Question

What is the cell potential for the reaction Mg(s) + Fe2+ (aq)—Mg²+ (aq) + Fe(s) at 53 °C when [Fe2+] = 3.80 Mand (Mg2+] = 0.310 M. Express your answer to three significant figures and include the appropriate units. ► View Available Hint(s) E = 2.002

Answer 1

. The given cell reaction is: Mac) + feta a motor + Fecs) . The cell representation for the above cell reaction is, (ng/mg) || (Fet?/ Fe) Anode cathode * Cell potential for a cell at temperature T can be Calculated from Nernst equatin. Nemst equations : Excel Ergin - RT in.Q Ecent = Cell potential Ecent - Standard Cell potential R = Universal gas constant = 8.314 Jmoletky T = Abscute temperative ne Number of electrons participated in - cell reaction F = Faraday constant - 96487 C.moly Q = Equilibrium constant, [meta][Fe] Equilibrium constant a= - [Mg] (fer concentrations of Metals = 1 = [Fe] = [Mg 1Q = [Mgt2] . [feta ne number of electrons Pasticipated = lē

LO et * Econa Ecotecode Formele) = [Ecritten) Fengering] Standard reduction potential of Fe Eº = -0.44 v Standard electrode reduction (Felfe) Potential of Mg Eemotyna) = -2.36 V * Een = 60.44)–(-2.36) - 60.44+2.36) ce= +1.92 V * Teiperature = 63.+273) = 326K * Eces = (+1.92)-(3-314) (326) An [Mgt2 L QF * Ecell = 41.92)- |(8 314)(326)] 3 loq/mgt?) 96487) Liety) = 6191) (60-032) og (et [mgta] = 0.310 M ; [fet'] = 3.20M 0.310 M

3. EO Ecell = 41.92) – (6.0323) log (-360 =(+1.92) -(0.0323) log (0.08157) = 41.92) -(0.0323)(-1.083) =(+1.92) + 0.0356) Egy = +1.955V :: Ecell = +1.95V