Question

Dichromate ion react with ferrous ion to form chromium ion and ferric ion in acidic medium. Balance the equation by using oxidation number method.​

Answer 1

Answer:

Explanation:Write the skeletal equation of the given reaction

Cr2O2−7(aq)+Fe2+(aq)→Cr3+(aq)+Fe3+(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 Cr2O2−7 to +3 Cr3+ and that of Fe increases form + 2 in Fe2+→+3∈Fe3+ therefreo Cr2O2−7 gets reduced while Fe2+ gets oxidised Thus the above sketetal eqation (i) can be divided in to the following two half reaction equation

Oxidation half equation Fe2+(a)→Fe3+(q)

Reduction half equation : Cr2O2−7(aq)→Cr3+(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 Cr2O2−7 on the L.H.S of Eq (iii) and only on eon R.H.S therefoere multiple Cr3+ by 2 we have Cr2O2−7(aq)→2Cr3+(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 H2O 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 H2O 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

6Fe2+(aq)→6Fe3++6e−

Cr2O2−7(aq)+6Fe2+(aq)+14+(aq)→2Cr3+(aq)+6Fe3+(aq)+7H2O(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