Question

Balance: H2C2O4(aq) +K2CrO4(aq) + HCl(aq) →CrCl3(aq) + KCl(aq) + H2O(l) +CO2(g)?

Just by oxidation method or ion-electron method.

Answer 1

Well oxalic acid is oxidized to carbon dioxide....and chromate ion is reduced to chromic ion....

Explanation:

`"Oxidation:"`

`HO(O=)stackrel(""^+III)C-C(=O)OH rarr 2stackrel(""^+IV)CO_2+2H^+ + 2e^(-)`

Charge and mass are balanced as required....

`"Reduction:"`

`stackrel(""^+VI)CrO_4^(2-) +8H^+ +3e^(-)rarr Cr^(3+)+4H_2O`

And so we take THREE of the former, and TWO of the latter to eliminate the electrons as virtual particles of convenience....

`3HO(O=)C-C(=O)OH+2CrO_4^(2-) +16H^+ +6e^(-) rarr 2Cr^(3+)+8H_2O+6CO_2+6H^+ + 6e^(-)`

...and cancel away....

`"3HO(O=)CC(=O)OH"+2"CrO"_4^(2-)"+10H"^+ rarr "2Cr"^(3+)"+8H"_2"O+6CO"_2`

The which is balanced with respect to mass and charge, as is absolutely required....

Answer 2

WARNING! Long answer! The balanced equation is

`"3H"_2"C"_2"O"_4 + "2K"_2"CrO"_4 + "10HCl" → "2CrCl"_3 + "4KCl" + "8H"_2"O" + "6CO"_2`

Explanation:

The unbalanced equation is

`"H"_2"C"_2"O"_4 + "K"_2"CrO"_4 + "HCl" → "CrCl"_3 + "KCl" + "H"_2"O" + "CO"_2`

Step 1. Identify the atoms that change oxidation number

Start by determining the oxidation numbers of every atom in the equation.

`"H"_2stackrelcolor(blue)("+3")("C")_2"O"_4 + "K"_2stackrelcolor(blue)("+6")("Cr")"O"_4 + "HCl" → stackrelcolor(blue)("+3")("Cr")"Cl"_3 + "KCl" + "H"_2"O" + stackrelcolor(blue)("+4")("C")"O"_2`

We see that the oxidation number of `"C"` increases from +3 to +4 and the oxidation number of `"Cr"` decreases from +6 to +3.

The changes in oxidation number are:

`"C:"color(white)(ll) "+3 → +4"; "Change ="color(white)(l) "+1 (oxidation)"`
`"Cr: +6 → +3"; "Change ="color(white)(ll) "-3 (reduction)"`

Step 2. Equalize the changes in oxidation number

We need 3 atoms of `"C"` for every 1 atom of `"Cr"` or 6 atoms of `"C"` for every 2 atoms of `"Cr"`.

Step 3. Insert coefficients to get these numbers

`color(red)(3)"H"_2"C"_2"O"_4 + color(red)(2)"K"_2"CrO"_4 + "HCl" → color(red)(2)"CrCl"_3 + "KCl" + "H"_2"O" + color(red)(6)"CO"_2`

Step 4. Balance `"K"`

`color(red)(3)"H"_2"C"_2"O"_4 + color(red)(2)"K"_2"CrO"_4 + "HCl" → color(red)(2)"CrCl"_3 + color(blue)(4)"KCl" + "H"_2"O" + color(red)(6)"CO"_2`

Step 5. Balance `"Cl"`

`color(red)(3)"H"_2"C"_2"O"_4 + color(red)(2)"K"_2"CrO"_4 + color(purple)(10)"HCl" → color(red)(2)"CrCl"_3 + color(blue)(4)"KCl" + "H"_2"O" + color(red)(6)"CO"_2`

Step 6. Balance `"O"`

Add `"H"_2"O"` molecules to the deficient side.

`color(red)(3)"H"_2"C"_2"O"_4 + color(red)(2)"K"_2"CrO"_4 + color(purple)(10)"HCl" → color(red)(2)"CrCl"_3 + color(blue)(4)"KCl" + color(green)(8)"H"_2"O" + color(red)(6)"CO"_2`

Every formula now has a coefficient. The equation should be balanced.

Step 6. Check that all atoms are balanced.

`ulbb("Atom"color(white)(m)"On the left"color(white)(m) "On the right")`
`color(white)(ml)"H"color(white)(mmmmll)16color(white)(mmmmmml) 16`
`color(white)(ml)"C"color(white)(mmmmml)6color(white)(mmmmmmll) 6`
`color(white)(ml)"O"color(white)(mmmmll)20color(white)(mmmmmml) 20`
`color(white)(ml)"K"color(white)(mmmmml)4color(white)(mmmmmmll) 4`
`color(white)(ml)"Cr"color(white)(mmmmm)2color(white)(mmmmmmll) 2`
`color(white)(ml)"Cl"color(white)(mmmmll)10color(white)(mmmmmm) 10`

The balanced equation is

`color(blue)("3H"_2"C"_2"O"_4 + "2K"_2"CrO"_4 + "10HCl" → "2CrCl"_3 + "4KCl" + "8H"_2"O" + "6CO"_2)`