2 Questions about an already solved problem. a) In the oxidation half-reaction,

Question

2 Questions about an already solved problem.

a) In the oxidation half-reaction, why do we not flip the sign of Ered (from +1.07 to -1.07) when in the table of the standard reaction potentials the reaction of Br is written as a REDUCTION reaction (Br2 + 2e- --> 2Br-) and had an Ered for a READUCTION half-reaction? We flip it to turn it into an oxidation reaction for our problem, but we keep the sign for the potential??

b) We find the concentrations of the ions using the inverse log and the pH, but where does the concentration of 0.1M of "other ionic compounds" come into play (it's a given in the problem)? And why was 5Br2 left out of the Nernst equation?? Thanks!

24. Consider the reaction below at 25 °C 2MnO4(aq) + 16H+ (aq) + 10Br(aq) 2Mn2+(aq) + 5Br2() + 8H20( Oxidation Half Reaction (anode): Reduction Half Reaction (cathode): n- 10 electrons Is the reaction spontaneous at a pH of 2.00 with all other ionic species at 0.100 M? 10B - 5Br2 + 10e. Ered = 1.07 V 10e. 16H, 2MnO,--) 2Mn2+ + 8H20 Ered = 1.51 V E = 1.51 V-1.07 V = 0.44 V cell 0.0592 E cell= E0 cell E cell= E0 cell [H+1 = 10-pH-10-2.00-0.010 M Ece,-(0.44 V Ece,-0.44 V-(0.00592)(log 1 x 1042)-0.19 V log Q 0.0592 Mn0H+16[Br 10 (0.100)2 (0.100)2(0.010)16 (0.100)10 0.0592 lo Ecell 0-spontaneous reaction

Explanation / Answer

The overall reaction is obtained by summing the above two equations which results in other ions.

the other ionic species are included in the expression shown Mn+2 and MnO4- and they are 0.12 and cancel out. This is an ionic equation

Br2 is liquid and the equation is ionic equation.

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