Prob set 5[1]

# Prob set 5[1] - Chemical Fate and Transport Problem Set#5...

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Unformatted text preview: Chemical Fate and Transport Problem Set #5 Solution Set (17) (a) In writing a half-reaction, the following steps should he performed: 0) Write the species of interest. HSe03' ~ Seof' (ii) Determine the oxidation state of the element of interest. For HSe03‘, oxygen is assumed to be in the (—11) oxidation state and hydrogen is in the (+1) oxidation state. Let the oxidation state of Se in HSeOa' be x. Then, (+1) + (x) ,+ (-II - 3) = ~1 x = om For SeOf‘, the oxidation state of Se is: x+(-II-4) é —2 . _- x=(+VI) (iii) Write the number of electrons lost or gained from one species to the other species, according to the oxidation states determined in the previous step. Two electrons are lost in the conversion from Se(+IV) to Se(+VI): HSe03' .a Seof' + 2.2- (iv) Add H10 to the side of the equation which loses electrons. The number of moles of H20 added must balance the number of oxygen molecules. Here, the number of oxygen molecules on the right side of the equation is 4. Therefore, 1 mol of [-120 must be added to the left side of the equation to balance the moles of oxygen: Hseo3‘ + H20 a Se042' + 242' . (v) Add H+ on the side of the equation which gains electrons. Determine the number of moles of H+ to be added by balancing. themoles of hydrogen on either side of the equation. ' Here, the left-hand side of the equation shows 3H“. Therefore, 3I-I+ is needed on the right-hand side: H5203" + H20 « Seof‘ + 2e' + 3H’ 1 in (Note that steps (iv) and (v) may be done in reverse order if desired.) . (vi) By convention, the half-reaction is expressed in terms of the transfer of one electron. Therefore, divide the above reaction by the number of electrons. l - 1 1 2- - 3 t — —-H0-—SeO +e+——H 2 H5603 + 2 2 2 “ 2 This is an oxidation reaction because H5603“ loses electrons. (b) The given reaction can be written as follows: i {steﬁ {123.9203}E {H*}3 {e‘}3 = 10l33 Given a pH of 7 and equal concentrations of { HZSeos} and {HZSe}: {6-}3 = 10-l8.3/{IO—7}3 {er = to” o {e-} = 100.9 —log{e'} = 40.97 = pe If the E11 were —0. IV, the pe, estimated from Eq. [2-59], would be: Eh pa = —— = «1.59 0.059 At 3 PH 0f 7, neglecting activity coefﬁcient corrections, the ratio of HZSeO3 to HZSe would be: _] _ [1:35.203]2 _ 10-183 - —’ _ {10‘7}3 {101-6531 = M x 10 3 ‘ .[HgSeF [#25903] = L7 x 104 [Hz-Se] Prob\em a*\% 30m mlmomacknksm 00m w 00m) cu cm ongong w! Om 0L3 O~ CMW smug (402“ «ﬁe ~—-=) LAOL oat—globw kg; +11: +1: “0' COMPEL W rm “4 Jaw: Ox'xck. 0’6 msomic maxim- LLJQ Much: ‘qu C01 c9 vm +ef :2 Va owe * 1C: “L0 K: [Cy—bl. 060; ‘(LT - “L?— 1 —— .oomox 139:} 10.55%: JIM IO Moi L ~ '-"- M03 keg-:13. moa‘ " I 0- -— \LC‘c—R iuolu ﬁght-Oi. A6 “ “3‘17- m‘ ' __ o__ ... @610 * who -"-> "$02. *‘W/‘re/ 56 “’ “‘03 a "’ guofwhcm$9.5m.» liULO-LHL‘COL 1.3.6 = was a. \ -’r\(\¥- “‘0 L05; Mk know whoa: ﬂu: *‘ﬁj‘C‘L’Q (Luzl 50 W mgr CQ’QC'AG)’ nggpm 569’s) Mu!) man MKS. v; MPEM ’(o Emu mmw VQOM % 50d“ 0mm MOUSWLQ reéox 394%)oavxcx. ' (b) (34) = 1020 [H T [6"] (a) At a pe of? and a pH of 4.5: [Pu 4i] [P1103] = 1020 - (10-45)“ - 10*7 = 10'5 Therefore, 13110;” is the dominant plutonium species. Under sulfidic conditions, p6 is approximately —4, from Fig. 2—21. Therefore: [PM 4‘] [Puog] = 1020 - (1045)“ 104 = 106 Pu4+ is the dominant plutonium species. From Fig. 2-21, aerated water has a p6 ofcapproximately 14. Therefore: 4+ [Pu ] L. 1020 _ (10-4.5)4 . 10-14 10-12 ll PuOZ+ is the dominant plutonium species. ...
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