You connect a 0.75M AgCl solution with a silver electrode to a solution of 0.52M FeCl2...

You connect a 0.75M AgCl solution with a silver electrode to a solution of 0.52M FeCl2 with an iron electrode and allow it to react spontaneously. (4pts) a. Give the cell notation b. What is the cell potential?

Expert Solution

A = 0.75 mg 11 given Agcl = 0.750 [agt Fecha = 0-52mg [reth] = 0.52m standard reduction potentials of fe and Ag ane t2 Fe Fe(s) -0.444 +2e Eo fety/fe (aq) Ago) Eyrargatas)= to.gov Caz) Agt té : reduction potential of Ag & more than fe iso Agt act as oxidising agent and fe act as reducing agent Oxidation at to Electrode. Canode) spontaneous reduction at Ag Electrode (cathode.) at anode: oxidation half reaction is fe → fetotze i Eo to.440 - Es +0.80V 2 Agtt ze zag retreaction: Fe + 2Agt Pertaag: E' = +1-24V for cell notation: It s represent oxidation reaction Canode) and Cathode. (reduction reaction) seperated by two bars anode at left site and Catherde sight side, two barn represent salt bridge. Cell notation fe colfe || Agt! Ag co Coisam)" (0.75m)

to. by usong nerast Eruation calculate Ecell (cell potential ) 0,059 v -E E cell cell n=2 Ecell sa log Ou059 Ecell = Ecell log (Ag +32 0.52 0.059 v logo Ecell = +1.24 V (0,75) Ecell 1.24 V + 0.00101 V ~ l-zur 1.241 Ecull

queen_honey_blossom answered 2 years ago

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