Question

Can someone please explain to me in detail how to balance this unbalanced equation?: `I^(-)(s) + H_3AsO_4(aq) -> IO_3(aq) + AsO_3 (aq) +H_2O (l)`

Answer 1

You use the method of half equations,,,,and you must make a meal of them...I think you mean to reduce the arsenate to `As(OH)_3`

Explanation:

Iodide is OXIDIZED to iodate....`IO_3^(-)`, a 6 electron oxidation...

`I(-I)rarrI(+V):`

`I^(-) +3H_2O rarrIO_3^(-)+6H^+ +6e^(-)` `(i)`

Charge and mass are balanced so this is kosher...

Arsenate (`As(V+)`) is reduced to `As(III+)` in `As(OH)_3` a 2 electron reduction......this is not what you have got in your question...so I suggest you check it...

`As(V+)rarrAs(III+)`

`H_3AsO_4 +2H^+ +2e^(-) rarr As(OH)_3+H_2O` `(ii)`

We add `(i)` and `(ii)` in such a way that the electorns are eliminated...`(i)+3xx(ii)`

`3H_3AsO_4 +6H^+ +6e^(-)+I^(-) +3H_2O rarrIO_3^(-)+6H^+ +6e^(-)+3As(OH)_3+3H_2O`

And we cancel out the common reagents...

`3H_3AsO_4 +I^(-)rarrIO_3^(-) +3As(OH)_3`

Charge and mass are balanced, so this is kosher.

AS to a more full treatment of redox processes, I pasted this in from an earlier answer. We conceive of `"oxidation"` as the loss of electrons, and when an oxidation number, an imaginary number, or rather a conceptual number, goes up, this corresponds to the loss of electrons. And please note that this is an entirely CONCEPTUAL exercise....

Oxygen is the quintessential oxidant .... it is conceived to GAIN electrons to form oxide ions...as shown....

`1/2stackrel(0)O_2(g) + 2e^(-) rarr O^(2-)` `(i)`

...dioxygen gas is said to be `"REDUCED"`; it has gained electrons...and its oxidation number has been reduced from `0` to `-II`; the `-II` is the charge on the oxide anion; and again if mass and charge are not balanced, if stoichiometry is NOT preserved, then the reaction cannot be accepted as a representation of chemical reality.

In the given reduction equation, the which represents the reduction of dioxygen gas, the electrons are conceived to come from somewhere. Some other species MUST LOSE electrons, and this process is dubbed `"oxidation"`...

For example, we could write the oxidation of dihydrogen gas as shown....

`stackrel(0)H_2(g) rarr 2H^+ +2e^(-)` `(ii)`

We adds `(i)` and `(ii)` in such a way that the electrons are removed from the final equation.....i.e. `(i) + (ii)` gives...

`H_2+1/2O_2rarrunderbrace(2H^+ +O^(2-))_"one water molecule"` or....

`H_2(g)+1/2O_2(g) rarr H_2O(l)`

And this is a simple example of a redox equation. Confused yet? But all I have done is assign some oxidation numbers, and conserved mass and conserved charge....

Formal rules of assignment of oxidation numbers are given here....

`1.` `"The oxidation number of a free element is always 0."`

`2.` `"The oxidation number of a mono-atomic ion is equal"` `"to the charge of the ion."`

`3.` `"For a given bond, X-Y, the bond is split to give "X^+` `"and"` `Y^-`, `"where Y is more electronegative than X."`

`4.` `"The oxidation number of H is +1, but it is -1 in when"` `"combined with less electronegative elements."`

`5.` `"The oxidation number of O in its"` compounds `"is usually -2, but it is -1 in peroxides."`

`6.` `"The oxidation number of a Group 1 element"` `"in a compound is +1."`

`7.` `"The oxidation number of a Group 2 element in"` `"a compound is +2."`

`8.` `"The oxidation number of a Group 17 element in a binary compound is -1."`

`9.` `"The sum of the oxidation numbers of all of the atoms"` `"in a neutral compound is 0."`

`10.` `"The sum of the oxidation numbers in a polyatomic ion"` `"is equal to the charge of the ion."`

Confused yet? Note that these are for reference...and not for learning off by heart....

Answer 2

WARNING! Long answer! The balanced equation is
`"I"^"-" + "3H"_3"AsO"_4 + "9OH"^"-" → "IO"_3^"-" + "3AsO"_3^"3-" + "9H"_2"O"`

Explanation:

There are problems in your equation.

• `"I"^"-"` is not a solid,
• You have two oxidation half reactions but no reduction half-reactions.

I will assume that your intended equation is

`"I"^"-""(aq)" + "H"_3"AsO"_4"(aq)" → "IO"_3^"-""(aq)" + "AsO"_3^"3-""(aq)" +"H"_2"O(l)"`

If iodate and arsenite ions are the products, we must balance in basic solution.

We should use the ion-electron method because the hydroxide ions come in as part of the balancing procedure.

I find it easiest to balance the equation as if it were in acid solution and then convert it to basic medium.

Step 1. Write the skeleton equation

Omit `"H"^"+", "OH"^"-"`, and `"H"_2"O"`

`"I"^"-" + "H"_3"AsO"_4 → "IO"_3^"-"+ "AsO"_3^"3-"`

Step 2. Separate into two half-reactions.

`"I"^"-" → "IO"_3^"-"`
`"H"_3"AsO"_4 → "AsO"_3^"3-"`

Step 3. Balance all atoms other than `"H"` and `"O"`.

Done.

Step 4. Balance `"O"` by adding `"H"_2"O"` molecules to the deficient side.

`"I"^"-" + "3H"_2"O" → "IO"_3^"-"`
`"H"_3"AsO"_4 → "AsO"_3^"3-" + "H"_2"O"`

Step 5. Balance `"H"` by adding `"H"^"+"` ions to the deficient side.

`"I"^"-" + "3H"_2"O" → "IO"_3^"-" + "6H"^"+"`
`"H"_3"AsO"_4 → "AsO"_3^"3-" + "H"_2"O" + "H"^"+"`

Step 6. Balance charge by adding electrons to the deficient side.

`"I"^"-" + "3H"_2"O" → "IO"_3^"-" + "6H"^"+" + "6e"^"-"`
`"H"_3"AsO"_4 + "2e"^"-" → "AsO"_3^"3-" + "H"_2"O" + "H"^"+"`

Step 7. Multiply each half-reaction by a number to equalize the electrons transferred.

`1×["I"^"-" + "3H"_2"O" → "IO"_3^"-" + "6H"^"+" + "6e"^"+"]`
`3×["H"_3"AsO"_4 + "2e"^"-" → "AsO"_3^"3-" + "H"_2"O" + "H"^"+"]`

Step 8. Add the two half-reactions.

`"I"^"-" + color(red)(cancel(color(black)("3H"_2"O"))) → "IO"_3^"-" + "6H"^"+" + color(red)(cancel(color(black)("6e"^"+")))`
`ul("3H"_3"AsO"_4 + color(red)(cancel(color(black)("6e"^"-"))) → "3AsO"_3^"3-" + color(red)(cancel(color(black)("3H"_2"O"))) + "3H"^"+")`
`"I"^"-" + "3H"_3"AsO"_4 → "IO"_3^"-" + "3AsO"_3^"3-" + "9H"^"+"`

Step 9. Convert to basic solution by adding the appropriate multiples of

`"H"^"+" + "OH"^"-" → "H"_2"O"` or `"H"_2"O" → "H"^"+" + "OH"^"-"`.

We want to cancel `"9H"^"+"`, so we add nine of the first equation.

`"I"^"-" + "3H"_3"AsO"_4 → "IO"_3^"-" + "3AsO"_3^"3-" + color(red)(cancel(color(black)("9H"^"+")))`
`ul(color(red)(cancel(color(black)("9H"^"+"))) + "9OH"^"-" → "9H"_2"O"color(white)(mmmmmmmmmmmm))`
`color(blue)("I"^"-" + "3H"_3"AsO"_4 + "9OH"^"-" → "IO"_3^"-" + "3AsO"_3^"3-" + "9H"_2"O")`

Step 10. Check that all atoms are balanced.

`ulbb("Atom"color(white)(m)"On the left"color(white)(m)"On the right")`
`color(white)(m)"I"color(white)(mmmmmml)1color(white)(mmmmmmm)1`
`color(white)(m)"H"color(white)(mmmmml)18color(white)(mmmmmml)18`
`color(white)(m)"As"color(white)(mmmmmll)3color(white)(mmmmmmll)3`
`color(white)(m)"O"color(white)(mmmmmll)21color(white)(mmmmmm)21`

Step 11. Check that charge is balanced

`ulbb("On the left"color(white)(m)"On the right")`
`color(white)(mmm)"10-"color(white)(mmmmml)"10-"`

Everything checks. The equation is balanced!