An electrochemical cell is based on these two half-reactions:
Ox: Sn(s) → Sn²⁺(aq, 2.00 M) + 2 e^–
Red: ClO₂(g, 0.100 atm) + e^– → ClO₂^–(aq, 2.00 M)
Calculate the cell potential at 25 °C.

Question

An electrochemical cell is based on these two half-reactions:

Ox: Sn(s) → Sn²⁺(aq, 2.00 M) + 2 e^–

Red: ClO₂(g, 0.100 atm) + e^– → ClO₂^–(aq, 2.00 M)

Calculate the cell potential at 25 °C.

Accepted Answer

Hey everyone in this example, we need to consider the following half reactions that occur in an electrochemical cell. We have our oxidation half reaction here and our reduction half reaction. And we need to determine the cell potential at 25°C. So right now we can see that all of our atoms and both half reactions balanced. However, we have different numbers of electrons in these two half reactions. We have four electrons in the in the reduction half, sorry, and two electrons in the oxidation half. So we're going to correct this by taking our oxidation half reaction and multiplying it by two. So that we now have a new reaction which I'll actually right here below where we have two moles of solid manganese producing two moles of the manganese, two plus cat ion plus two electrons. And so now that we have the electrons and sorry, that would be four electrons where now we would have equal amounts of electrons for both half reactions. So we can add up these two reactions. So we have added to this our reduction half reaction, where we have oxygen gas plus two moles of liquid water plus four electrons, yields four moles of hydroxide. And adding these up, we're going to get a final equation here where we have two moles of manganese added to one mole of oxygen gas added to two moles of water, produces two moles of manganese, cat ion Plus four moles of hydroxide and we didn't include the electrons because since they appear in both half reactions on opposite sides of these half reactions, we can just cancel them out. So this completes the addition of these two reactions. We should also recognize that because this is our oxidation half reaction. This will occur at our node. And because this is our reduction half reaction, we should recall that this occurs at our cathode of our electoral voltaic cell. And so because we know that our oxidation half reaction occurs that are an ode, we want to refer to our standard reduction potential table. We can find this table online or in our textbooks and we would see that for the oxidation of our manganese to Pluskat ion. We have a standard cell potential value equal to negative 1.18 volts. Likewise with our cathode or reduction half reaction. We would see that for the reduction of oxygen gas, we have a electrode potential value equal to 0.40 volts. And so now that we have these values from our textbooks or online, we can go ahead and calculate our standard cell potential E degree cell And recall that we can calculate this by taking our cell potential of our cathode, subtracted from the cell potential of our an ode. And so plugging in what we know from above. We can say that our standard cell potential for reaction is equal to the difference between our self potential of our cathode which we set occurs at our reduction being 0.040V. So we'll plug that in 0.40V. This is then subtracted from our self potential of our an ode which we set above occurs at our oxidation of the manganese two plus catty on being negative 1 18 volts. So we'll plug that in below. And when we get the difference of these values in our calculator we would get that are standard cell potential for reaction is equal to a value of 1.58V. So our next step is to calculate our cell potential at 25°C. To answer this prompt. And so we're going to recall our first equation where we take our or we find our cell potential. We set that equal to the standard cell potential for reaction. And this is subtracted from our constant value being 0.0592. And sorry, that's a two there. This has units of volts. And this is divided by N. Where we recall that N is going to be our number of electrons transferred in our reduction. So we can already say that N is equal to four electrons. Because above we showed that in our reduction half reaction, we have four electrons that were transferred. Given in the prompt. So to continue our nursed equation, we should then recall that This question here is multiplied by the log of our equilibrium constant Q. So we should recall that Q. R equilibrium constant is equal to the concentration of products divided by our concentration of reactant. So this is going to be equal to the concentration of the manganese two plus caddy on which is our product here. And according to our reaction here we have a coefficient of two. So we're going to raise this to a power of two and then looking at our second product, this is going to be multiplied by our concentration of hydroxide. Where according to our reaction, we have a coefficient of four. So we're going to raise this to a power of four. Then we have divided by our reactant. So recall that this only includes either a liquid or gaseous reactant. And according to our equation we only have oxygen gas as the only applicable reactant for our Equilibrium constant. And so we can say this is divided by the pressure of oxygen since it's a gas or gaseous reactant. So now that we have our nursed equation and values for N&Q outlined, we're going to go ahead and solve for our cell potential at 25°C. So we can say that our self potential is equal to Our standard cell potential which above we calculated as 1.58V. So we're going to plug that in below. This is then subtracted from Our quotient where we have our constant value 0.0592V divided by N, which we agreed is four electrons for our electrons transferred in our reduction half reaction. This is then multiplied by the log of our equilibrium constant. Where in our numerator we're going to have our concentration of the manganese two plus carry on in the prompt that is given to us as 2.50. And this is going to be squared because we have a coefficient of two in our reaction above. Then we have multiplied by this. Our concentration of hydroxide which given in the prompt is also 2.50. This is then raised to the fourth power because we have a coefficient of four in front of hydroxide in our reaction. So just to make this clear, we're dividing this by our concentration of our reactant which we stated would be the pressure of oxygen given in the prompt as 0.150. And we have units of a T. M. Given to us. So we're going to simplify this so that we have our standard cell potential equal to 1.58V. Which is our standard cell potential of our reaction above. This is then subtracted from the right hand side which we can simplify to 0.1 or sorry, 0.0148V. When we take this quotient here divided by the four electrons. And then when we take the value of the log of this quotient we're going to get a value which we will multiply being 3. And so continuing on to simplify. We would get that our cell potential for reaction at 25°C is equal to 1.58 subtracted from 0.05V for the right hand side or rather for this portion of our multiplication here and so for our final value we would get that are sorry that our self potential At 25°C is equal to a value of 1.53V. So this would actually complete this example as our final answer for the cell potential of our reaction at 25°C. So I hope that everything I reviewed was clear. If you have any questions, just leave them down below and I will see everyone in the next practice video.

An electrochemical cell is based on these two half-reactions:
Ox: Sn(s) → Sn²⁺(aq, 2.00 M) + 2 e^–
Red: ClO₂(g, 0.100 atm) + e^– → ClO₂^–(aq, 2.00 M)
Calculate the cell potential at 25 °C.

Verified Solution

Key Concepts

Electrochemical Cells

Nernst Equation

Standard Reduction Potentials

An electrochemical cell is based on these two half-reactions: Ox: Sn(s) → Sn2+(aq, 2.00 M) + 2 e– Red: ClO2(g, 0.100 atm) + e– → ClO2–(aq, 2.00 M) Calculate the cell potential at 25 °C.

Verified Solution

Key Concepts

Electrochemical Cells

Nernst Equation

Standard Reduction Potentials

An electrochemical cell is based on these two half-reactions:
Ox: Sn(s) → Sn²⁺(aq, 2.00 M) + 2 e^–
Red: ClO₂(g, 0.100 atm) + e^– → ClO₂^–(aq, 2.00 M)
Calculate the cell potential at 25 °C.