Use cell potential to calculate an equilibrium constant.
Calculate the cell potential and the equilibrium constant for
the following reaction at 298 K:
Cu2+(aq) +
Hg(l) =
Cu(s) +
Hg2+(aq)
Hint: Carry at least 5 significant figures during intermediate
calculations to avoid round off error when taking the
antilogarithm.
Equilibrium constant: -------
Calculate the Nernst potential for each ion, the membrane
potential, and driving force for each of
the following examples:
Intracellular ions: [Na+]= 12 mM, [K+]=145 mM, [Cl-]=15 mM,
[Ca2+]=250 nM
Extracellular ion: [Na+]= 150 mM, [K+]=2.9 mM, [Cl-]=150 mM,
[Ca2+]=7 mM
Derive an expression of the van Deemter equation that shows how
to calculate the minimum value of H based upon the values for A, B
and C and not flow rate. (Assume the C terms are combined into one
C term).
Derive and explain the relationship between the Nernst equation,
the equilibrium constant and other thermodynamic quantities such as
enthalpy, Gibbs energy and entropy of reaction.
Using E° = .697 V and the Nernst equation, calculate the
concentration of Hg2+ in a sample of pond water if [Sn2+] = 0.500
M, [Sn4+] = 0.305 M and Ecell = 0.711 V.
(i) Use the information in the Resource section to calculate the
standard potential of the cell Ag(s)|AgNO3(aq)||Cu(NO3)2(aq)|Cu(s)
and the standard Gibbs energy and enthalpy of the cell reaction at
25°C.
(ii) Estimate the value of AG at 35 °C.
a. Show with the aid of a form of the Nernst equation that when
using a pH sensitive glass electrode and reference electrode that
the measured potential is proportional to pH.
b. Sketch a diagram of a pH sensitive glass electrode, label all
parts and indicate the function of each part.
(A) Derive Bernoulli's Equation from Pressure, Kinetic Energy and Potential Energy
(B) Express Bernoulli's equation:
- Per unit mass
-Per unit volume
- In terms of heads
Calculate the cell potential for a reaction in a electrolytic
cell with the following half-reactions if: [U 3+] = 0.10 M, [MnO4 -
] = 0.20M, [Mn2+], and [H+ ] = 0.20 M
U 3+ + 3e -> U o Ecell = - 1.642 V
MnO4 - + 8H+ + 5e- -> Mn2+ + 4H2O Ecell = +1.51 V