Looking for an experience chem expert to answer the follow question. I\'m having
ID: 882581 • Letter: L
Question
Looking for an experience chem expert to answer the follow question. I'm having difficulty with it so do explain the process. Use the chart provided to help answer the question.
Balence the following equation in acid solution and determine if it is spontaneous when all ion concentrations are 0.010 M
BrO3^1-(aq) + Sn^+2(aq) ---> Br2(l) + Sn^+4(aq)
* = degrees
Cl2 + 2e^- ---> 2Cl^-
MnO4^- + 8H^+ +5e^- ---> Mn^+2 + 4H2O
Half-Reaction E* (volts) Half-Reaction E* (volts) Ag^+ + 1e^- ---> +0.799 Hg2^+2 + 2e^- ---> 2Hg +0.789 Al^+3 + 3e^- ---> Al -1.66 2Hg^+2 +2e^- ---> 2Hg2^+2 +0.920 H3AsO4 + 2H^+ +2e^- ---> H3AsO3 + H2O +0.559 Hg^+2 + 2e^- ---> Hg +0.584 Ba^+2 + 2e^- ---> Ba -2.90 I2 + 2e^- ---> 2I^- +0.536 Br2 + 2e^- ---> 2Br^- +1.065 IO3^- + 6H^+1 + 5 e^- ---> 1/2 I2 + 3 H2O +1.195 2BrO3^- + 12H^+1 + 10e^- --->Br2 + 6H2O +1.48 Mg^+2 +2e^- ---> Mg -2.37 Ca^+2 + 2e^- ---> Ca -2.87 Mn^+2 + 2e^- ---> Mn -1.18Cl2 + 2e^- ---> 2Cl^-
+1.359 MnO2 + 4 H^+ + 2e^- ---> Mn^+2 +2H2O +1.23 ClO- + H2O + 2e^- ---> Cl^- + 2OH^-1 +0.89MnO4^- + 8H^+ +5e^- ---> Mn^+2 + 4H2O
+1.51 ClO3^- + 6H^+1 +5e^- ---> 1/2Cl2 +3H2O +1.47 MnO4^- + 2H2O +3e^- ---> MnO2 + 4OH^- +0.59 Co^+2 +2e^- ---> Co -0.277 HNO2 + H^+ + 1e^- ---> NO + H2O +1.00 Co^+3 + 1e^- ---> Co^+2 +1.842 NO3^- + 4H^+ + 3e^- ---> NO + 2H2O +0.96 Cr^+3 + 3e^- ---> Cr -0.74 Ni^+2 + 2e^- ---> Ni -0.28 Cr^+3 + 1e^- ---> Cr^+2 -0.41 O2 + 4H^+ +3e^- ---> 2H2O @ pH 0 +1.23 Cr2O7^-2 +14H^+1 +6e^- ---> 2Cr^+3 + 7H2O +1.33 O2 + 2H2O + 4e^- ---> 4 OH^- @ pH 7 +0.40 Cu^+ + 2e^- ---> Cu +0.337 O2 + 2H^+ +2e^- ---> H2O2 +0.68 Cu^+2 + 1e^- ---> Cu^+1 +0.153 Pb^+2 + 2e^- ---> Pb -0.126 Cu^+1 + 1e^- ---> Cu +0.521 PbO2 + HSO4^- + 3H^+ +2e^- ---> PbSO4 + 2H2O +1.685 F2 + 2e^- ---> 2F^-1 +2.87 PbSO4 + H^+ +2e^- ---> Pb + HSO4^- -0.356 Fe^+2 + 2e^- ---> Fe -0.440 H2SO2 + 4H^+ +4e^- ---> S + 3 H2O +0.45 Fe^+3 +1e^- ---> Fe^+2 +0.771 HSO4^-1 + 3H^+ + 2e^- ---> H2SO3 + H2O +0.17 2H^+ +2e^- ---> H2 @ pH 0 0.000 Sn^+2 + 2e^- ---> Sn -0.136 2H2O + 2e^- ---> H2 + 2 OH^- @pH 7 -0.83 Sn^+4 + 2e^- ---> Sn^+2 +0.154 HO2^- + H2O + 2e^- ---> 3OH^- @pH 7 +0.88 VO2^+ + 2H^+ +1e^- ---> VO^+2 + H2O +1.00 H2O2 + 2H^+ 2e^- ---> 2H2O @ pH 0 +1.776 Zn^+2 + 2e^- ---> Zn -0.763Explanation / Answer
The reaction is
BrO3-(aq) + Sn+2(aq) ---> Br2(l) + Sn+4(aq)
Reducion Half cell : ( cathode)
2BrO3- + 12H+ + 10e --> Br2 + 6H2O
Oxidation half cell (anode)
Sn+2 --> Sn+4 + 2e ] X 5
Overall reaction = 2BrO3- + 12H+ + 5Sn+2 --> 5Sn+4 + Br2 + 6H2O ( balanced)
E0cell = E0cathode - E0anode
From table E0cathode = 1.48
E0anode = 0.154
So E0cell = 1.48- 0.154 = 1.326 V
Ecell = E0cell - 0.0592 / 10 log Q
Q = [Sn+4]5 / [Sn+2]5 [H+]12 [BrO3-]2
Q = [0.01]5 / [0.01]5 [0.01]12 [0.01]2
Q = (0.01)-14 = 10^-28
So Ecell = 1.326 - 0.00592 log 10^-28 = 1.326 - 0.165 = 1.161
So the reaction will be spontaneous as Ecell is positive
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.