The cell notation for a redox reaction is given as the following at (T= 298 K). Calculate the cell potential for the reaction at 25ºC. Zn (s) | Zn 2+ (aq, 0.37 M) || Ni 2+ (aq, 0.059 M) | Ni (s) Standard Reduction Potentials Zn 2+ (aq) + 2 e – →. Zn (s) E° red = - 0.7621 Ni 2+ (aq) + 2 e – → Ni (s) E° red = - 0.2300

Here it says the sub notation for reactant reaction is given as the following at a temperature of 298 kelvin. Now, they want us to calculate the cell potential for the reaction at 25 degrees Celsius, which is the 298 Kelvin. Now here, if we take a look, this is cell notation, so we're gonna say that this is our phase boundary, our physical boundary, and our other phase boundary. It is as easy as a, b, c, so here we have our anode, here we have our physical break, here we have our cathode. Now they're asking for cell potential, but if you're paying attention, we have these concentrations for the ions. They are not equal to 1 molar, which means that the cell potential we're being asked to find is not standard cell potential, but non standard cell potential. Remember, if your molarities are not equal to 1, we're gonna have to utilize the Nernst equation. So here, use utilizing the Nernst equation, that is our nonstandard cell potential equals our standard cell potential minus 0.05916 divided by the moles of electrons transferred times log of our reaction quotientq. Alright. So if we take a look here, what do we have? Well, we have in our anode compartment, we have zinc involved, and then in our cathode compartment, we have nickel involved. Let's first figure out our standard cell potential, which is cathode minus atom. So we're gonna say the cathode is the nickel compartment, so that's negative point 2300, which is this value here, minus a minus 0.7621. So remember, minus and a minus really means you're adding, so it's negative point 23 plus point 7621. Plus 0.7621. So it's gonna give me positive 0.5321 volts. So that is our standard cell potential, we're going to plug that in. The moles of electrons transferred both half reactions have 2 moles of electrons, so 2 is the moles of electrons transferred. And then we do log of q. Now here q represents products over reactants. To do this, we need the overall redox reaction. So how do we do that? Well, here for the anode, it is written as it needs to be written as an oxidation because that the anode is where oxidation occurs. All that means is we'd have to flip this equation here to have our electrons as products. So here we have zinc solid, gives me zinc 2+2electrons. The cathode compartment is reduction occurs. The half reaction is already written as a reduction, so keep it that way. So Ni 2+ aqueous+2electrons gives me an I solid. Here, our 2 electrons cancel out. Our overall reaction is zinc solid +nickel2+ion, gives me zinc ion+nickelsolid. Now that we have the overall reaction, we can determine what q is. Q is equal to products over reactant, so it's equal to zinc 2+ion divided by nickel 2+ion. Remember, we ignore solids and liquids. Alright. So now what's zinc's concentration? It's 0.37 molar. And what is nickel's concentration? It's 0.059 molar. So now we've filled out everything that we need to fill out, so we're gonna plug this in. When I do that I get 0.5085 volts as the non standard cell potential for this particular question. Again, we know that we're utilizing the Nernst equation because the concentrations given to us for each of the ions is not equal to 1 molar. That means we're dealing with nonstandard concentrations, and therefore the Nernst equation needs to be used.

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20. Electrochemistry

Cell Notation

General Chemistry

20. Electrochemistry

Cell Notation

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Multiple Choice

The cell notation for a redox reaction is given as the following at (T= 298 K). Calculate the cell potential for the reaction at 25ºC.
Zn (s) | Zn²⁺ (aq, 0.37 M) || Ni²⁺ (aq, 0.059 M) | Ni (s)
Standard Reduction Potentials
Zn²⁺ (aq) + 2 e^– →. Zn (s) E°_red = - 0.7621
Ni²⁺ (aq) + 2 e^– → Ni (s) E°_red = - 0.2300

0.3130 V

0.4033 V

0.5085 V

0.1199 V

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Verified step by step guidance

Identify the half-reactions from the cell notation: Zn (s) | Zn2+ (aq) and Ni2+ (aq) | Ni (s).

Determine the standard reduction potentials for each half-reaction: Zn2+ + 2e⁻ → Zn (E°red = -0.7621 V) and Ni2+ + 2e⁻ → Ni (E°red = -0.2300 V).

Calculate the standard cell potential (E°cell) using the formula: E°cell = E°cathode - E°anode. Here, Ni is the cathode and Zn is the anode.

Use the Nernst equation to calculate the cell potential at non-standard conditions: Ecell = E°cell - (RT/nF) * ln(Q), where Q is the reaction quotient.

Substitute the given concentrations into the reaction quotient Q = [Zn2+]/[Ni2+] and solve for Ecell using the Nernst equation at 298 K (25°C).