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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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TopNewestHere you go:

$\underline{4}CuSCN+\underline{7}KIO_3+\underline{14}HCl=\underline{4}CuSO_4+\underline{7}KCl+\underline{4}HCN+\underline{7}ICl+\underline{5}H_2O$

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If you want to solve this problem you have to write out the redox half equations for three reactions:

The oxidation of Copper I to copper II

The oxidation of SCN- to Sulphate

The reduction of iodate V to iodine I (this is the trap as we usually think of Iodine as -1, but in ICl it is behaving like I+)

You need to keep the oxidation equation coefficients equal as Cu=SCN

The common factor for eliminating electrons from the oxidation and reduction equations is 28

Hope this helps...

Periodic Player

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By redox equations do you mean like $2H_2+O_2=2H_2O$? Ok so I need to balance

$\underline{\hspace{2mm}} CuSCN+\underline{\hspace{2mm}}KIO_3+\underline{\hspace{2mm}}HCl=\underline{\hspace{2mm}}CuSO_4+\underline{\hspace{2mm}}KCl+\underline{\hspace{2mm}}HCN+\underline{\hspace{2mm}}ICl+\underline{\hspace{2mm}}H_2O$

Sounds interesting. I shall give it a try.

P.S. here's the latex for the equation:

\underline{\hspace{2mm}}CuSCN+\underline{\hspace{2mm}}KIO

3+\underline{\hspace{2mm}}HCl=\underline{\hspace{2mm}}CuSO4+\underline{\hspace{2mm}}KCl+\underline{\hspace{2mm}}HCN+\underline{\hspace{2mm}}ICl+\underline{\hspace{2mm}}H_2O\Log in to reply

If you want to solve this problem you have to write out the redox half equations for three reactions:

The oxidation of Copper I to copper II

The oxidation of SCN- to Sulphate

The reduction of iodate V to iodine I (this is the trap as we usually think of Iodine as -1, but in ICl it is behaving like I+)

You need to keep the two oxidation equation coefficients equal so that n(Cu) = n(SCN)

The common factor for eliminating electrons from the oxidation and reduction equations is 28

Hope this helps...

Periodic Player

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Plz give the solution to balance the rxn

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any eqn in chemistry can be balanced by assuming coefficients as 1,a,b,c,d.....simply equate no of each atoms on both sides.

CuSCN + aKIO3 + bHCl === cCuSO4 + dKCl + eHCN + fICI + gH2O

u may get fractions but multiply by lcm of denominators.

But this doesnt mean i can do any Q from chem.... i am very bad at it just like oil with water(does not go together)

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Are all they in aqeuous state??

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2CuS- +3IO3- + 2H+======2CuSo4 + 3I- + H2O

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the question have 2 answer

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It is hard but I remember having such equation by the last year in the high school

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Hey there... aCuSCN+bKIO3+cHCl = dCuSO4 +eKCl+fHCN+gICl+ hH2O

I got a=d=f=1 b=7/4 c=7/2 e=b=g=7/4 H=5/4

Multiplying ever coefficient by 4 A=4 B=7 C=14 D=4 E=7 F=4 G=7 H=5 So balanced chemical equation=

4CuSCN+7KIO3+14HCl== 4CuSO4+7KCl+4HCN+7ICl+5H2O

Me of class 10:- Ayush Raj.. Is it right...??

Done by classic algebra

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1 HCl + KIO3 + CuSCN → H2O + KCl + CuSO4 + HCN + ICl 2 HCl + KIO3 + Cu(SCN) → H2O + KCl + CuSO4 + HCN + ICl

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2CuSCN+3KIO3+6HCl====)2CuSO4+3KCl+2HCN+3ICl+2H2O

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