Galvanic Cell - Online Tutor, Practice Problems & Exam Prep

• A galvanic cell, also known as a voltaic cell, is a spontaneous electrochemical cell that produces electricity by converting stored chemical energy into electrical energy through redox reactions, where oxidation occurs at the anode (losing electrons) and reduction at the cathode (gaining electrons).
• The major components of a galvanic cell include the negatively charged anode where oxidation occurs, the positively charged cathode where reduction occurs, a salt bridge that connects both half-cells and allows the flow of neutral ions to maintain charge balance, and a voltmeter to measure the voltage produced.
• Over time, the anode in a galvanic cell loses mass due to the loss of electrons and dissolution into ions, while the cathode gains mass as it attracts cations from the solution, which plate out on its surface.
• The salt bridge plays a crucial role in maintaining electrical neutrality by allowing the flow of ions between the two half-cells, preventing the buildup of charge that would otherwise stop the flow of electrons.
• Galvanic cells operate under spontaneous redox reactions with a positive standard cell potential, indicating a release of energy (negative Gibbs free energy) and an increase in entropy.
• When a galvanic cell has discharged all its electricity, it reaches equilibrium, characterized by zero standard cell potential, zero change in Gibbs free energy, and an equilibrium constant (K) equal to 1, effectively rendering the cell 'dead' in chemical terms.

The Galvanic Cell

Now recall there are two types of electrochemical cells, one that produces electricity and one that consumes it. In this video we're going to take a look at galvanic cells. Remember, galvanic cell, or sometimes called a voltaic cell, is a spontaneous electrochemical cell that produces electricity. And since it's making electricity, it's basically a battery.

Here we're going to say the way it does this is it uses stored chemical energy and converts it into electrical energy. I remember it utilizes this through the through redox reaction. We have oxidation. We're losing electron reduction where we're gaining an electron. The transferring of this electron through a conductive wire is what produces our electricity.

So just remember our galvanic cell is spontaneous, it produces electricity. So in essence it's a battery. It's utilizing the chemical energy of a redox reaction and converting it into a electrical energy source.

Galvanic Cell Example

Here it says the purpose of a galvanic cell is 2A purify solids, B allow for only oxidation, C generate electricity, D to consume electricity. All right. So first of all, we never talked about purifying solids in any way. So we know that's not going to be an answer.

Galvanic saw is a type of electrochemical cell where both oxidation and reduction occur. And then remember we have two types of electrochemical cells, one that can produce electricity and one that consumes it. Galvanic cells happen to be the ones that produce electricity or generate electricity in this case.

So C would be our answer, D would not be our answer. So our final option here is choice C.

Galvanic Cell Components

Now the major components of a galvanic cell are given as the following. So first we have our atom. Our anode for galvanic cell is negatively charged. This is the metal electrode and compartment where oxidation occurs. Remember oxidation means we are losing electrons. Then we have our cathode compartment. This is our positively charged electrode. So this is a metal electrode in compartment where reduction occurs. Reduction means that we are gaining electrons.

So if we take a look here, we have our Galvan Excel. We have our annual compartment which is the anode. Here is this metal rod. In this case it's a zinc metal rod. The cathode here is positively charged, Represents our copper electrode, this copper rod. Now oxidation occurs at the anode. Reduction occurs at the cathode. So if we're losing electrons, electrons are literally leaving this metal zinc electrode and they are heading towards my copper electrode.

Now besides the anode and the cathode, we have a salt bridge. This is a tube that connects both half cells to one another and allows for the flowing of neutral ions. Now, neutral ions may not make sense, right? Apina ion be neutral has a positive and a negative charge. Well here when we say neutral ions, they're just a special type of spectator ions within the solution that possess no acidic or basic properties. Ions can be acidic, basic or neutral depending on their origin. Here we're only using the ions that are neutral in nature, they're not acidic or basic.

Now here, what's the whole purpose of the salt bridge? Its purpose is to neutralize the buildup of cations within the anion half cell. So here this tube would represent my salt bridge. This tube is what connects both electrolytic well electrolyte solutions to one another. It's connecting both half cells to each other. Within this we have neutral ions. Typical neutral ions are sodium ions, potassium ions, bromide ions. Here we're going to say that the negative negatively charged ones like bromide ions would travel over here to the atom compartment and then the NA ones, Na or K positive would go towards this side towards the cathode compartment.

Later on when we talk about electrodes a little bit more specifically, we'll talk about why this is necessary. But for right now I just realize that the salt bridge is that tube that connects both half cells to each other. Within it we have neutral ions. Finally, we have our voltmeter, the device that record the amount of electricity generated by the galvanic cell. It's a voltmeter, so it measures voltage which is capital V And here's the voltmeter here. This little circular thing here we're going to put off because it's measuring voltage O as electrons, as electrons travel from the anode to the cathode, So on this conductive wire we have electrons traveling, we're generating electricity. This voltmeter will give us a reading of value of the amount of electricity that's being produced, right? So here we just have the fundamental and major components of any given galvanic slash. Voltaic so.

Galvanic Cell Example

Which of the following statements is true for a salt bridge with the following redox reaction, right? So here we're talking about what side would my bromide ions flow. C contains neutral atoms that interact with the ions in both half cell compartments. We know that the salt bridge itself doesn't contain atoms which are neutral, it deals with neutral ions. So this is not correct. It's not atoms, it's ions, it's sodium ions flow to the magnesium cell half cell.

All right. So we know that the negatively charged ones like bromide ions would flow towards the add on compartment and the positive ones, in this case sodium ions would flow to the Catholic compartment. Remember the anode is where oxidation occurs and the cathode is where reduction occurs. Here, if we take a look at our overall redox reaction, we have magnesium going from zero to +2, so it's oxidation number increased. So it's being oxidized, which means it is the anode. OK, so this will be the magnesium compartment, or half some.

Cadmium goes from +2 to 0, so its oxidation number reduces. Therefore, it's the cathode. So we have our cathode compartment being the compartment being the captive. So if we look here, it's bromide ions will flow with the magnesium half cells. So bromide is in fact, yes, it should flow to the magnesium side because the magnesium compartment represents our anode. So this is true.

It wouldn't flow to the cadmium half cell because that's where reduction occurs. It's sodium ions flow. It should flow to the magnesium half cell. No, sodium should be flowing towards the cathode side which deals with the cadmium half some. So out of all our options, only A is correct.

Galvanic Cell Electrodes

In this video we're going to take a look at the galvanic cell electrodes. Here we're going to say recall all electrochemical cells possess 2 half cell compartments, one for oxidation and the other for reduction. Remember the anode equals oxidation and a way to remember that is an Aux and then the cathode equals reduction. A way to remember that is red cat.

Now, before we start filling in the rest of this, let's just talk about some key things. We're going to say for a galvanic cell, in terms of its electrodes, the anode is negatively charged and the cathode is positively charged. What effect will this have on the electrodes over time? All right, so we know the anode is where oxidation occurs. Here we're looking at the zinc electrode much more closely and we're going to say that zinc is being oxidized to zinc on. Now what's happening here is electrons are literally leaving the electrode and traveling towards the cathode.

The cathode is positively charged, it's gaining electrons. Here copper 2 plus would be reduced to copper solid. Now what effect is this going to have? We're going to say here we see our zinc solid surface here and what's happening is we have electrons traveling up the electrode leaving the electrode to go to the cathode. What effect does this have? Well my Zinc is losing electrons. As a result of this it starts to produce Zinc 2 plus ions which become more and more dissolved within my solution. So electrons don't weigh very much. But overtime if you keep losing electrons those little masses do add up.

Eventually, you're going to lose a good chunk of this anime, and let's say that this is what's left. The rest of it's gone. You've been losing electrons consistently over time, so you're you're losing the size of your atom. So here we'd say that the anode is our negative electrode. It loses electrons and overtime it causes a decrease in mass. Here we say that the anode dissolves away. What's happening to the captain? Well, the cathode is our positive electrode, yes, but over time it's gaining more and more electrons. These electrons are depositing themselves onto the surface of the cathode.

It is in a solution of its own ions, in this case copper 2 plus ions. The surface is becoming more and more negatively charged with the electrons and gains, which is going to attract these positive ions floating around. They're going to move towards the surface. Two plus ion seeing electrons on the surface connects to them. Negative charge, positive charge neutralize each other. So the copper ions are going to start to coat the surface of this electrode. So over time, the cathode is going to get bulkier, it's going to get bigger, it's gaining mass. We're going to say the cathode gains electrons, causing the surface to become more negatively charged, attracting cations to it. And this over time causes an increase in mass.

Here we would say that the cathode plates out, one gets skinnier, 1 gets bigger. Now here we talked about our salt bridge in earlier videos. Remember a salt bridge there just has neutral ions within it. So we know that the negative ions here would go towards the anode compartment and positive ions go here towards the Catholic apartment. But why exactly are these negative ions going towards the anode side? Well, what's happening? More and more positive ions are coming off of the anode, so your solutions can become saturated with these positive ions. So much so that if too many of them are hanging around, the electrolyte solution is super positive. These negative electrons won't want to leave.

They'll be like why should I leave if my solution has all these dissolved positive ions around opposite charges attract. That's why we need a salt bridge. The salt bridge has these negative ions go into the solution, neutralizing these positive cations that are being produced, keeping their concentration down, and in that way the electrons are not conflicted. Do I go towards the cathode, which is becoming less positive over time as it gains more and more electrons on its surface? Or should I stick around and hang out with these dissolved cat ions in my solution? Here electrons don't have to worry about that because again, the negative ions within the salt bridge are neutralizing. These excess cations are being produced in the anode chamber.

Now here if we're talking about reactions, remember zinc is being oxidized to zinc. So what would that look like? We'd have zinc solid benoxidized to zinc 2 plus aqueous and the two electrons that are lost in the Catholic compartment, copper 2 plus is reduced, which means it gains 2 electrons to produce copper solid. We can't slot intermediates, which in this case are the electrons, and what's left behind gives us our overall reaction. So here we have zinc 2 plus or zinc solid plus. Copper 2 plus gives me zinc 2 plus plus copper solid. So this would be the overall reaction that occurs from this electrochemical cell dealing with our zinc electrode and copper electrodes.

Zn 2 + + Cu 2 + → Zn 2 + + Cu 2 +

Galvanic Cell Example

How many electrons will be transferred between a sodium and gallium electrode from the following? So here we have gallium ion absorbing 3 electrons to become gallium solid. Gallium goes from +3 to 0 so it's oxidation number was reduced so it represents the catheter. Remember we know that this is the cathode because the catheters were a reduction occurs. Also, we know that this is a reduction because the electrons are reactants.

Sodium on the other hand goes from a neutral charge to +1. Its oxidation number increased, so it's been oxidized, which means it's the anode. We also know it's an oxidation because your electrons are products. Now remember your electrons are intermediates and they have to cancel out with each other. Here they can't because one says 3 electrons and one says only one.

You would multiply this equation by three so that both half reactions have the same number of electrons. Doing that tells me that three electrons are transferred between the sodium and gallium electrodes. 3 electrons are lost by the anode half cell and given over to the cathode half cell, right? So 3 electrons will be transferred within this particular example question.

Galvanic Cells and Spontaneity

Now, in terms of spontaneity, we can say that a galvanic cell uses spontaneous redox reactions to produce and discharge electricity. So Galvanic cells again are batteries. Now here they can make electricity. Discharging means that the electron, the energy that we've made, the electricity that we've made, we start to use it and that means we're discharging it.

Now here we're going to say all spontaneous reactions have a positive standard cell potential value. So here we're going to talk about galvanic cells. They use spontaneous redox reactions and how does that relate to our different variables? We're also going to talk about what happens when we're at equilibrium. So the variables we're going to look at here are changes in your standard Gibbs free energy, changes in your standard entropy total, so energy of the universe, our equilibrium constant K, equilibrium constant versus our reaction quotient Q, and of course our standard cell potential.

We already said that galvanic cells use spontaneous redox reactions and that all spontaneous redox reactions have a positive standard cell potential. So that means that this would have to be greater than 0. We also talked about these other variables in other chapters. For it to be spontaneous, the change in your standard Gibbs free energy has to be less than 0. The change in your standard total entropy has to be greater than 0, your equilibrium constant K has to be greater than one. And then if we're talking about K versus Q here, we'd have to say that our equilibrium constant K would have to be greater than Q.

Now what happens if our galvanic cell makes all this electricity and decides to use it, to use all of it? It's completely discharged all of its electricity. Well, in that case, what happens when you use all the electricity of a battery? It goes dead. But in chemistry, we don't like to say dead, we like to say that it's reached equilibrium. So at equilibrium, our galvanic cell has used up all of its electricity that's produced. So in essence, it's a dead battery.

Here, that would just mean that each of these variables are equal to these same values. So change in standard Gibbs free energy is equal to 0. Change in my standard entropy total is equal to 0, your equilibrium constant K is equal to 1, K is equal to Q, and your standard cell potential is equal to 0. So just remember, a galvanic cell represents a spontaneous electrochemical cell because it utilizes spontaneous redox reactions. If it were to use all of the electricity that it's produced, it would reach equilibrium where this is true for each one of these variables.

Galvanic Cell Example

A reduction reaction with an equilibrium constant of 4.8×10^2. All right, so remember your equilibrium constant here is K, and we're talking about which of the following statements is true. K here is greater than one, which means that it is a spontaneous reaction. Is non spontaneous. We don't need to continue reading. We know that this is not true.

Since K is greater than one, it's spontaneous, has a negative change in my standard Gibbs free energy, and produces electricity. That is true. If it's spontaneous, it's a galvanic cell. Galvanic cells utilize a redox reaction in order to produce electricity. Here the Gibbs free energy would have to be less than 0, making it negative.

Has discharged all electricity and has a dead battery. That only be true if K was equal to 1, which it's not. Is spontaneous. OK, and has a negative standard cell potential. So the first part is true, it is spontaneous, but its standard cell potential should be positive and not negative. So here only option B is correct.

​​​​​Which of the following is false about a Voltaic cell?

• A. Anode electrode dissolves while cathode electrode plates out.
• B. It changes chemical energy into electrical energy.
• C. Anode half-cell accumulates positive charge and cathode half-cell accumulates negative charge.
• D. Half reaction with more negative reduction potential attracts electrons and undergoes reduction.

For the redox reaction label: the anode, cathode, half-reactions occurring at each half-cell, direction of electron flow, and direction of neutral ions flow.

$Ni\left(s\right)+Zn^{2+}\left(aq\right)\rightarrow Ni^{2+}\left(aq\right)+Zn\left(s\right)$

For the following redox reaction, label the anode, cathode, half reactions occurring at each half cell, direction of electron flow, and direction of ion movement. Alright. So that's a lot to do. Here we have the initial reaction: nickel in a zero oxidation state going to a +2 state. So its oxidation number is increasing, which means it's been oxidized, so that means it is the anode. Remember, "anode equals oxidation" and an easy way to remember is "AN OX". So here we'd say that this is the anode. Here, we're going to say the anode is negatively charged. Our electrode is nickel. The half reaction involved is: Ni(s)→Ni2+(aq)+2e- We're going to have nickel 2+ ions being created here. And we're going to say, remember, in our salt bridge, the negative ions from the salt bridge, such as bromide ions, are moving towards the anode compartment. Electrons are leaving the anode and they are going towards the cathode. We'll say that this side is the cathode, which is our positive electrode. Here we have zinc 2+, so its charge is 2+, its oxidation number is 2+, and it changes to its neutral form. Its oxidation number has decreased, so it's been reduced, therefore, it is the cathode. Remember, "reduction equals cathode", and an easy mnemonic is "RED CAT". Here, the electrode is zinc. We have zinc 2+ ions, which are already dissolved within the solution. In the salt bridge, our positive ion, which is typically sodium, is moving towards the cathode compartment. Again, electrons are flowing towards the cathode. Here the equation will be: Zn2+(aq)+2e-→Zn(s) So here we have determined who was the anode, which is the nickel, and who was the cathode, which is the zinc. We have outlined the half reactions for both half cells, discussed the direction of electrons, which is always from anode to cathode, and identified the movement of ions within the salt bridge. These are all the components needed to answer this question.