Redox III - Examples

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Transcript Redox III - Examples

ΔG of an Electrochemical Cell
The change in Gibbs Energy ΔG is the
maximum non-PV work* that can be obtained
from a chemical reaction at constant T and P:
ΔG = wmax
For an electrochemical cell, wmax = -nFE
ΔG = -nFE
E = E° - RT ln Q
*or minimum energy needed for the nonspontaneous reaction
The Nernst Equation – Example 1
E = E° - 0.0592 V log Q
What is the emf of the following cell at 298 K? This is
called a concentration cell.
Zn2+(aq, 0.500 M)  Zn2+(aq, 0.150 M)
Q = 0.150
= 0.300 M
E = 0 - 0.0592 log .300 = 0.015 V
The Nernst Equation – Example 2
Find the solubility product for silver thiocyanate
at 25°C, given
AgSCN(s) + e-  Ag(s) + SCN-(aq)
E°(25°C) = 0.0895 V
and using the standard reduction potential for Ag+
found in Appendix E.
Answer: KSP = 1.0 x 10-12
The Nernst Equation – Example 3
The concentration of potassium ions in the
intracellular fluid (ICF) is 135 mM. In the
extracellular fluid (ECF), it is 4 mM. What
potential must be applied to the cell to keep the
K+ from diffusing out of the cell?
Answer: The ECF must have a potential of 94 mV
relative to the ICF in order to keep the K+ from
diffusing out of the cell.
The Nernst Equation – Example 4
In a galvanic cell with two hydrogen electrodes,
both at 298 K, the first electrode has PH2 = 1.00
bar and an unknown concentration of H+. The
second hydrogen electrode is the SHE. Ecell is
found to be 0.211 V and the electron flow is from
the first electrode to the second. What is the pH
of the solution in the first hydrogen electrode?
Answer: pH = 3.56
The Nernst Equation – Example 5
In a galvanic cell at 25°C, the cathode halfreaction is Ag+(aq, 1M) + e-  Ag(s).
The anode is a hydrogen electrode with PH2 = 1.00
bar. It is in a buffer containing 0.10 M benzoic
acid and 0.050 M sodium benzoate.
Ecell(25°C) = 1.030 V. What is the pKa of the
benzoic acid?
Answer: pKa = 4.20
The Nernst Equation – Example 6
4Fe2+(aq) + O2(g)+ 4H+(aq)  4Fe3+(aq) +2H2O(l)
What is the cell emf at 25°C when [Fe2+]=1.3M,
[Fe3+]=0.010M, PO2 = 0.5 bar, and pH=3.50?
E = Eº - .0592 log Q …need n, Q, and Eº
Answer: E = 0.37V
The Nernst Equation – Example 7
2Fe3+(aq) + H2(g)  2Fe2+(aq) + 2H+(aq)
What is the cell emf at 25°C when [Fe2+]=0.0010M,
[Fe3+]=2.50M, PH2 = 0.85 bar, and pH=5.00?
E = Eº - .0592 log Q …need n, Q, and Eº
Answer: E = 1.27V