A voltaic cell employs the following redox reaction: Sn²⁺(aq) + Mn(s) → Sn(s) + Mn²⁺(aq) Calculate the cell potential at 25 °C under each set of conditions. c. [Sn²⁺] = 2.00 M; [Mn²⁺] = 0.0100 M

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Hey everyone in this example, we're given the following redox reaction occurring in a galvanic cell and we need to determine the cell potential at 25 degrees Celsius. When the concentration of copper two plus is equal to 2.50 moller. And the concentration medical two plus is equal to 0.150 moller. Our first step is to write out our two half reactions. So first showing for the formation of solid copper, we have the copper two plus Catalan which will gained two electrons to produce solid copper as a product. For our second half reaction, we're showing the formation of the nickel two plus carry on where our solid nickel will break down into the nickel two plus carry on, which will gain two electrons to form solid nickel. We want to refer to our standard electrode reduction potential table in our textbooks or online and we would see that for the reduction of copper two plus we have a Electro potential value equal to 0.34V. Now looking up in the same table for the oxidation of solid nickel, we would have an electrode potential value equal to negative 0.23V. Now because we recognize that this value here is negative. This means that this occurs at our anodes as an oxidation half reaction. And because this value here is positive, we recall that this occurs at our cathode and our half reaction is a reduction here. Our next step is to recognize that we have different numbers or rather the same numbers of electrons in each of our half reactions, which is good. So we don't have to manipulate our reactions and we can just go ahead and add them together to write out one full reaction. So we can go ahead and cancel out the two electrons here in both sides of our equation. We have one occurring here on the react inside and the other electrons occurring on the product side here. So when we bring everything down we would have that are copper two plus, carry on Reacting with solid, Nickel will produce the copper solid as a product and Nickel two plus as an ion product. And we would say that the electrons transferred n is equal to a value of two. Our next step is to calculate our standard cell potential E degree cell which we can recall is calculated by taking the cell potential of our cathode and subtracting that from the cell potential of our an ode. So what we would get is for our cell potential of our cathode above. We stated that that is 0.34 volts subtracted from our cell potential of our an ode which above the stated is negative 0.23 volts from our textbooks. And what this difference would give us is a value equal to 0.57 volts. So now we have our standard cell potential value for reaction. We have our electrons transferred. And now we want to make note of our reaction quotient Q which we recall represents our concentration of our products divided by our concentration of our reactant. And this involves only our ion products and ion reactant. So this would be equal to the concentration of our ion product being nickel two plus, divided by our concentration of our ion reactant being copper two plus. And we are given the concentrations for each of these ions are pumped. So for nickel two plus we have a concentration of 0.150 moller. And then in our denominator from the prompt were given the concentration of copper two plus as 2.50 moller. So for our value of our reaction quotient Q. We should get a value of six times 10 to the negative third power. So next we're gonna recall our first equation to calculate our cell potential E cell which we should recall is equal to our standard cell potential degree, sell subtracted from our equation constant 0.592 votes divided by our electrons transferred N. Which is multiplied by the log of our reaction quotient Q. So plugging in what we know to calculate for the cell potential sl we would get our standard cell potential which above we stated is 0.57V subtracted from our nurse equation constant 0.0592V divided by the electrons transferred which we stated is two electrons here. So too is in our denominator multiplied by the log of our reaction quotient, which we stated is equal to six times 10 to the negative third power. So simplifying this inner calculators, we would get that our self potential S. L. Is equal to 0.57V subtracted from 0.0296V. When we take the quotient in the middle here And then this is going to be multiplied by our quotient where we take the log of six times 10 to the negative third power, which is going to give us a value equal to negative 2.2-185. So what we would simplify too is that our self potential is equal to 0.57V added to 0.065767. And this is in units of volts which is equal to 0.63V. So this would be our final answer to complete this example. As the standard cell potential for our reaction at 25°C. So I hope that everything I reviewed was clear. But if you have any questions just leave them down below and I will see everyone in the next practice video

A voltaic cell employs the following redox reaction: Sn²⁺(aq) + Mn(s) → Sn(s) + Mn²⁺(aq) Calculate the cell potential at 25 °C under each set of conditions. c. [Sn²⁺] = 2.00 M; [Mn²⁺] = 0.0100 M

Verified Solution

Key Concepts

Redox Reactions

Nernst Equation

Cell Potential

A voltaic cell employs the following redox reaction: Sn2+(aq) + Mn(s) → Sn(s) + Mn2+(aq) Calculate the cell potential at 25 °C under each set of conditions. c. [Sn2+] = 2.00 M; [Mn2+] = 0.0100 M

Verified Solution

Key Concepts

Redox Reactions

Nernst Equation

Cell Potential

A voltaic cell employs the following redox reaction: Sn²⁺(aq) + Mn(s) → Sn(s) + Mn²⁺(aq) Calculate the cell potential at 25 °C under each set of conditions. c. [Sn²⁺] = 2.00 M; [Mn²⁺] = 0.0100 M