Question

Exercise 19.106 Part A The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt(s) H2 (9,1 atm) H+(aq, 1.0 M) Au'+(aq, ?M) Au(s). What is the concentration of Au”+ in the solution if Ecell is 1.24 V? Express your answer using two significant figures. O AIO O O ? [Aud+] = Submit Previous Answers Request Answer X Incorrect; Try Again

Answer 1

. . Forov anode reaction Coxidation ) Ha 199 - 2H+ + 21 Cathode peaction ( Recluction.) 2 Au²t can) + 3e Aucs). . E pel = pour The above E oxidation and to reduction value. are the stand andest values at 25°c. 1. so full of oxidation + Sponducton = 0 + loua Emi = 4 lounni veriall Beaction ( By multiplying and de reaction with 3. and multiplying cathode Reaction by a and add then such that no te is there in over all equation ) ( Hz Cy) - 2H+ + ze ) 3. ( Aust tzé Rucs) :) . x2 .: 34,69) + 2A438 — 644 +60- + 2auls 342 (9) + 2A u*Cam) — 6H tas ir zau (s) 76€ Q = [auiss] ? (1476 . [ Au?+]?1 PH2) - as it is in solid form so it activity is estaw always 1, PH₂ = 1 atm Auls) E

and (4t] = loom pa man rautoga 2- - using NERST equation. Whend Eau = Eau - Rt In Q where R = 8.314 5/mobile. T = 2a8k. Or0591 F = 965ooc. So buy alay - loy Q . ( By putting all values and Changing enQ to log Q. n=6. 10240.21.45 va 0.059!log [Au 34 ) 10 24V = 1045 V + 0-0 halaman raut] 6 x 280g [Au 3+] -0-21 X 3. O-0591 4 n = [05 (Au34). -10-659 = log (434) s [A434] = 10 Tongapo Trau / pola 710 /m} [Aust) = 10- 10.65.98 - 2.188.410-19 M Aut] =..2-2x10-1m (in two significant degits) .2.188 is pounded of to two signi freemt digits answers i n pule ton round of het a numbers ao be so : 0 it cy : then b. btl it can then b b 1 if can then b s btl (if b is odd) bob (if bis even)