Question

TIL ELECTROCHEMISTRY Calculating standard reaction free energy from standard... Using standard reduction potentials from the ALEKS Data tab, calculate the standard reaction free energy AGº for the following redox reaction. Round your answer to 3 significant digits. 2MnO, (5) +80H" (aq) + 3Br, (1) - 2 Mno, (aq) +4H20 (1) +6Br" (ag) E х ?

Answer 1

First we calculate the standard electrode potential of the cell. On combining and balancing the charges we get that in this redox reaction 6 electrons are being exchanged.

Also we have to use Latimar diagram to calculate the the standard reduction potential for the anode half cell that is E(Mn+7/Mn+4).

delta G = - n Ecell F

F= 96.5 kJ per mol.

Put in equation : 2 MnO (1) + Bor (as) + 3 Boy (1) 2 MnO (21) + 4Holes + 685 () = 0.8o v gr is a Redox Reaction Cell = (-0.29) + 1.01 E cell = +0. Bov +4 Muty oxidation of Reduction of (4 to +1) (o to -1) D F - n n{call oxidation Half Cell AGO = - 6x0.80 V X F Мае тайр BIXED 12 F = Faraday constant F = 96.5 KJ molt MnO, +2HO Myog + 4H+ + Reduction Half cell aço = - 6x0.80x96.5 KJ Malt 463.2 KJ/mol 6 Br 3 Brz. tbe → Multiply By 2 in epuer'oy and add we Have used To Celalete it2 BO Diagrem. mnt/ mit (3 My az +240 + 6 Bry MnO4 + 4H+ + 12 Br Letimes 42 = 1.5v nabel bé exchanged during MM mity 12 a My oxidation & Reduction process +4 rest Myt2 o b E cell 6° mita/m2 = 1.23v • Reduction potential at Cethode (Reduction porential at anode 6 no we E cell we get mythmuta o Ma / Mat2 Byg/ Br +1.07 mith 6° Cell mutl/rente mutt tº: (115-1.23) = 0.217 2 Mno My My Oz Mytz MOT bent om te entrate you = (1.5-1.23)v Per ty mata. 0:24V CS Scanned with CamScanner