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)#
2 Answers
You use the method of half equations,,,,and you must make a meal of them...I think you mean to reduce the arsenate to
Explanation:
Iodide is OXIDIZED to iodate....
Charge and mass are balanced so this is kosher...
Arsenate (
We add
And we cancel out the common reagents...
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
Oxygen is the quintessential oxidant .... it is conceived to GAIN electrons to form oxide ions...as shown....
...dioxygen gas is said to be
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
For example, we could write the oxidation of dihydrogen gas as shown....
We adds
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....
Confused yet? Note that these are for reference...and not for learning off by heart....
WARNING! Long answer! The balanced equation is
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
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
Step 2. Separate into two half-reactions.
Step 3. Balance all atoms other than
Done.
Step 4. Balance
Step 5. Balance
Step 6. Balance charge by adding electrons to the deficient side.
Step 7. Multiply each half-reaction by a number to equalize the electrons transferred.
Step 8. Add the two half-reactions.
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
Step 10. Check that all atoms are balanced.
Step 11. Check that charge is balanced
Everything checks. The equation is balanced!