A Galvanic cell (also known as a voltaic cell) uses the difference in reduction potentials of two ionic species in order
Posted: Wed May 04, 2022 4:03 pm
A Galvanic cell (also known as a voltaic cell) uses the
difference in reduction potentials of two ionic species in order to
generate an electric current. For example, the cell at the right
involves two metals with the following reduction potentials:
Zn2+(aq) + 2e → Zn(s) (Eored = -0.760 V) Al3+(aq) + 3e → Al(s)
(Eored = -1.660 V) What is the voltage generated by this cell?
(Eocell).
Note that the cell potential is independent of the number of
electrons involved in the reaction at either electrode. The number
of electrons involved in the reaction come into play, however, in
the Nernst equation, which shows the relationship of the cell
potential to the actual concentration of ions: Ecell = Eocell
-(RT/nF)lnQ where R is the gas constant (8.314 J/K-mol), F is the
Faraday constant (9.6485×104 Coulombs/mol or J/V-mol), T is the
absolute temperature, Q is the reaction quotient, and n is the
number of electrons exchanged in the overall balanced equation. The
number of electrons also comes into play in the relationship
between the cell potential and the free energy change of the
reaction: ΔGo = -nFEocell (for standard conditions), and ΔG =
-nFEcell (for ion concentrations at other than the standard state).
(Note also that ΔG is negative, indicating a spontaneous reaction,
when E is positive). Use the Reaction Editor to write the overall
balanced equation occurring in the galvanic cell described above.
(Indicate the physical states of reactants and products, i.e.
Al(s), Zn2+(aq), etc).
For this reaction, Q=[Al3+(aq)]2[Zn2+(aq)]3, because the pure
solids are in their standard states (activity = 1) and are
therefore not included in the reaction quotient expression.
Calculate Ecell at 25°C when [Zn2+(aq)] = 3.7 M and [Al3+(aq)] =
0.075 M.
Calculate ΔGo, the standard free energy change for this
reaction
Calculate ΔG, free energy change at 25°C when [Zn2+(aq)] = 3.7 M
and [Al3+(aq)] = 0.075 M.
difference in reduction potentials of two ionic species in order to
generate an electric current. For example, the cell at the right
involves two metals with the following reduction potentials:
Zn2+(aq) + 2e → Zn(s) (Eored = -0.760 V) Al3+(aq) + 3e → Al(s)
(Eored = -1.660 V) What is the voltage generated by this cell?
(Eocell).
Note that the cell potential is independent of the number of
electrons involved in the reaction at either electrode. The number
of electrons involved in the reaction come into play, however, in
the Nernst equation, which shows the relationship of the cell
potential to the actual concentration of ions: Ecell = Eocell
-(RT/nF)lnQ where R is the gas constant (8.314 J/K-mol), F is the
Faraday constant (9.6485×104 Coulombs/mol or J/V-mol), T is the
absolute temperature, Q is the reaction quotient, and n is the
number of electrons exchanged in the overall balanced equation. The
number of electrons also comes into play in the relationship
between the cell potential and the free energy change of the
reaction: ΔGo = -nFEocell (for standard conditions), and ΔG =
-nFEcell (for ion concentrations at other than the standard state).
(Note also that ΔG is negative, indicating a spontaneous reaction,
when E is positive). Use the Reaction Editor to write the overall
balanced equation occurring in the galvanic cell described above.
(Indicate the physical states of reactants and products, i.e.
Al(s), Zn2+(aq), etc).
For this reaction, Q=[Al3+(aq)]2[Zn2+(aq)]3, because the pure
solids are in their standard states (activity = 1) and are
therefore not included in the reaction quotient expression.
Calculate Ecell at 25°C when [Zn2+(aq)] = 3.7 M and [Al3+(aq)] =
0.075 M.
Calculate ΔGo, the standard free energy change for this
reaction
Calculate ΔG, free energy change at 25°C when [Zn2+(aq)] = 3.7 M
and [Al3+(aq)] = 0.075 M.