w SG2216 Assignmer X c Quantitative Chem X CI2.2s2032-21- x G consider the follo
ID: 1059264 • Letter: W
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w SG2216 Assignmer X c Quantitative Chem X CI2.2s2032-21- x G consider the follo x C Chemist uestior X C chegg study IGui x CIOX SG2216 Ass x x C www.webassign.net/web/Student/Assignment-Responses submit?dep 14204282 O E 0.268 59 CaHRO Right half-cell: E 0.268 0.05916 log IC E- 0.05916 0.700 6 16 2. +12 E- 0.05916 0.700 CCHRO. 0.05916 0.268 CCI-12 E 0.700 0.05916 log( 1 0.05916 6h6 0.268 6 14 2 E 0.700 0.05916 log([Cl-12 E 0.268 0.05916 CaHRO (b) Ignoring activities, rearrange the Nernst equation for the net reaction to the form E(cell) A (B. pH), where A and B are constants. Calculate A and B at 25oC. (Assume that the Nernst potential is 0.05916 V (c) If the pH were 4.49, in which direction would electrons flow through the potentiometer? O Electrons would flow from left to right through the potentiometer. O Electrons would flow from right to left through the potentiometer. Supporting Materials Physical Constants Supplemental Periodic Table A 4 Ask me anything /21/2016Explanation / Answer
In the left hand electrode quinone is reduced to hydroquinone.Half cell reaction is as follows:
E = 0.700 - 0.059/2 log [C6H6O2/]/[C6H4O2][H+]
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Right half cell reaction is 2Hg(s) + 2Cl^- <==> Hg2Cl2 + 2e
E = 0.268 - 0.059/2 log 1/[Cl-]^2
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Overall raction : quinone + 2H+ + 2Hg(s) + 2Cl^- <==> hydroquinone + Hg2Cl2
E = E^0 - 0.059/2 log [hydroquinone][Hg2Cl2 ]/[quinone][Hg][Cl-]^2[H+]^2
[quinone] = [hydroquinone] and [Cl-] = 0.5 M
E = E^0 - 0.059/2 log 0.5/[H+]^2
E^0 = 0.700+ (-0.268) = 0.432 V
Substitute the value of E^0 . E =0.432 - 0.059/2 log 0.5/[H+]^2
= 0.44+0.059log[H+] = 0.44 - 0.059pH
A = 0.44
B = 0.059
electrons will flow from right to left
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