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blsea [12.9K]
3 years ago
5

Part V Write the (a) half reaction of oxidation, (b) half reaction of reduction, & (c) the balanced redox reaction for all o

f the chemical reactions listed below. 26. 2Sr + O2 → 2SrO (a) O0 + 2e- → O-2 (b) Sr0 → Sr+2 + 2e- (c) O0 + Sr0 → O-2 + Sr+2
Chemistry
1 answer:
Eduardwww [97]3 years ago
8 0

Answer:

Explanation:

a. Oxidation : 2O + 4e^- ------> 2O^2-

b. Reduction: 2Sr - 4e- -------> Sr^2+

c. Balanced redox reaction

2Sr + O2 ------------> 2Sr O

Oxidation and reduction can be defined by various means, addition of oxygen, removal of hydrogen, removal of electrons. For this reaction, this definition is used, oxidation is the loss of electrons while reduction is the gaining of electrons.

In (a) oxidation half reaction, the valency of oxygen is zero and then moves into lossing two electrons resulting into -2 valency.

In (b) reduction half reaction, the valency of Sr is zero and gains electrons resulting into valency of 2.

In the overall redox reaction, Sr and O2 with valency of 0 each reacts together and form SrO with valency of 2 and -2 respectively, which gives 0 and then balances the equation.

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Answer:

0.001199 year^{-1} is the rate constant for this reaction.

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Explanation:

A decomposition reaction follows first order kinetics:

Half life of the reaction = t_{1/2}=578 years

Rate constant of the reaction = k

For first order reaction, half life and rate constant are linked with an expression :

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{578 years}=0.001199 year^{-1}

0.001199 year^{-1} is the rate constant for this reaction.

Initial concentration of reactant =[A_o] =  x

Final concentration of reactant after time t =[A] =  12.5% of x = 0.125x

The integrated law of first order reaction :

[A]=[A_o]\times e^{-kt}

0.125x=x\times e^{-0.001199 year^{-1}\times t}

t = 1,734.31 years =1.73\times 10^3 years

It will take 1.73\times 10^3 years to concentration to reach 12.5% of its original value.

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3 years ago
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P1V1 = P2V2

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