Chromium metal in basic medium is oxidised in air to give chromic tetrahydroxide anion.

Explanation

a) This ionic skeleton equation is written as
`Cr+O_(2)overset(OH^(-))rarrCr(OH)_(4)^(-)`
b) Writing oxidation numbers
`overset(0)(Cr)+overset(0)(O_(2))rarr[overset(+3)(Cr)(overset-2)(O)overset(+1)(H))_(4)]^(-)`
c) Locating atoms undergoing change in oxidation numbers
`overset(0)(Cr)+overset(0)(O_(2))rarr[overset(+3)(Cr)(overset(-2)(O)H)_(4)]^(-)`
d) Dividing the reaction into two halves and balancing in acidic medium. separately
Oxidation half - reaction : `Crrarr[Cr(OH)_(4)]^(-)`
Step 1 : Balance oxygen atoms
`Cr+4H_(2)Orarr[Cr(OH)_(4)]^(-)`
Step 2 : Balance hydrogen atoms
`Cr+4H_(2)O+4OH^(-)rarr[Cr(OH)_(4)]^(-)+4H_(2)O`
Step 3 : Balance the charge
`Cr+4H_(2)O+4OH^(-)rarr[Cr(OH)_(4)]^(-)+4H_(2)O+3e^(-)" ....(a)"`
Reduction half - reaction : `O_(2)rarr[Cr(OH)_(4)]^(-)`
Step 1 : Balance chromium atoms
`O_(2)+Crrarr [Cr(OH)_(4)]^(-)`
Step 2 : Balance oxygen atoms
`O_(2)+Cr+2H_(2)Orarr[Cr(OH)_(4)]^(-)`
Step 3 : Balance hydrogen atoms
`O_(2)+Cr+2H_(2)rarr[Cr(OH)_(4)]^(-)`
Step 4 : Balance charnge
`O_(2)+Cr+2H_(2)O+e^(-)rarr[Cr(OH)_(4)]^(-)" ....(b)"`
Step 5 : Equalising the electrons and adding the two half reactions
`"eq (a)"xx1+"eq (b)"xx"3, we get"`
`{:(Cr+4H_(2)O+4OH^(-)rarr),(" "[Cr(OH)_(4)]^(-)+4H_(2)O+3e^(-)),(Cr+4H_(2)O+4OH^(-)rarr),(" "[Cr(OH)_(4)]^(-)+4H_(2)O+3e^(-)),(3O_(2)+3Cr+6H_(2)O+3e^(-)rarr2[Cr(OH)_(4)]^(-)),(bar(4Cr+3O_(2)+6H_(2)O+4HO^(-)rarr4[Cr(OH)_(4)]^(-))):}`
This is the balnced equation.