I didn\'t get an answer to this question from the previous expert, so I\'m posti

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I didn't get an answer to this question from the previous expert, so I'm posting this again: why did we not flip the sign for this half-reaction? In the table of standard potentials the reaction Br2 + 2e --> 2Br- has the standard reduction potential of 1.07. In our problem, we flip this half-reaction to turn it into an oxidation reaction. Why didn't we flip the sign for 1.07? Why does it stay positive?

24. Consider the reaction below at 25 °C 2MnO4(aq) + 16H+ (aq) + 10Br(aq) 2Mn2+(aq) + 5Br2() + 8H20( Is the reaction spontaneous at a pH of 2.00 with all other ionic species at 0.100 M? Oxidation Half Reaction (anode): 10B - 5Br2 + 10e. tion Half Reaction (cathode): 10e. 16H, 2MnO,--) 2Mn2+ + 8H20 Ered = 1.07 V educt Ered = 1.51 V n- 10 electrons 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 cell Ecell 0-spontaneous reaction

Explanation / Answer

If we use the following formula to calculate the Ecell:

Ecell = Ecathode – Eanode

Then no need to change the sign.

E cell = E reduction + E oxidation

Then we need to change the sign .

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