Dichromate ion in aqueous acidic medium reacts with ferrous ion give ferric and chormium ions write th balanced chemical equation corresponding to the reaction

Explanation

Write the skeletal equation of the given reaction
`Cr_(2)O_(7)^(2-)(aq)+Fe^(2+)(aq)rarrCr^(3+)(aq)+Fe^(3+)(aq)`
Step 2 Write the O.N of all the elements above their respective symbols

Step 3 Find out hte specied whihc hve been oxidised nad reduced and spilt the given skeltal equation in to two half reactions
Since the O.N of Cr decreases form + 6 `Cr_(2)O_(7)^(2-)` to +3 `Cr^(3+)` and that of Fe increases form + 2 in `Fe^(2+) to +3 in Fe^(3+)` therefreo `Cr_(2)O_(7)^(2-)` gets reduced while `Fe^(2+)` gets oxidised Thus the above sketetal eqation (i) can be divided in to the following two half reaction equation
Oxidation half equation `Fe^(2)+(a) rarr Fe^(3+) (q)`
Reduction half equation : `Cr_(2)O_(7)^(2-)(aq) rarr Cr^(3+)(aq)`
Step 4 To balance oxidaiton half equation (ii)
(a) Balacne all atoms other than O and H
Not neeeded because Fe by adding electrons
Step 5 To balance the reducation half eqaution (iii) (a)Balance all atoms other than H and O
Since there ar two Cr atoms in `Cr_(2)O_(7)^(2-)` on the L.H.S of Eq (iii) and only on eon R.H.S therefoere multiple `Cr^(3+)` by 2 we have `Cr_(2)O_(7)^(2-)(aq) rarr 2 Cr^(3+)(aq)`
(b)Balacne the O.N by adding elctrons
Balance charge by adding `H^(+)` ions sice hte reaction occurs in the acidic medium
The total charge on L.H.S of Eq (vi) is -8 while on the R.H.S it is +6 Therefore add 14 `H^(+)` to L.H.S of Eq (vi) we have
(d) Balance O atoms by adding `H_(2)O` molecules since there are seven O atoms on the L.H.S of Eq (vii) but no O atom on the R.H.S therefore and 7 `H_(2)O` to the R.H.S of Eq (vii) we have
the H atoms get automatically balanced
Thus Eq (viii) represent the balaced reduction half equation
Step 6 To balance the electrons lost in Eq (vi) and gained in Eq (viii) multiply Eq (vi) by 6 and add to Eq (viii) We have
`6Fe^(2+)(aq)rarr6Fe^(3+)+6 e^(-)`
`Cr_(2)O_(7)^(2-)(aq)+6 Fe^(2+)(aq)+14^(+)(aq) rarr 2 Cr^(3+)(aq)+6 Fe^(3+)(aq)+7 H_(2)O(l)`
This gives the final balanced redox equation
Step 7 Verification Total charge on L.H.S of Eq (ix) =-2+6(+2)+14(+1)=+24
Total charge on R.H.S of Eq (ix) =2(+3)+6(+3)=+24
Since the magnituede if charge on both sides of eq (ix) equal therefore eq (ix) represent the correct balaced redox equation