This question is clearly not closeable as homework because there is effort provided. If you want to close it, please describe your reasoning in a comment.
$endgroup$
– M.A.R. ಠ_ಠ
May 27 at 7:28
add a comment |
$begingroup$
Well, I get what you got. Iodate to iodide is +6e balanced with protons and water. Cupric to cuprous is + 1e. A total of +7e for reduction. Two iodide to iodine is -2e oxidation. Multiply reduction by 2 and the oxidation by 7 and it works out. Can't speak for the "answer scheme".
$endgroup$
– user55119
May 27 at 1:51
$begingroup$
This question is clearly not closeable as homework because there is effort provided. If you want to close it, please describe your reasoning in a comment.
$endgroup$
– M.A.R. ಠ_ಠ
May 27 at 7:28
$begingroup$
Well, I get what you got. Iodate to iodide is +6e balanced with protons and water. Cupric to cuprous is + 1e. A total of +7e for reduction. Two iodide to iodine is -2e oxidation. Multiply reduction by 2 and the oxidation by 7 and it works out. Can't speak for the "answer scheme".
$endgroup$
– user55119
May 27 at 1:51
$begingroup$
Well, I get what you got. Iodate to iodide is +6e balanced with protons and water. Cupric to cuprous is + 1e. A total of +7e for reduction. Two iodide to iodine is -2e oxidation. Multiply reduction by 2 and the oxidation by 7 and it works out. Can't speak for the "answer scheme".
$endgroup$
– user55119
May 27 at 1:51
$begingroup$
This question is clearly not closeable as homework because there is effort provided. If you want to close it, please describe your reasoning in a comment.
$endgroup$
– M.A.R. ಠ_ಠ
May 27 at 7:28
$begingroup$
This question is clearly not closeable as homework because there is effort provided. If you want to close it, please describe your reasoning in a comment.
$endgroup$
– M.A.R. ಠ_ಠ
May 27 at 7:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In your eq. (4), the copper/iodate ratio was adjusted to match the stoichiometry present in copper(II) iodate. 1 copper(II) ion to 2 iodate ions equals 2 copper(II) ions to 4 iodate ions.
edited May 27 at 12:16
orthocresol♦
41.7k7125256
answered May 27 at 1:00
Oscar LanziOscar Lanzi
17.1k22853
$endgroup$
add a comment |
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$begingroup$
In your eq. (4), the copper/iodate ratio was adjusted to match the stoichiometry present in copper(II) iodate. 1 copper(II) ion to 2 iodate ions equals 2 copper(II) ions to 4 iodate ions.
edited May 27 at 12:16
orthocresol♦
41.7k7125256
answered May 27 at 1:00
Oscar LanziOscar Lanzi
17.1k22853
$endgroup$
add a comment |
$begingroup$
In your eq. (4), the copper/iodate ratio was adjusted to match the stoichiometry present in copper(II) iodate. 1 copper(II) ion to 2 iodate ions equals 2 copper(II) ions to 4 iodate ions.
edited May 27 at 12:16
orthocresol♦
41.7k7125256
answered May 27 at 1:00
Oscar LanziOscar Lanzi
17.1k22853
$endgroup$
add a comment |
$begingroup$
In your eq. (4), the copper/iodate ratio was adjusted to match the stoichiometry present in copper(II) iodate. 1 copper(II) ion to 2 iodate ions equals 2 copper(II) ions to 4 iodate ions.
edited May 27 at 12:16
orthocresol♦
41.7k7125256
answered May 27 at 1:00
Oscar LanziOscar Lanzi
17.1k22853
$endgroup$
In your eq. (4), the copper/iodate ratio was adjusted to match the stoichiometry present in copper(II) iodate. 1 copper(II) ion to 2 iodate ions equals 2 copper(II) ions to 4 iodate ions.
edited May 27 at 12:16
orthocresol♦
41.7k7125256
answered May 27 at 1:00
Oscar LanziOscar Lanzi
17.1k22853
edited May 27 at 12:16
orthocresol♦
41.7k7125256
edited May 27 at 12:16
orthocresol♦
41.7k7125256
edited May 27 at 12:16
orthocresol♦
41.7k7125256
41.7k7125256
answered May 27 at 1:00
Oscar LanziOscar Lanzi
17.1k22853
answered May 27 at 1:00
Oscar LanziOscar Lanzi
17.1k22853
answered May 27 at 1:00
Oscar LanziOscar Lanzi
17.1k22853
17.1k22853
add a comment |
add a comment |
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Well, I get what you got. Iodate to iodide is +6e balanced with protons and water. Cupric to cuprous is + 1e. A total of +7e for reduction. Two iodide to iodine is -2e oxidation. Multiply reduction by 2 and the oxidation by 7 and it works out. Can't speak for the "answer scheme".
$endgroup$
– user55119
May 27 at 1:51
$begingroup$
This question is clearly not closeable as homework because there is effort provided. If you want to close it, please describe your reasoning in a comment.
$endgroup$
– M.A.R. ಠ_ಠ
May 27 at 7:28