Correct answer: In a voltaic cell, the electrochemical reaction is spontaneous. In an electrolytic cell, it is nonspontaneous.
Why?:
A voltaic cell uses the energy produced by a chemical reaction to generate electricity. The chemical reaction in the voltaic cell is spontaneous, and produces an electrical current, converting chemical energy to electrical energy.
In contrast the electrolytic cell used in the Hall process consumes electrical energy to drive a chemical reaction. An amount of energy is still needed to drive the reaction process. It is not spontaneous.
Answering this question:
In answering this question, it is important to have knowledge on how voltaic and electrolytic cells work. A voltaic cell is spontaneous because it occurs without the "push" of an outside force. Moreover, it produces energy instead of consuming it. An electrolytic cell on the other hand needs a driving force before it can start. So it is nonspontaneous.
To sum it up, an electrochemical reaction is spontaneous in a voltaic cell while nonspontaneous in an electrochemical cell.
A half-reaction shows the change in oxidation state of an ion. Reduction is a process wherein an atom or ion gains electrons. If an atom or ion gains electrons, the oxidation state decreases. The number of electrons gained is the same as decrease in the oxidation state.
Answering this question:
Since reduction involves the gain of electrons, there should be a decrease in the oxidation state. And since electrons are gained, they should appear on the reactant side. Write first the equation representing the transformation of Al+3 to Al.
Al3+→ Al
This equation is not charge-balanced. We start with a +3 charge on the left side and end up with no charge on the product side. We need to add electrons to the left side to make this equation balanced.
Al3+ + 3e-→ Al
This is the complete half-reaction equation.
So the correct balance equation is Al+3 + 3e- → Al
Cryolite is an ionic compound with the formula Na3AlF6. Of these elements, fluorine is by far the most electronegative. In fact, it is the most electronegative element. In ionic compounds it will accept one additional electron to fill its octet and achieve a noble gas configuration. This extra electron will give fluorine a -1 charge. This is equivalent to saying that fluorine has a -1 oxidation state.
Sodium on the other hand is a metal in the first column of the periodic table. It loses an electron to form a Na +1 cation. This is equivalent to saying that Na has a +1 oxidation state.
Aluminum belongs to group IIIA or group 13 of the periodic table. It has 5 valence electrons. This means that aluminum is capable of accepting 3 more electrons for a complete octet. Therefore it has an oxidation state of +3.
The molecular formula as a whole is uncharged or neutral. It has an overall charge of zero. We also know that it contains three positive charges from its sodium cations and six negative charges from its fluorine anions. This means that aluminum must have a +3 charge to make the molecule neutral.
Answering this question:
Sodium, aluminum and fluorine are all representative elements. This means that they belong to the s-block and/or p-block of the periodic table. The oxidation states of the representative elements can be predicted through their group numbers. Elements under the group IA or 1 of the periodic table has an oxidation state of +1. Group IIA or 2 has +2. Group IIIA or 13 has +3. Group IVA or 14 has +4 or -4. Group VA or 15 has -3. Group VIA or 16 has -2. Group VIIA or 17 has -1.
Sodium is under group 1 so it has an oxidation state of +1. There are 3 sodium atoms in Cryolite so it would be a total of a +3 oxidation state for all the sodium atoms. Fluorine is under group 17 so it has an oxidation state of -1. There are 6 fluorine atoms, so it would be a total of -6 oxidation state for fluorine. Aluminum is under group 13 so it has an oxidation state of +3. There is only one aluminum atom in Cryolite. To verify, we add all the oxidation state. Since Cryolite is a neutral compound, the sum of the oxidations states must be zero.
So the correct answer is Na = +1, F = -1, and Al = +3.
The electrochemical cell involves the movement of charged species through a solution and the movement of electrons through a conductor. Generally, electrochemical cells involve the gain or loss of electrons and not of protons.
The anode is where oxidation takes place in the electrochemical cell. Reduction occurs in the cathode.
Answering this question:
It would be helpful to remember the phrase "red cat" in studying electrochemical cells. This could give you the idea that reduction occurs in the cathode. And since there are two electrochemical processes occurring in an electrochemical cell, the other one, which is oxidation, occurs in the anode. With the word reduction, this means reduction of charges. An example would be reduction of Ag+ ion to Ag metal. Here, the +1 oxidation state of Ag+ is reduced to 0. It has gained an electron. The opposite, which is the loss of electrons, is true with oxidation. This will enable us to conclude that (2) is the correct answer. Furthermore, electrochemical cells are all about the gaining and losing of electrons. Protons are not being gained or lost.
Redox stands for reduction-oxidation reactions. This also refers to changes in oxidation states in different elements. A change in an oxidation state would be through the transfer of electrons from one atom or ion to another.
Answering this question:
Obtain the oxidation states of each of the atoms on the reactant side. After which, obtain the oxidation states of the atoms on the product side. If it is a redox reaction, then there should be a change in the oxidation state of at least one atom. The only answer choice that involves a change in oxidation state is (3). The oxidation state of copper changes from 2+ to 0. The oxidation state of carbon changes from +2 to +4. Thus, (3) is the only acceptable answer. All the other reactions involve different ions swapping places with each other in acid-base reactions or double displacement reactions. However, none of them involve changes in oxidation states.
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