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Dichromate ion in aqueous acidic medium reacts with ferrous ion give ferric and chormium ions write th balanced chemical equation corresponding to the reaction |
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Answer» Solution :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 speciedwhihc 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 thanH 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)+6Fe^(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 |
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