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olga nikolaevna [1]
3 years ago
7

Identify the oxidizing and reducing agents in the following: 2H+(aq) + H2O2(aq) + 2Fe2+(aq) → 2Fe3+(aq) + 2H2O(l)

Chemistry
1 answer:
schepotkina [342]3 years ago
7 0

Answer :  The oxidizing and reducing agents are, H_2O_2 and Fe^{2+}.

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Reducing agent : It is defined as the agent which helps the other substance to reduce and itself gets oxidized. Thus, it will undergo oxidation reaction.

Oxidizing agent : It is defined as the agent which helps the other substance to oxidize and itself gets reduced. Thus, it will undergo reduction reaction.

The given redox reaction is:

2H^+(aq)+H_2O_2(aq)+2Fe^{2+}(aq)\rightarrow 2Fe^{3+}(aq)+2H_2O(l)

The half oxidation-reduction reactions are:

Oxidation reaction : Fe^{2+}\rightarrow Fe^{3+}+1e^-

Reduction reaction : O^-+1e^-\rightarrow O^{2-}

The oxidation state of oxygen in H_2O_2 and H_2O is, (-1) and (-2) respectively.

In this reaction, 'Fe' is oxidized from oxidation (+2) to (+3) and 'O' is reduced from oxidation state (-1) to (-2). Hence, 'Fe^{2+}' act as a reducing agent and 'H_2O_2' act as a oxidizing agent.

Thus, the oxidizing and reducing agents are, H_2O_2 and Fe^{2+}.

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820 g Li2SO4 is dissolved into 2500 mL of solution. What is the molar concentration?
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Molarity is defined as the moles of solute per liter of solution. M=\frac{n}{V}. Where M is molarity, n is the number of moles and V is the volume. First we must find the molar mass of Li_2SO_4 which is 109.98 g/mol

Li_2SO_4= 2 \times Li_{Ar} + S_{Ar} + 4 \times O_{Ar}\\= (2 \times 6.941 + 32.1 + 4 \times 16.0) = 109.98 g/mol

820\ g\ Li_2SO_4 \times \frac{mol}{109.98\ g}= 7.46\ mol\ Li_2SO4

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