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Verizon [17]
3 years ago
5

Calculate the equilibrium constant for the following reaction: Co2+ (aq) + Zn(s> CO (s) + Zn2+ (aq)

Chemistry
1 answer:
Simora [160]3 years ago
8 0

<u>Answer:</u> The K_{eq} of the reaction is 1.73\times 10^{16}

<u>Explanation:</u>

For the given half reactions:

Oxidation half reaction: Zn(s)\rightarrow Zn^{2+}+2e^-;E^o_{Zn^{2+}/Zn}=-0.76V

Reduction half reaction: Co^{2+}+2e^-\rightarrow Co(s);E^o_{Co^{2+}/Co}=-0.28V

Net reaction: Zn(s)+Co^{2+}\rightarrow Zn^{2+}+Co(s)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=-0.28-(-0.76)=0.48V

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 = 0.48 V

R = Gas constant = 8.314 J/K.mol

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

K_{eq} = equilibrium constant of the reaction = ?

Putting values in above equation, we get:

2\times 96500\times 0.48=8.314\times 298\times \ln K_{eq}\\\\K_{eq}=1.73\times 10^{16}

Hence, the K_{eq} of the reaction is 1.73\times 10^{16}

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vovikov84 [41]

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\rule[225]{225}{2}

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<h3>~AH1807</h3>
8 0
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