What is the calculated value of the cell potential at 298 K for
an electrochemical cell with the following reaction, when the
H2 pressure is
5.00×10-3 bar, the
H+ concentration is
1.46 M, and the
Zn2+concentration is
3.42×10-4 M ?
2H+(aq)
+ Zn(s) ---->
H2(g) +
Zn2+(aq)
Answer: _____ V ?
The cell reaction as written above is product-favored (spontaneous)
for the concentrations given: _______(true/false)?
i)
0.780V
Explanation
2H+(aq) + Zn(s) ------> H2(g) + Zn2+(aq)
Reactant quotient, Q = PH2 × [Zn2+]/[H+]2
Q = 0.005bar × 0+.000342M/(1.46)2
Q = 8.022×10-7
Oxidation half reaction (at anode)
Zn(s) -------> Zn2+(aq) + 2e E°red = - 0.762V
Reduction half reaction (at cathode)
2H+(aq) + 2e --------> H2(g) E°red = 0.000V
E°cell = E°red,cathode - E°red,anode = 0.000V - (- 0.762V) = 0.762V
Number of electron transfer ,n = 2
at 25°C , Nernst equation is as follows
Ecell = E°cell - (0.0592V/n)logQ
Ecell = 0.762V - (0.0592V/2)log(8.022×10-7)
Ecell = 0.762V + 0.18V
Ecell = 0.780V
ii)
True
Get Answers For Free
Most questions answered within 1 hours.