Question

Electrochemical Cells and Thermodynamics Key (calculated) Shorthand cell designation ΔΟ (calculated) E cell (measured) Temperature ( ) and Keq for an exemplary pair. For Pb | Pb2+ Isn2+ Sn: Show calculations for E half-cell (calculated) Half-cell equation Effect of Temperature on Cell Potential Cell designation: E (measured Tempegature fc) Temperature (kK) AG (calculated) 5 c 1134 228 K 2315 AS determined from the slope of a plot of AG versus T (show calculations) AfP calculated at 298 K 233

Answer 1

CALCULATION OF ΔGº AND K_eq

For a given electrochemical cell, Gibbs energy is related to its cell potential according to the equation:

$\Delta G^{\circ}=-nFE^{\circ}_{cell}\;\;\;\;(1)$

Where:

ΔGº is the standard Gibbs energy of the cell

n is the number of moles of electrons transferred in the reaction (mol e^-)

F is the Faraday constant (F = 96485 coulomb/mol e^-)

Eº_cell is the standard cell potential (V)

For the cell:

Sn +2

The involved reduction reaction is:

Sn+2 +2e - Sn(o)

Notice that 2 moles of electrons are transferred on this equation. Then: n = 2 mol e^-. Substituting values in equation (1), for this cell, the standardg Gibbs energy is:

i8s coulomb(0.717V) --(2mol e-)(96485 mole

AG138, 359.5J

In equilibrium, the relationship between cell potential and equilibrium contant K_eq, is given by:

$E^{\circ}_{cell}=\frac{RT}{nF}LnK_{eq}\;\;\;\,(2)$

Where

R is the ideal gas constant (R = 8.314 J/mol*K).

T is the absolute temperature (K)

Solving equation (2) for K_eq, we have:

$K_{eq}=e^{\frac{nFE^{\circ}_{cell}}{RT}}\;\;\;\;(3)$

Substituting known values for cell (1) in equation (3), for an absolute temperature of T = (25+273.15) = 298.15K, the equilibrium constant of the chemical reaction is:

29648500.717 e (8.314)(298.15) eq e

Keg 1.74 1024

For the cell:

2. Zn)ZnSnE 0.395V

We have the reduction reaction:

Sn+2 +2e - Sn(o)

Notice that 2 moles of electrons are transferred on this equation. Then: n = 2 mol e^-. Substituting values in equation (1), for this cell, the standardg Gibbs energy is:

$\Delta G^{\circ}=-(2mol\, e^-)(96485\frac{coulomb}{mol\, e^-})(0.395V)$

$\Delta G^{\circ}=-76,223.2\,J$

Substituting known values for this cell in equation (3), the equilibrium constant for this reaction is:

$K_{eq}=e^{\frac{(2)(96485)(0.395)}{(8.314)(298.15)}}$

$K_{eq}=2.26*10^{13}$

For the cell:

We have the reduction reaction:

$Cu^{+2}+2e^-\rightarrow Cu_{(s)}$

Notice that 2 moles of electrons are transferred on this equation. Then: n = 2 mol e^-. Substituting values in equation (1), for this cell, the standardg Gibbs energy is:

coulomb --(2mol e-)(96485 (1.134V)

$\Delta G^{\circ}=-218,828\,J$

Substituting known values for this cell in equation (3), the equilibrium constant for this reaction is:

$K_{eq}=e^{\frac{(2)(96485)(1.134)}{(8.314)(298.15)}}$

$K_{eq}=2.18*10^{38}$