Chemistry, 19.03.2020 06:48 JohnStewart714

Compute the equilibrium constant, K, at 25 ∘C for the reaction between Ni2+(aq) and Zn(s), which form Ni(s) and Zn2+(aq)

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Answer from: tytianadyson74

The answer to the question is;

The equilibrium constant,K, at 25 °C for the reaction between Ni²⁺(aq) and Zn(s), which form Ni(s) and Zn²⁺ is 2.04×10¹⁷.

Explanation:

The half reactions are as follows

Ni²⁺ (aq) + 2e⁻ -> Ni (s)

Zn (s) -> Zn²⁺ (aq) + 2e⁻

For the Ni²⁺/Ni system we have the potential given as -0.23V (Reduction)

For the Zn²⁺/Zn sytem, the potential is -0.76. Here however, we should note that the zinc is being oxidized and therefore the potential is positive, that is;

Zn/Zn²⁺ = 0.76

Therefore the voltage for the sum of the reactions on both sides of the process is

-0.23 V + 0.76 V = 0.53 V

We then call upon the Nernst equation to calculate the equilibrium constant as follows

E⁰ $_{cell}$ = $\frac{0.0592}{n} logK$

Where:

E⁰ $_{cell}$ = Standard cell potential = 0.53 V

n = Number of moles of electrons = 2 moles of e⁻

K = Equilibrium constant

Therefore we have

0.53 V = $\frac{0.0592}{2} logK$

Therefore log K = 17.905

and K = $10^{17.9054}$ = 2.04×10¹⁷.

Answer from: Quest

the formula for the conversion of temperature from degree celsius to kelvin is:

$k = ^{o}c + 273.15$

the sample of water at $38^{o}c$ is converted to kelvin temperature by using the above formula as:

$k = 38^{o}c + 273.15 = 311.15 k$

on comparing the two samples of water at $38^{o}c$ or we can say that at $311.15 k$ and at $295 k$ respectively. the difference between the temperatures of two samples of water is:

$311.15 k - 295 k = 16.15 k$

hence, one can say that the sample of water at $38^{o}c$ is greater than the sample of water at $295 k$ by $16.15 k$ .

Answer from: Quest

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