Question

# Consider the following redox reaction : S4O2−6(aq)+Al(s)→H2S(aq)+Al3+(aq) What will be the net balanced equation in an acidic medium (by oxidation number method).

A
2S4O26(aq)+6Al(s)+15H+2H2S(aq)+6Al3+(aq)+6H2O(l)
B
S4O26(aq)+6Al(s)+20H+4H2S(aq)+6Al3+(aq)+6H2O(l)
C
3S4O26(aq)+6Al(s)+2H+3H2S(aq)+6Al3+(aq)+6H2O(l)
D
S4O26(aq)+5Al(s)+20H+4H2S(aq)+5Al3+(aq)+4H2O(l)

Solution

## The correct option is B S4O2−6(aq)+6Al(s)+20H+→4H2S(aq)+6Al3+(aq)+6H2O(l)+2.5S4O2−6(aq)+0Al(s)→H2−2S(aq)++3Al3+(aq) S4O2−6 is oxidising agent Al is reducing agent. nf=(|O.S.Product−O.S.Reactant|×number of atom +2.5S4O2−6(aq)→H2−2S(aq)  oxidation nf=(|−2−2.5|×4=18 0Al(aq)→+3Al3+(aq)  reduction nf=(|3−0|×1=3  Ratio of nf  for oxidation to reduction is 6:1 Cross mutiply the oxidising and reducing agents with ratio of n-factors.  S4O2−6(aq)+6 Al(s)→H2S(aq)+Al3+(aq) Balancing elements except O and H S4O2−6(aq)+6Al(s)→4H2S(aq)+6Al3+(aq) Balance atoms oxygen by adding H2O. S4O2−6(aq)+6Al(s)→4H2S(aq)+6Al3+(aq)+6H2O(l) Balance hydrogen by adding H+ S4O2−6(aq)+6Al(s)+20H+→4H2S(aq)+6Al3+(aq)+6H2O(l) Balancing charge:   charge in reactant side = -2+20=+18 charge in product side = +18 so the balanced equation is  S4O2−6(aq)+6Al(s)+20H+→4H2S(aq)+6Al3+(aq)+6H2O(l)

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