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3 years ago

9

S2- has the same electron configuration as which neutral element?

1 answer:

Llana [10]3 years ago

6 0

Answer:

ANSWER OS HERE

Explanation:

THE O2- has two more extra electrons due to presence of 2 negative charges so a total of 10 electrons and the element that has 10 electrons in the periodic table in its neutral state is Neon (Ne) ( atomic number (10) = no.of protons = no. of electrons) so neutral element with same electron configuration as O2- is Ne. Log in or register to post comments.

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

Simora [160]

<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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