this is one of the hardest redox equations.

the problem is:

CuSCN + KIO3 + HCl === CuSO4 + KCl + HCN + ICI + H2O

Acid Or Base Does Not Matter.

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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

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

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

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Are all they in aqeuous state??

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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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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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