Question

Part A: What substance is produced at the cathode during the electrolysis of molten calcium bromide, CaBr2? Assume standard conditions. Express your answer as a chemical formula.

Part B: What substance is produced at the anode during the electrolysis of molten calcium bromide, CaBr2? Assume standard conditions.

Part C: What substance is produced at the cathode during the electrolysis of a mixture of molten calcium bromide, CaBr2(l), and molten magnesium iodide MgI2(l),? Assume standard conditions. Express your answer as a chemical formula.

Part D: What substance is produced at the anode during the electrolysis of a mixture of molten calcium bromide, CaBr2(l), and molten magnesium iodideMgI2(l),? Assume standard conditions.

 

Answer 1

Concepts and reason

Generally, the electrochemical cell contains two electrode half-cells. One is the cathode (negative electrode), and another one is the anode (positive electrode). In any chemical reaction, cations and anions are present. The element by losing electrons to form cation and element by gaining of electrons form anion. Since the anode is a positive electrode, and the cathode is a negative electrode in an electrochemical cell, cations move towards the cathode, and anions move towards the anode. Electrons always are moving from anode to cathode. The moment of electrons from one element to another in a reaction is known as oxidation-reduction reaction ("redox reaction"). The loss of electrons is called oxidation, and the gain of electrons is called reduction. Therefore, cathode reduction takes place, and anode oxidation takes place.

Fundamentals

In the given chemical reaction, we must know cations and anions. Since cations are moving towards the cathode (negative electrode), anions move towards the anode (positive electrode). 

Oxidation-half cell reaction: \(\mathrm{X} \rightarrow \mathrm{X}^{+}+\mathrm{e}^{-}\) 

Reduction-half cell reaction: \(\mathrm{X}+\mathrm{e}^{-} \rightarrow \mathrm{X}\) 

From this, we can explain the substance formed at the cathode and anode.

 

First consider the reactant species and write the chemical reaction. Write the ionization reaction of molten calcium bromide at STP conditions. STP conditions means at standard temperature \((273 \mathrm{~K})\) and standard pressure (1 \(\mathrm{atm}\) ). The ionization of calcium bromide is, \(\mathrm{CaBr}_{2}(a q) \rightarrow \mathrm{Ca}^{+2}(a q)+2 \mathrm{Br}^{-}(a q)\)

since cations are moving towards cathode and anions are moving towards anode, \(\mathrm{Ca}^{+2}\) is towards cathode and \(\mathrm{Br}^{-}\) is towards anode. At cathode reduction takes place. Thus, \(\mathrm{Ca}^{+2}\) undergo reduction itself acts as a oxidizing agent. \(\mathrm{Ca}^{+2}(a q)+2 \mathrm{e}^{-} \rightarrow \mathrm{Ca}(s)\)

Therefore, the substance formed at cathode is calcium metal (Ca).

Part A The substance formed at the cathode during the electrolysis of molten calcium bromide is Calcium (Ca) metal.

Since at cathode always reduction takes place and at the anode, oxidation occurs in a chemical reaction.

 

Consider the two cations formed in the two ionization reactions of two substances are \(\mathrm{Ca}^{+2}\) and \(\mathrm{Mg}^{+2}\). Write the possible reactions of two cations at cathode as follows:

\(\mathrm{Ca}^{+2}(a q)+2 \mathrm{e}^{-} \rightarrow \mathrm{Ca}(s) \mathrm{E}_{\mathrm{Ca}^{+2} / \mathrm{Ca}}^{0}=-2.87 \mathrm{~V}\)

\(\mathrm{Mg}^{+2}(a q)+2 e^{-} \rightarrow \mathrm{Mg}(\mathrm{s}) \mathrm{E}_{\mathrm{Mg}^{+2} / \mathrm{Mg}}^{0}=-2.37 \mathrm{~V}\)

since \(\mathrm{Mg}^{+2}\) has low electronegative electrode potential, \(\mathrm{Mg}^{+2}\) will be reduced easily. Therefore, the substance produced at the cathode is magnesium (Mg) metal.

Part \(C\)

The substance formed at the cathode during the electrolysis of mixture of molten calcium bromide ( \(\mathrm{CaBr}_{2}\) ) and molten magnesium iodide ( \(\mathrm{MgI}_{2}\) ) at STP conditions is magnesium (Mg) metal.

The less negative reduction potential cation acts as a strong reducing agent compared to that of at cation with high negative reduction potential. And at the cathode, reduction always takes place, and oxidation takes place in a chemical reaction at the anode.

 

Consider the two anions formed in the two ionization reactions of two substances are \(\mathrm{Ca}^{+2}\) and \(\mathrm{Mg}^{+2}\). Write the possible reactions of two anions at anode as follows:

\(2 \mathrm{Br}^{-}(\mathrm{aq}) \rightarrow \mathrm{Br}_{2}(\mathrm{~g})+2 \mathrm{e}^{-} \mathrm{E}_{\mathrm{Br}_{2} / \mathrm{Br}^{-}}^{0}=1.09 \mathrm{~V}\)

\(2 \mathrm{I}^{-}(a q) \rightarrow \mathrm{I}_{2}(\mathrm{~s})+2 \mathrm{e}^{-} \mathrm{E}_{\mathrm{I}_{2} / \mathrm{I}^{-}}^{0}=0.54 \mathrm{~V}\)

since \(\mathrm{I}^{-}\) has low electropositive reduction potential, \(\mathrm{I}^{-}\) will be oxidized easily. Therefore, the substance produced at the cathode is iodide ( \(\mathrm{I}_{2}\) ) gas.

Part D The substance formed at the anode during the electrolysis of mixture of molten calcium bromide ( \(\mathrm{CaBr}_{2}\) ) and molten magnesium iodide ( \(\mathrm{MgI}_{2}\) ) at STP conditions is iodine \(\left(\mathrm{I}_{2}\right.\) ) gas.

The less positive reduction potential anion acts as a strong oxidizing agent compared to that of the anion with a high positive reduction potential. And at the cathode, reduction always takes place, and oxidation takes place in a chemical reaction at the anode.