Consider the following pair of half cells in an electrochemical cell. Cr^3+ + 3e
ID: 527430 • Letter: C
Question
Consider the following pair of half cells in an electrochemical cell. Cr^3+ + 3e^- rightarrow Cr (s) E^= -0.74 V Ag^+ + e^- rightarrow Ag(s) E^= +0.80 V Which electrode would be the cathode and which would be the anode in a spontaneous (galvanic) electrochemical cell? An electrochemical cell is constructed with 1 times 10^-3 M concentrations of the respective ion (Cr^3+ or Ag^+) in each of the Cr and Ag half cells. Determine E_cell for this cell. In this lab we had two metrics to determine whether an electrochemical cell follows Nernstian behavior. If we evaluated the Cr^3+ half cell to determine whether it follows Nernstian behavior, what are the two metrics we would be looking for? Be as specific as you can.Explanation / Answer
Q1.
a)
cathode --> reduces, the higher potential is Ag+, therefore choose Silver
anode --> oxidizes, the lowest potential, that is Cr3+, then Chromium
b)
Apply E°Cell
3Ag+(aq) + Cr(s) --> Cr3+ + 3Ag(s)
Q = [Cr+3] / [AG+]^3
Q = (10^-3)/ ( 10^-3)^3 = 1000000
E° = Ered- Eox = 0.80 - -0.74 = 1.54 V
n = 3 electrons
E = E - 0.0592/n * log(Q)
E = 1.54 - 0.0592/3* log(1000000)
E = 1.4216 V
Q3.
- metrics are --> concentration & Voltage
if it follows both, then, this should follow Enerst equation