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faltersainse [42]
3 years ago
13

In the reaction: pb + 2ag+ → pb2+ + 2ag, the oxidizing agent is

Chemistry
2 answers:
wariber [46]3 years ago
4 0

<u>Answer:</u> The oxidizing agent for the given reaction is Silver.

<u>Explanation:</u>

Oxidizing agent is defined as the agent which helps the other substance to get oxidized and itself gets reduced. It undergoes reduction reaction in which a substance gains electrons. The oxidation state of this substance gets reduced and the substance gets reduced.

Reducing agent is defined as the agent which helps the other substance to get reduced and itself gets oxidized. It undergoes oxidation reaction in which a substance looses its electrons. The oxidation state of this substance gets increased and the substance gets oxidized.

For the given chemical reaction:

Pb+2Ag^+\rightarrow Pb^{2+}+2Ag

The half reactions for the given above chemical reaction is:

Oxidation half reaction:  Pb\rightarrow Pb^{2+}+2e^-

Reduction half reaction: 2Ag^++2e^-\rightarrow 2Ag

As, lead is loosing electrons. So, it is getting oxidized and is considered as a reducing agent and silver is gaining electrons. So, it is getting reduced and is considered as an oxidizing agent.

Thus, the correct answer is silver.

shutvik [7]3 years ago
3 0
With reference to present question, following things may be noted
1) Oxidation is process of lose of electron. Atom//ion undergoing oxidation is refereed as reducing agent
2) Reduction is process of gain of electron. Atom/ion undergoing reduction is refereed as oxidizing agent.

in present system, Ag+ gain an electron to reduced to Ag. Therefore, Ag+ is an oxidizing agent. 
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Fluorine (F) and bromine (Br) are in the same group on the periodic table. How do atoms of these elements compare when they form
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3 years ago
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A chemist adds 0.60L of a 0.20/molL sodium thiosulfate Na2S2O3 solution to a reaction flask. Calculate the millimoles of sodium
Serggg [28]

Answer:

1.2×10² mmole of Na₂S₂O₃

Explanation:

From the question given above, the following data were obtained:

Volume = 0.6 L

Molarity = 0.2 mol/L

Mole of Na₂S₂O₃ =?

Molarity is simply defined as the mole of solute per unit litre of water. Mathematically, it is expressed as:

Molarity = mole /Volume

With the above formula, we can obtain the number of mole of Na₂S₂O₃ in the solution as illustrated below:

Volume = 0.6 L

Molarity = 0.2 mol/L

Mole of Na₂S₂O₃ =?

Molarity = mole /Volume

0.2 = Mole of Na₂S₂O₃ / 0.6

Cross multiply

Mole of Na₂S₂O₃ = 0.2 × 0.6

Mole of Na₂S₂O₃ = 0.12 mole

Finally, we shall convert 0.12 mole to millimole (mmol). This can be obtained as follow:

1 mole = 1000 mmol

Therefore,

0.12 mole = 0.12 mole × 1000 mmol / 1 mole

0.12 mole = 120 = 1.2×10² mmole

Thus, the chemist added 1.2×10² mmole of Na₂S₂O₃

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