Normal 0 false false false EN-US JA X-NONE /* Style Definitions */table.MsoNormalTable{mso-style-name:"Table Normal";mso-tstyle-rowband-size:0;mso-tstyle-colband-size:0;mso-style-noshow:yes;mso-style-priority:99;mso-style-parent:"";mso-padding-alt:0in 5.4pt 0in 5.4pt;mso-para-margin:0in;mso-para-margin-bottom:.0001pt;mso-pagination:widow-orphan;font-size:10.0pt;font-family:Cambria;} During electrolysis the concentration of I 2 increases to 8.3 x 10 -3 M, while all other concentrations remain unchanged. If the electrical resistance is 1.8 ohms, the current is 71 mA, the anode overpotential is 0.013 V and the cathode overpotential is 0.115 V, what is the voltage needed?

So this is a continuation of the example problem we had up above. Here it says during electrolysis, the concentration of i2 increases to 8.3 times 10 to the negative 3 molar, while all other concentrations remain unchanged. If the electrical resistance is 1.8 ohms, the current is 71 milliamps, the anode overpotential is 0.013 volts and the cathode overpotential is 0.115 volts. What is the voltage needed? Alright. So, we're going to say here that our new equation becomes our non standard cell potential equals cathode minus anode minus your ohmic potential minus your overpotentials. So remember, this part here deals with your concentration polarization. Here, your ohmic potential. Here, this would equal your current times your resistance and here, your overpotentials would be cathode minus anode as well. Here, they say that only the concentration of I two is changing. Everything else stays the same. So for your cathode, it'd be the same exact value because we don't change either one of these concentrations. So we'd still get this negative 0.76 volts. The anode would change though because for the anode, we're changing this value here which is going to give us a new value overall for our anode. So we're going to have to calculate what that will be. So if we take a look, we're going to say here our anode would equal standard cell potential minus 0.05916volts divided by n times log of I minus to the 4th divided by i2 squared. So that is 0.54 volts minus 0.05916 volts divided by 4 electrons times log of 0.15 to the 4th. Now, we have our new concentration for I 2, which is 8.3 times 10 to the minus 3 squared. Here, when you work that out, that gives you 0.54 minus 0.013. So it's going to be 0.527 volts. So notice here that we increased the concentration here on the bottom which overall gave us a larger value for my anode. So at this point, we have negative 0.76 volts minus 0.527 volts minus So remember, your ohmic potential, which I'll just say is o p, equals I times r. I here is your ohms, which is 1.8 ohms, and we use Omega for the symbol. Your resistance needs to be in amps, so we divide this by a 1000 to give us 0.071 amps. So that goes here, minus the cathode overpotential which is point 115 volts minus the anode overpotential. So when we subtract all of those, that's going to give me a value of negative 1.54 volts. So as we said earlier, when the current is not small and it's a sufficient amount, that means that we're gonna have to take into account concentration polarization here. We're gonna to take into account ohmic potential as well as overpotentials. The effect that these all have is they make your overall cell potential a smaller, more negative value, meaning your reaction becomes even harder to accomplish. Here, we have to apply even more current in order to drive the reaction forward. That's what this is showing us. By incorporating ohmic potential with over potential, it just drags down the overall cell potential. So you're gonna have to invest more voltage in order to force this reaction to occur. Remember, electrolysis itself is a non spontaneous process. And because of that, we have to supply this outside energy, which is why our cell potential overall is less than 0.

16. Electroanalytical Techniques

Fundamentals of Electrolysis

Analytical Chemistry

16. Electroanalytical Techniques

Fundamentals of Electrolysis

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Multiple Choice

During electrolysis the concentration of I₂ increases to 8.3 x 10 ^-3 M, while all other concentrations remain unchanged. If the electrical resistance is 1.8 ohms, the current is 71 mA, the anode overpotential is 0.013 V and the cathode overpotential is 0.115 V, what is the voltage needed?

-1.48 V

-1.04 V

-1.54 V

-0.0228 V

-0.846

Verified step by step guidance

Identify the components of the total cell voltage needed for electrolysis. The total voltage (E_total) is the sum of the thermodynamic cell potential (E_cell), the overpotentials at the anode and cathode, and the ohmic potential drop (IR).

Calculate the ohmic potential drop using Ohm's Law: V_ohmic = I * R, where I is the current (71 mA or 0.071 A) and R is the resistance (1.8 ohms).

Add the anode overpotential (0.013 V) and the cathode overpotential (0.115 V) to the ohmic potential drop calculated in the previous step.

Determine the thermodynamic cell potential (E_cell). This value is typically derived from standard reduction potentials, but since it is not provided, assume it is included in the options given.

Sum all the components: E_total = E_cell + V_ohmic + anode overpotential + cathode overpotential. Compare this calculated total voltage with the options provided to identify the correct answer.

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Struggling with Analytical Chemistry?

During electrolysis the concentration of I2 increases to 8.3 x 10 -3 M, while all other concentrations remain unchanged. If the electrical resistance is 1.8 ohms, the current is 71 mA, the anode overpotential is 0.013 V and the cathode overpotential is 0.115 V, what is the voltage needed?

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Fundamentals of Electrolysis practice set

During electrolysis the concentration of I₂ increases to 8.3 x 10 ^-3 M, while all other concentrations remain unchanged. If the electrical resistance is 1.8 ohms, the current is 71 mA, the anode overpotential is 0.013 V and the cathode overpotential is 0.115 V, what is the voltage needed?