An electrochemical cell consists of a half-cell in which a piece of platinum wire is dipped into a solution that is 2.5 M in KI and 0.046 M in I2. The other half-cell consists of magnesium metal immersed in a 0.37 M Mg2+ solution. Assume that the temperature is 25°C. (a) Which electrode is the anode and which is the cathode? anode: O Pt|12, 1- O Mg2+|Mg cathode: O Pt|I2, I~ O Mg2+|Mg (b) What is the emf of the cell? 2.869 x V (c) What is the spontaneous cell reaction? chemPad O Help Greek |-563 -563 Your answer differs too much from the expected answer to provide useful feedback. (d) What is the equilibrium constant of the cell reaction? 6.09e98

The standard reduction potential values are: Mg2+(aq) + 2e- ---> Mg(s) ; Eo = - 2.37 V I2(s) + 2e- -----> 2I-(aq) ; Eo = 0.54 V #(b): Since I2(s) has a higher reduction potential, Pt|I2, I- acts as cathode, whereas Mg2+|Mg acts as anode. The half-reactions are: Anode: Mg(s) ----> Mg2+(aq) + 2e- ; Eoanode = - 2.37 V Cathode: I2(aq) + 2e- ----> 2I-(aq) ; Eocathode = 0.54 V Since 2e- are transferred, n = 2 Standard emf, Eocell = Eocathode - Eoanode = 0.54 V - (-2.37V) = (0.54+2.37) V = 2.91 V EMF of the cell can be calculated using Nernst equation: Ecell = Eocell - 0.0591n*log[Mg2+(aq)]*[I-(aq)]2[I2(aq)]⇒Ecell =2.91V - 0.05912*log0.37*(2.5)20.046⇒Ecell =2.86 V EMF of the cell = 2.86 V (Answer)#(c): Spontaneous cell reaction can be found by adding the anode and cathode half-reactions. Anode: Mg(s) ----> Mg2+(aq) + 2e- + + Cathode: I2(aq) + 2e- ----> 2I-(aq) ------------------------------------------- Spontaneous cell reaction: Mg(s) + I2(aq) -----> Mg2+(aq) + 2I-(aq) (Answer) #(d): ∆Go = - n*F*Eocell => ∆Go = - 2*96485*2.91 = 561542.7 J Also, ∆Go = - RT*ln(K) => - 561542.7 J = - 8.314 * 298.15 * ln(K) => 561542.7 J = 8.314 * 298.15 * ln(K) ⇒ ln(K) = 561542.78.314 * 298.15⇒ln(K) = 226.5⇒ K = e226.5 = 2.42*1098 Equilibrium constant = 2.42*1098 (Answer)