Write the balanced ½ reactions and calculate E°cell for the reaction of the foll

Question

Write the balanced ½ reactions and calculate E°cell for the reaction of the following species; is this reaction spontaneous? 2 Bi(s) +3 O3(g) 2 H+ >2 BiO*(aq)3 02(g)+ H20 A Reduction: O3(g) + 2 H+(aq) + 6?? o2(g) + H20(1) Ecell-2.076 V Oxidation: Bi(s)H2Bio (ag) 2 H (a) +6e Ecel 0.320 v Ecell- 2.396 V; nonspontaneous Ecell = 2.076 V Ecell = 0.320 V B. Reduction: 3 Oz(a)6Ht(aa) +6e- >3 O2(a)+3 H20U) Oxidation: 2 Bi(s) + 2 H2O(l) ? 2 Bio+(aq) + 4 H+(aq) + 6 e- Ecell = 2.396 V; nonspontaneous Ecell 2.076 v Ecell 0.320 v C. Reduction: O3(9) +2 H(a)6O2g)+ H20 Oxidation: Bi(s) + H2O(l) ? BiO+(aq) + 2 H+(aq) + 6 e- Ecell 1.756 V; spontaneous D Reduction: 03(g) 2 H (aq)6eO2(g)+H200) Ecell2.076 V Oxidation: Bi(s)H2Bio (ag) 2 H (a) +6e Ecel 0.320 v Ecell- 2.396 V; spontaneous Ecell 2.076 v Ecell = 0.320 V E. Reduction: 3 Oz(a)6Ht(aa) +6e- >3 O2(a)+3 H20U) Oxidation: 2 Bi(s) + 2 H2O(l) ? 2 Bio+(aq) + 4 H+(aq) + 6 e- Ecell = 2.396 V; spontaneous Eceli = 2.076 V cell0.320 V F Reduction: 3 03(g)+6H(aq)+6e>3 02(g)3 H200) Oxidation: 2 Bi(s)2 H200) 2 Bio*(aq) + 4 H*(aq)+6 e" Egel,-1.756 V; nonspontaneous G. Reduction: O3(g) + 2 H+(aq) + 6e-? O2(g) + H20(1) Ecell = 2.076 V Oxidation: Bi(s) + H20(I) ? BiO+(aq) + 2 H+(aq) + 6 e- Ece,--0.320 V Ecell 1.756 V; nonspontaneous Reduction: 3 03(g)+ 6H(aq)6e3 O2(g) 3 H20(l) Oxidation: 2 Bi(s)2 H200) 2 Bio (aq) +4 H*(aq)+6 e" Ecell 1.756 V; spontaneous Ecell 2.076 v Ecell0.320 V

Explanation / Answer

Here Bi undergoes oxidation by losing 3 electron

So, half reaction at anode:

2Bi(s) + 2 H2O(l) - - - - - >2BiO+(aq) + 4H+(aq) + 6e-

And, O3 undergoes reduction, so half reaction at cathode

3O3(g) + 2H+(aq) + 6e- - - - > 3O2(g) + 3H2O(l)

Since, E° cell = E° red E°oxd

=( 2.076 + (-0.320) )V

= 1.756 V

Since E°cell is positive, it is a spontaneous reaction..

So, option H is the correct choice.

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