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How would you balance the following redox reaction in acidic medium?

$$\ce{KSCN + H2O + I2 -> KHSO4 + HI + ICN}$$

In this reaction, it seems the sulfur is being oxidized but iodine is undergoing both (disproportion?). The tricky part about the problem is the carbon. The carbon is going from a +4 to a +2 but still is not considered a redox atom. Why is this?

Oxidation half reaction:

$\ce{KSCN -> KHSO4}$ (sulfur being oxidized from -2 to +6)

$\ce{I2 -> ICN}$ (iodine has +1 charge)

Reduction half reaction:

$\ce{I2 -> HI}$ (iodine has -1 charge)

$\ce{KSCN -> ICN}$ (here the carbon is going from +4 to +2)

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    $\begingroup$ Did you check this explanation? periodni.com/… $\endgroup$
    – M. Farooq
    Jul 12 at 5:19
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    $\begingroup$ Can you give a source for the reaction itself? I mean, a source that shows that the reaction actually occurs when $\ce{KSCN}$ and $\ce{I2}$ are mixed? I'm not doubting you, just that this reaction is important for an exam of mine and a reliable source showing the same would be helpful :) Thank you. $\endgroup$
    – TRC
    Jul 14 at 9:59
  • $\begingroup$ I did find an ACS journal (its an old journal though) that talks more about the kinetics of this reaction. If you're an ACS member you should be able to view the entire journal. Hope this helps! Here's the link: :pubs.rsc.org/en/Content/ArticleLanding/TF/1935/… $\endgroup$ Jul 14 at 20:48
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I would solve this problem by algebra, without thinking about oxidation numbers. It is quicker. Let the coefficients be $a, b, c, d, e, f$ in front of the reactants and products. It yields $$\ce{a KSCN + b H2O + c I2 -> d KHSO4 + e HI + f ICN}$$ The number of each atom must be the same on both sides of the equation arrow. Counting atoms by order of appearance, it is easy to see that, for potassium and sulfur : $a = d$. Counting $\ce{C}$ and $\ce{N}$ atoms yields also $a = f$. Now let's admit that $a = 1$, then $d = f = 1$.

Let's consider the remaining $\ce{H, O, I}$ atoms.

Counting $\ce{O}$ atoms yields : $b = 4d = 4$.

Counting $\ce{H}$ yields : $d + e = 2b = 8$. So $b = 4$ and : $e = 8 - 1 = 7$

Counting $\ce{I}$ atoms yields : $2 c = e + f = 7 + 1 = 8$. So : $c = 4$

So the final equation is $$\ce{ KSCN + 4 H2O + 4 I2 -> KHSO4 + 7 HI + ICN}$$

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Use molecular species in your half-reactions.

In your case, identify thiocyanate as the reducing agent and, since the cyano function combines with iodine, include enough iodine in that half-reaction to form the iodine cyanide/cyanogen iodide product. This gives

$\ce{SCN^- + (1/2)I2 -> S^{(+6)} + ICN +7 e^-}$

where the +6 oxidation state on sulfur in the bisulfate is rendered equivalent to six positive charges; thus seven electrons on the right.

If we include the hydrogen and oxygen atoms so that the bisulfate ion is a complete species, we would have

$\ce{SCN^- + (1/2)I2 -> HSO4^- + ICN +7 e^-}$ (hydrogen and oxygen unbalanced)

where we still have seven electrons based on the oxidation state differential noted previously. Adding hydrogen ions on the right to balance charges and water on the left to balance hydrogen or oxygen atoms then gives this form for the thiocyanate oxidation:

$\ce{SCN^- + (1/2)I2 +4H2O -> HSO4^- + ICN + 7 H^+ + 7 e^-}$

Combine either thiocyanate oxidation half-reaction with iodine reduction:

$\ce{(1/2)I2 +e^- -> I^-}$

Balancing electrons in the usual way then reveals that the redox reactant ratio is $\ce{(1)SCN^-:4I2}$. So leave the thiocyanate coefficient at $1$ and put a coefficient of $4$ before $\ce{I2}$, then use atom balances to handle the remaining elements. (If you used the second balanced equation given for thiocyanate oxidation, you have to add just the potassium ions.)

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