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For the last question, I believe the latter would be true because if the anions produced at the negative electrode react with the cations that are attracted to the electrode, the cations that are attracted to the electrode would not be available to accept electrons and hence the current would become increasingly restricted until it stops all together. Could someone negate or confirm this reasoning?

You need to do more research to understand what an electrolytic & galvanic cell is and how it works. Especially salt bridges. A couple of links to check out would be.

• Why the salts in a salt bridge?
• Voltaic and electrolytic cells
• Difficulties understanding how salt bridge works
• Why is it important to use a salt bridge in a voltaic cell? Can a wire be used?

Don't be limited to those above, keep searching until you solve your dilemma.

Now for the Hard Part

To best explain this I will work you through a practice problem. It is not a perfect algorithm to write down, but this way you will get a general idea.

Two electrodes are inserted into a solution of $\ce{NiF2}$ and a current of 2.2 A is run through them. A list of standard reduction potentials is as follows.

$$\ce{O2_{(g)} + 4H+ + 4e- -> H2O_{(l)}}~~~~~~~~~~~~~~~~~1.23V$$ $$\ce{F2_{(g)} + 2e- -> 2F-}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~2.87V$$ $$\ce{2H2O{(l)} + 2e- -> H2_{(g)} + 2OH-}~~~~~-0.83V$$ $$\ce{Ni2+ + > 2e- -> Ni_{(s)}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-0.25V$$

So generally I would usually start of with determining with what is going on, you know that $\ce{Ni^{+2}}$ & $\ce{2F-}$ are present in the solution ($\ce{H2O}$).

We can eliminate the third reaction since Fluorine already exists reduced, and it does not like being oxidized. Then you have Nickel and Water, which of them is most likely to be reduced? Nickel.

Now we must decide the compound being oxidized. Fluorine and Nickel are out of the question so that leaves us with Equation 1 and 3. As you can see even if we were to flip Eq. 3 it would make no sense since there is no $\ce{OH-}$ present in water. So that leaves us with Eq. 1. We must flip this Equation for the reaction to make sense. That leaves us with:

Equation 1(this one is flipped) and 4. $\ce{Ni^{+2}}$ is being reduced while $\ce{H2O}$ is being oxidized.

In General...

1. Eliminate which reaction cannot take place.
2. Look at what is left, and judge which reaction can best represent the reaction taking place in the cell.

With practice you will be able to solve these kinds of problems. If this is not what you are looking for, then hopefully you learned about these kinds of problems. :)

For the last question, I believe the latter would be true because if the anions produced at the negative electrode react with the cations that are attracted to the electrode, the cations that are attracted to the electrode would not be available to accept electrons and hence the current would become increasingly restricted until it stops all together. Could someone negate or confirm this reasoning?

You need to do more research to understand what an electrolytic and galvanic cell is and how it works. Especially salt bridges. A couple of links to check out would be.

• Why the salts in a salt bridge?
• Voltaic and electrolytic cells
• Difficulties understanding how salt bridge works
• Why is it important to use a salt bridge in a voltaic cell? Can a wire be used?

Don’t be limited to those above, keep searching until you solve your dilemma.

Now for the Hard Part

To best explain this I will work you through a practice problem. It is not a perfect algorithm to write down, but this way you will get a general idea.

Two electrodes are inserted into a solution of $\ce{NiF2}$ and a current of $2.2\ \mathrm{A}$ is run through them. A list of standard reduction potentials is as follows.

$$\begin{alignat}{2} \ce{O2_{(g)} + 4H+ + 4e- &-> H2O_{(l)}}\quad&&1.23\ \mathrm{V}\\ \ce{F2_{(g)} + 2e- &-> 2F-}\quad&&2.87\ \mathrm{V}\\ \ce{2H2O{(l)} + 2e- &-> H2_{(g)} + 2OH-}\quad&&{-}0.83\ \mathrm{V}\\ \ce{Ni^2+ + 2e- &-> Ni_{(s)}}\quad&&{-}0.25\ \mathrm{V} \end{alignat}$$

So generally I would usually start of with determining with what is going on, you know that $\ce{Ni^2+}$ and $\ce{F-}$ are present in the solution ($\ce{H2O}$).

We can eliminate the third reaction since fluorine already exists reduced, and it does not like being oxidized. Then you have nickel and water, which of them is most likely to be reduced? Nickel.

Now we must decide the compound being oxidized. Fluorine and nickel are out of the question so that leaves us with Equation 1 and 3. As you can see even if we were to flip Eq. 3 it would make no sense since there is no $\ce{OH-}$ present in water. So that leaves us with Eq. 1. We must flip this equation for the reaction to make sense. That leaves us with:

Equation 1  (this one is flipped) and 4. $\ce{Ni^2+}$ is being reduced while $\ce{H2O}$ is being oxidized.

In general …

1. Eliminate which reaction cannot take place.
2. Look at what is left, and judge which reaction can best represent the reaction taking place in the cell.

With practice you will be able to solve these kinds of problems. If this is not what you are looking for, then hopefully you learned about these kinds of problems. :)

For the last question, I believe the latter would be true because if the anions produced at the negative electrode react with the cations that are attracted to the electrode, the cations that are attracted to the electrode would not be available to accept electrons and hence the current would become increasingly restricted until it stops all together. Could someone negate or confirm this reasoning?

You need to do more research to understand what an electrolytic & galvanic cell is and how it works. Especially salt bridges. A couple of links to check out would be.

• Why the salts in a salt bridge?
• Voltaic and electrolytic cells
• Difficulties understanding how salt bridge works
• Why is it important to use a salt bridge in a voltaic cell? Can a wire be used?

Don't be limited to those above, keep searching until you solve your dilemma.

Now for the Hard Part

To best explain this I will work you through a practice problem. It is not a perfect algorithm to write down, but this way you will get a general idea.

Two electrodes are inserted into a solution of $\ce{NiF2}$ and a current of 2.2 A is run through them. A list of standard reduction potentials is as follows.

$$\ce{O2_{(g)} + 4H+ + 4e- -> H2O_{(l)}}~~~~~~~~~~~~~~~~~1.23V$$ $$\ce{F2_{(g)} + 2e- -> 2F-}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~2.87V$$ $$\ce{2H2O{(l)} + 2e- -> H2_{(g)} + 2OH-}~~~~~-0.83V$$ $$\ce{Ni2+ + 2e- -> Ni_{(s)}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-0.25V$$

So generally I would usually start of with determining with what is going on, you know that $\ce{Ni^{+2}}$ & $\ce{2F-}$ are present in the solution ($\ce{H2O}$).

We can eliminate the third reaction since Fluorine already exists reduced, and it does not like being oxidized. Then you have Nickel and Water, which of them is most likely to be reduced? Nickel.

Now we must decide the compound being oxidized. Fluorine and Nickel are out of the question so that leaves us with Equation 1 and 3. As you can see even if we were to flip Eq. 3 it would make no sense since there is no $\ce{OH-}$ present in water. So that leaves us with Eq. 1. We must flip this Equation for the reaction to make sense. That leaves us with:

Equation 1(this one is flipped) and 4. $\ce{Ni^{+2}}$ is being reduced while $\ce{H2O}$ is being oxidized.

In General...

1. Eliminate which reaction cannot take place.
2. Look at what is left, and judge which reaction can best represent the reaction taking place in the cell.

With practice you will be able to solve these kinds of problems. If this is not what you are looking for, then hopefully you learned about these kinds of problems. :)

For the last question, I believe the latter would be true because if the anions produced at the negative electrode react with the cations that are attracted to the electrode, the cations that are attracted to the electrode would not be available to accept electrons and hence the current would become increasingly restricted until it stops all together. Could someone negate or confirm this reasoning?

You need to do more research to understand what an electrolytic & galvanic cell is and how it works. Especially salt bridges. A couple of links to check out would be.

• Why the salts in a salt bridge?
• Voltaic and electrolytic cells
• Difficulties understanding how salt bridge works
• Why is it important to use a salt bridge in a voltaic cell? Can a wire be used?

Don't be limited to those above, keep searching until you solve your dilemma.

Now for the Hard Part

To best explain this I will work you through a practice problem. It is not a perfect algorithm to write down, but this way you will get a general idea.

Two electrodes are inserted into a solution of $\ce{NiF2}$ and a current of 2.2 A is run through them. A list of standard reduction potentials is as follows.

$$\ce{O2_{(g)} + 4H+ + 4e- -> H2O_{(l)}}~~~~~~~~~~~~~~~~~1.23V$$ $$\ce{F2_{(g)} + 2e- -> 2F-}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~2.87V$$ $$\ce{2H2O{(l)} + 2e- -> H2_{(g)} + 2OH-}~~~~~-0.83V$$ $$\ce{Ni2+ + > 2e- -> Ni_{(s)}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-0.25V$$

So generally I would usually start of with determining with what is going on, you know that $\ce{Ni^{+2}}$ & $\ce{2F-}$ are present in the solution ($\ce{H2O}$).

We can eliminate the third reaction since Fluorine already exists reduced, and it does not like being oxidized. Then you have Nickel and Water, which of them is most likely to be reduced? Nickel.

Now we must decide the compound being oxidized. Fluorine and Nickel are out of the question so that leaves us with Equation 1 and 3. As you can see even if we were to flip Eq. 3 it would make no sense since there is no $\ce{OH-}$ present in water. So that leaves us with Eq. 1. We must flip this Equation for the reaction to make sense. That leaves us with:

Equation 1(this one is flipped) and 4. $\ce{Ni^{+2}}$ is being reduced while $\ce{H2O}$ is being oxidized.

In General...

1. Eliminate which reaction cannot take place.
2. Look at what is left, and judge which reaction can best represent the reaction taking place in the cell.

With practice you will be able to solve these kinds of problems. If this is not what you are looking for, then hopefully you learned about these kinds of problems. :)

For the last question, I believe the latter would be true because if the anions produced at the negative electrode react with the cations that are attracted to the electrode, the cations that are attracted to the electrode would not be available to accept electrons and hence the current would become increasingly restricted until it stops all together. Could someone negate or confirm this reasoning?

You need to do more research to understand what an electrolytic & galvanic cell is and how it works. Especially salt bridges. A couple of links to check out would be.

• Why the salts in a salt bridge?
• Voltaic and electrolytic cells
• Difficulties understanding how salt bridge works
• Why is it important to use a salt bridge in a voltaic cell? Can a wire be used?

Don't be limited to those above, keep searching until you solve your dilemma.

Now for the Hard Part

To best explain this I will work you through a practice problem. It is not a perfect algorithm to write down, but this way you will get a general idea.

Two electrodes are inserted into a solution of $\ce{NiF2}$ and a current of 2.2 A is run through them. A list of standard reduction potentials is as follows.

$$\ce{O2_{(g)} + 4H+ + 4e- -> H2O_{(l)}}~~~~~~~~~~~~~~~~~1.23V$$ $$\ce{F2_{(g)} + 2e- -> 2F-}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~2.87V$$ $$\ce{2H2O{(l)} + 2e- -> H2_{(g)} + 2OH-}~~~~~-0.83V$$ $$\ce{Ni2+ + 2e- -> Ni_{(s)}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-0.25V$$

So generally I would usually start of with determining with what is going on, you know that $\ce{Ni+2}$ & $\ce{2F-}$ are present in the solution ($\ce{H2O}$).

We can eliminate the third reaction since Fluorine already exists reduced, and it does not like being oxidized. Then you have Nickel and Water, which of them is most likely to be reduced? Nickel.

Now we must decide the compound being oxidized. Fluorine and Nickel are out of the question so that leaves us with Equation 1 and 3. As you can see even if we were to flip Eq. 3 it would make no sense since there is no OH- present in water. So that leaves us with Eq. 1. We must flip this Equation for the reaction to make sense. That leaves us with:

Equation 1(this one is flipped) and 4. $\ce{Ni^{+2}}$ is being reduced while $\ce{H2O}$ is being oxidized.

In General...

1. Eliminate which reaction cannot take place.
2. Judging from what is left, look at what is left, and judge which reaction can best represent the reaction taking place in the cell.

With practice you will be able to solve these kinds of problems. If this is not what you are looking for, then hopefully you learned about these kinds of problems. :)

For the last question, I believe the latter would be true because if the anions produced at the negative electrode react with the cations that are attracted to the electrode, the cations that are attracted to the electrode would not be available to accept electrons and hence the current would become increasingly restricted until it stops all together. Could someone negate or confirm this reasoning?

You need to do more research to understand what an electrolytic & galvanic cell is and how it works. Especially salt bridges. A couple of links to check out would be.

• Why the salts in a salt bridge?
• Voltaic and electrolytic cells
• Difficulties understanding how salt bridge works
• Why is it important to use a salt bridge in a voltaic cell? Can a wire be used?

Don't be limited to those above, keep searching until you solve your dilemma.

Now for the Hard Part

To best explain this I will work you through a practice problem. It is not a perfect algorithm to write down, but this way you will get a general idea.

Two electrodes are inserted into a solution of $\ce{NiF2}$ and a current of 2.2 A is run through them. A list of standard reduction potentials is as follows.

$$\ce{O2_{(g)} + 4H+ + 4e- -> H2O_{(l)}}~~~~~~~~~~~~~~~~~1.23V$$ $$\ce{F2_{(g)} + 2e- -> 2F-}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~2.87V$$ $$\ce{2H2O{(l)} + 2e- -> H2_{(g)} + 2OH-}~~~~~-0.83V$$ $$\ce{Ni2+ + 2e- -> Ni_{(s)}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-0.25V$$

So generally I would usually start of with determining with what is going on, you know that $\ce{Ni^{+2}}$ & $\ce{2F-}$ are present in the solution ($\ce{H2O}$).

We can eliminate the third reaction since Fluorine already exists reduced, and it does not like being oxidized. Then you have Nickel and Water, which of them is most likely to be reduced? Nickel.

Now we must decide the compound being oxidized. Fluorine and Nickel are out of the question so that leaves us with Equation 1 and 3. As you can see even if we were to flip Eq. 3 it would make no sense since there is no $\ce{OH-}$ present in water. So that leaves us with Eq. 1. We must flip this Equation for the reaction to make sense. That leaves us with:

Equation 1(this one is flipped) and 4. $\ce{Ni^{+2}}$ is being reduced while $\ce{H2O}$ is being oxidized.

In General...

1. Eliminate which reaction cannot take place.
2. Look at what is left, and judge which reaction can best represent the reaction taking place in the cell.

With practice you will be able to solve these kinds of problems. If this is not what you are looking for, then hopefully you learned about these kinds of problems. :)