Question

What is the logarithm of the equilibrium constant, log K, at 25°C of the voltaic cell constructed from the following two half-reactions? Fe2+ (aq) + 2e¯ Fe(s); ​E​° = – 0.41V Ag+(aq) + e–Ag(s); E° = 0.80 V

Answer 1

Answer:

4.0921 is the logarithm of the equilibrium constant.

Step-by-step explanation:

$Fe^{2+} (aq) +2e^{-}\rightarrow Fe(s)$; ​E​° = - 0.41 V

$Ag^+(aq) + e^-\rightarrow Ag(s)$; E° = 0.80 V

Iron having negative value of reduction potential .So ,that means that it will loose electron easily and get oxidized.Hence, will be at anode.

$E^{o}_{cell}$=Reduction potential of cathode - Reduction potential of anode

$E^{o}_{cell}=E^{o}_c-E^{o}_a$

$=0.80 V-(-0.41 V)=1.21 V$

$Fe^{2+} (aq) + 2e^{-}\rightarrow Fe(s)$; ​E​° = - 0.41 V

$2Ag^+(aq) + 2e^-\rightarrow 2Ag(s)$; E° = 0.80 V

Net reaction: $Fe(s)+2Ag^{+}\rightarrow Fe^{2+}+2Ag(s)$

n = 2

To calculate equilibrium constant, we use the relation between Gibbs free energy, which is:

$\Delta G^o=-nfE^o_{cell}$

and,

$\Delta G^o=-RT\ln K_{eq}$

Equating these two equations, we get:

$nfE^o_{cell}=RT\ln K_{eq}$

where,

n = number of electrons transferred = 2

F = Faraday's constant = 96500 C

$E^o_{cell}$ = standard electrode potential of the cell = 1.21 V

R = Gas constant = 8.314 J/K.mol

T = temperature of the reaction = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$2\times 96500\times 1.21 V=8.314\times 298\times \ln K_{eq}$

$\ln K_{eq}=9.3478$

$\log K_{eq}=\frac{9.3478}{2.303}=4.0921$

4.0921 is the logarithm of the equilibrium constant.