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2 years ago

What was the resulting molarity of your primary standard solution of potassium iodate kio3?

1 answer:

8 0

Missing question: 0,535 gram of KIO₃ dissolved in 250 mL of de-ionized water to <span>make primary standard solution.
m(</span>KIO₃) = 0,535 g.
V(KIO₃) = 250 mL ÷ 1000 mL/L = 0,25 L.
n(KIO₃) = m(KIO₃) ÷ M(KIO₃).
n(KIO₃) = 0,535 g ÷ 214 g/mol.
n(KIO₃) = 0,0025 mol.
c(KIO₃) = n(KIO₃) ÷ V(KIO₃).
c(KIO₃) = 0,0025 mol ÷ 0,25 L.
c(KIO₃) = 0,01 mol/L = 0,01 M.

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lina2011 [118]

<u>Answer:</u> The above reaction is non-spontaneous.

<u>Explanation:</u>

For the given chemical reaction:

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Here, nickel is getting reduced because it is gaining electrons and iron is getting oxidized because it is loosing electrons.

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$E^o_{(Fe^{3+}/Fe^{2+})}=0.77V\\E^o_{(Ni^{2+}/Ni)}=-0.23V$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

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

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$E^o_{cell}=-0.23-0.77=-1.0V$

Relationship between standard Gibbs free energy and standard electrode potential follows:

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

As, the standard electrode potential of the cell is coming out to be negative for the above cell. Thus, the standard Gibbs free energy change of the reaction will become positive making the reaction non-spontaneous.

Hence, the above reaction is non-spontaneous.

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<h3>What is an exothermic reaction?</h3>

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