Question

Please show all work step by step and final answer.

Reference Electrodes The Eº value for the following reaction is 0.446 V relative to the standard hydrogen electrode (SHE): Ag2 CrO4(s) + 24 + 2Ag(s) + Cro42 (aq) A chemist wishes to determine the concentration of Cro42- electrochemically. A cell is constructed consisting of a saturated calomel electrode (SCE; which has a reduction potential of 0.242 V relative to the SHE) and a silver wire coated with Ag Cro4 and suspended in a Cro42 solution. What is the potential of this cell at 25°C when [Cro42-) = 1.00 M? 1 pts Submit Answer Tries 0/5 Calculate AG for the full-cell reaction at 25°C when [Cro42-) = 1.00 M? 1 pts Submit Answer Tries 0/5 What is the potential of this cell at 25°C when [Cro42-) = 3.40x10-4 M? 1pts Submit Answer Tries 0/5 For a solution of unknown (Cro42'), the measured potential for the cell at 25°C is 0.528 V. What is [Cro42"] (in mol/L)? Opts Submit Answer Tries 0/5

Answer 1

Agacooglesy + ze aagis & crop lag, E = 0.440 volt Standard Reduction potential of SCE wort S.HE = 0,242 volt cell Erell - standard Reduction potential of = (0-446 -0.242) v 0.204V The expression E = E - cell for Nennst Equation is shown belan 2.303RT log/ (@Ag ² x [cros nF dl o Ag, crosis) [crop-] = Concentration of cruf = 1.00M - Activity of pure solids and liquide is considered to be unity. R = 8.314]/ kmole I=2988 - f = 965000/mole n =No of è lost gaised Esel = B'cele - C-059 log (lk Icom = 0.204 volt - 0.059 xo 1 E cell = 0.204 volt Ecell = cell potential. I AG = AGO +Riing AG = Gibbs force Energy change for AGO = stand and Gibbs free Energy = Reaction Quotient Reaction change 8 co e ta Agtste * [com er Ag, croels)

AGO =-n FE cell = -a molex 96500c/molex AGO - - 39.327KI 0.204 v Answer

Given Concentration of crot = crop - = 3-40x10 AM a gesi (cro Eccp = E cell - 0.059 log ang croptst . 0.3.04wd+ - 0.058 log ** 3.40x104) E cell = 0. 363 volt Ecell = 0. 3063 volt Answer Ecell = cell potential D Given - Ecell = 0.528 volt, cell Erell potential =0-24 volt Fase Fixe -oooss log ages, & [cros? - agus a greater 0.5.28v = 0.204 volt -0.059, -0.059 los l 11 _C04 [crog - (0.528 - 0-204] V = -0.059 log[croott) Leg [cro?] = -10.983_ Taking antilog both sides we get [cr 0,2-) = 10^-10.983 (cro2-] = 1004x10lm Ansons