Consider a voltaic cell to study the reaction of chromium and zinc. The cell con
ID: 993215 • Letter: C
Question
Consider a voltaic cell to study the reaction of chromium and zinc. The cell consists of the usual array of a salt bridge, appropriate wiring, and two half-cells: (1) a chromium metal electrode dipping into an aqueous solution of Cr^2+ ions; and (2) a zinc metal electrode dipping into an aqueous solution of Zn^2+ ions. Calculate the cell potential at standard conditions. Calculate delta_T Gdegree_cell. The concentration of metal ions is changed to Cr^2+ = 0.050 M and Zn^2+ = 0.010 M. Calculate delta_T G degree_cell and E_cell.Explanation / Answer
consider the reduction potentials
Eo Cr+2/Cr = -0.91 V
Eo Zn+2/Zn = -0.76 V
we know that
half cell with higher reduction potential is cathode
so
the one with zinc is cathode and the one with chromium is anode
a)
now
Eo cell = Eo cathode - Eo anode
Eo cell = Eo Zn+2/Zn - Eo Cr+2/Cr
Eo cell = -0.76 + 0.91
Eo cell = 0.15 V
b)
dGo = -nFEo
consider the anode reaction
Cr ---> Cr+2 + 2e-
cathode reaction :
Zn+2 + 2e- --> Zn (s)
here
n = 2 as two electrons are transferred
so
dGo = -2 x 96485 x 0.15
dGo = -28945.5 J/mol
so
value of dGo is -28.9455 kJ/mol
3)
now
dG = dGo + RT lnK
overall reaction is
Zn+2 + Cr --> Zn + Cr+2
K = [Cr+2] / [Zn+2]
so
dG = dGo + RT ln [Cr+2/Zn+2]
dG = ( -28945.5 ) + (8.314 x 298 x ln (0.05 / 0.01)
dG = -24958 J /mol
dG = -24.958 kJ/mol
now
dG = -nFE
-24958 = -2 x 96485 x E
E = 0.13
so
the value of Ecell is 0.13 V