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blsea [12.9K]
2 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]2 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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What mass of copper is deposited when a current of 10.0a is passed through a solution of copper(ii) nitrate for 30.6 seconds?
asambeis [7]
Data Given:

Time = t = 30.6 s

Current = I = 10 A

Faradays Constant = F = 96500

Chemical equivalent = e = 63.54/2 = 31.77 g

Amount Deposited = W = ?

Solution:
             According to Faraday's Law,

                                          W  =  I t e / F

Putting Values,
 
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                            W  =  0.100 g

Result:
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Answer:

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