Consider a voltaic cell with Cr(s) and Cr^3+(aq) in one compartment and Zn(s) an
ID: 972127 • Letter: C
Question
Consider a voltaic cell with Cr(s) and Cr^3+(aq) in one compartment and Zn(s) and Zn^2+(aq) in the other compartment. Draw and label a diagram of this cell. Calculate the standard cell potential (Edegree) and the equilibrium constant (K) for the overall reaction. Calculate the cell potential if [Zn^2+] = 5.00 M and [Cr^3+] = 0.50 M? What docs this result tell you? How would your diagram change? How does Q compare to K? If zinc is the anode and [Cr^3+] = 0.010 M. what must the [Zn^2+] be in order to achieve 0.050V?Explanation / Answer
(b) Eocell = Eored - Eooxid
Zn acts as anode and Cr as cathode
Eocell = (- 0.72) - (- 0.76) = 0.04V
(c) Ecell = Eocell- (0.059 /n )log[Zn2+] / [Cr3+], Nrnest equation
n = 6 , as 6 electrons are involved in the balanced chemical equation
3Zn ---------> 3Zn2+ + 6e-
2Cr3+ + 6e- ---------> 2 Cr
Ecell = 0.04V - (0.059 / 6 )log[5.0] / [0.5] = 0.03V
(d) 0.05 = 0.04 - (0.059 / 6) log [Zn2+] / [Cr3+]
log [Zn2+] / [Cr3+] = - 1.02
[Zn2+] / [Cr3+] = antilog(-1.02)
[Zn2+] / [0.01] = 0.095
[Zn2+] = 0.01 X 0.095 = 0.0095
(a) The Zn electrode will act as ANODE HALF CELL and Cr as CATHODE HALF CELL. The two half cells will be joined by salt bridge. It is not possible to draw it in QA board although it is easy.