What is the logarithm of the equilibrium constant, log K, at 25°C of the voltaic cell constructed from the following two half-re

Question

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actions? Fe2+ (aq) + 2e¯ Fe(s); E° = – 0.41V Ag+(aq) + e–Ag(s); E° = 0.80 V

1 answer:

lawyer [12.5K] · Answer 1 · 2026-03-09T04:43:22+03:00

Answer:

4.0921 reflects 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 has a negative reduction potential, indicating its tendency to lose electrons and undergo oxidation, and thus it will be at the 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 determine the equilibrium constant, we utilize the correlation with Gibbs free energy, as follows:

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

and,

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

Aligning these two equations yields:

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

where,

n = 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 = reaction temperature = $25^oC=[273+25]=298K$

Substituting values into the equation, we arrive at:

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

$\ln K_{eq}=9.3478$

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

4.0921 represents the logarithm of the equilibrium constant.