# PS4 - -|AgCl|Ag Relate its E Cell to the equilibrium...

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CHEMISTRY 15 Fall 2009 Problem Set #4: Galvanic Cells due on Tuesday, Oct. 13 1. For the following galvanic cells, calculate the cell voltage at 25°C, identify the anode, write a balanced equation for the overall cell reaction, and indicate the direction of the spontaneous reaction. a) Pt|VO 2 + (10 mM), VO 2+ (1.0 mM), H + (0.10 mM)||Fe 2+ (0.10M), Fe 3+ (1.0 mM)|Pt VO + 2 + 2H + + e - = VO 2+ + H 2 O E° = +1.00 V; Fe 3+ + e - = Fe 2+ E° = +0.77 V [Ans: E cell = 0.066 V] b) Sb|SbO + (5.0 ! 10 -2 M), H + (1.0 ! 10 -2 M)||Pb 2+ (1.0 ! 10 -4 M)|Pb SbO + + 2H + + 3e - = Sb(s) + H 2 O E° = +0.212 V. Pb +2 + 2e - = Pb(s) E° = -0.126 V. [Ans: E cell = 0.352 V] 2. Calculate the solubility product for AgCl from the following data: Ag + + e - = Ag(s) E° = +0.799 V. AgCl(s) + e - = Ag(s) + Cl - E° = +0.222 V. [Ans: 1.79 ! 10 -10 ] Hint : Construct the following galvanic cell: Ag|Ag
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Unformatted text preview: -|AgCl|Ag. Relate its E Cell to the equilibrium constant for the cell reaction. 3. Given the following half-reactions: Sn 2+ + 2e-= Sn(s) E° = -0.136 V; Pb 2+ + 2e-= Pb(s) E° = -0.126 V, calculate the equilibrium constant for the reaction: Sn(s) + Pb 2+ = Sn 2+ + Pb(s) [Ans: K = 2.18] 4. The voltage of the cell Zn(s)|Zn 2+ (xM)||Cu 2+ (0.1M)|Cu(s) was measured to be 1.150 V. Calculate the Zn 2+ concentration. [Ans: 2.18 ! 10-3 M] 5. Cells of the type: Cd||Cd 2+ (0.1M)||Cd 2+ (0.001M)|Cd are called concentration cells. a) Write the Nernst equation for each of the two half-cells and calculate the overall cell voltage. [Ans: a) E cell = 0.059 V] b) Which electrode is the cathode? c) What is the value of Cell E for this cell?...
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