Question

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Reactions at Electrodes during Electrolysis Answer the following questions by selecting the appropriate half-reaction. Assume that in each case the experiment is performed in an electrolysis cell having electrodes made of non-reacting material. Which half-reaction will take place at the anode during the electrolysis of 1.0 M H2O2 solution containing 1.0 M H2SO4? O 02 + 4H+ + 4e - 2H20 H202 0 2 + 2H+ + 2e H2O2 + 2H+ + 2e- 2H20 2H20 + 2 H2 + 20H Which half-reaction will take place at the anode during the electrolysis of molten MgCl2? Cl2 + 2e 20 201 Cl2 + 2e Mg Mg2+ + 2e Mg2+ + 2e Mg Which half-reaction will take place at the cathode during the electrolysis of molten KF? OF F2 + 2e OK-> K+ + e F2 + 2e - 2F OK+ + e- K Which half-reaction will take place at the cathode during the electrolysis of 1.0 M AgNO3? 2H20 - 02 + 4H+ + 4e* Ag- Ag+ + e Ag+ + e Ag 2H20 + 2 H2 + 2OH Submit Answer Tries 0/5

Answer 1

In an electrolysis cell, Oxidation half-reaction occurs at anode and Reduction half-reaction occurs at cathode. This is the key concept used in answering all the 4 questions below.

1. H₂O₂ under acidic conditions, can undergo oxidation as follows:

H₂O₂$\rightarrow$ 2H⁺ +2e^- + O₂

This is the only reaction from the above given options that show change of oxidation state of Oxygen from (-1) in H₂O₂ to (0) in O₂.

2. For molten MgCl₂,

At cathode, the reduction of Mg⁺² ion will take place to Mg metal.

At anode, oxidation of Cl^- ions to chlorine gas will take place.

so, 2 Cl^-$\rightarrow$ Cl₂ + 2e^-

3. For molten KF,

At cathode, potassium metal will be obtained.

K⁺ + e^-$\rightarrow$ K

Electrolysis of fused halides is one of the ways to extract alkali and alkaline-earth metals.

But the reaction with fluoride is not a practical one, as fluorine produced at anode is highly corrosive.

4. For AgNO₃ solution, assuming by default that the 1.0 M AgNO₃ is present in aqueous medium.

Therefore,

AgNO₃ is dissociated as: AgNO₃$\rightarrow$ Ag⁺ + NO₃^-

Reduction reaction at cathode is: Ag⁺ + e^-$\rightarrow$ Ag

And, self ionization of water will take place, liberating oxygen gas at anode.

2 H₂O $\rightarrow$ 4H⁺ + O₂ + 4e^-

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