Electrolytic Cell

[bullsfan1986]

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This might be a dumb question, but bear with me.

The question reads:

An engineer attempts to recharge a Ni-Cd battery using a Zn-HgO cell. If the reduction potential at the anode of the Zn-HgO cell is -0.762 V, in order to successfully recharge the battery the reduction potential at the cathode must be: (Note: The Ni-Cd battery has a voltage of 1.4V.)


a) less than 0.638V

b) greater than 0.638V

c)less than 2.162 V

d) greater than 2.162 V


The answer is b. I understand how it can be b, since you need a total voltage to be greater than 1.4 V in order to power the electrolytic battery. What I don't understand is why the answer can't be d. Won't this have essentially the same effect and thus drive the cell as well. Thanks.

 

[fizzgig]

sure will. they asked what the pot'l MUST BE. it MUST BE over .638. it does not HAVE to be 2.162 to run. as in it could be 2.000 and still run.

 

[bullsfan1986]

Aha! Thanks for clearing that up.

 

[IamSuperDoctor]

To clear up some of the confusion: the reduction potential is 0.638, but the cathode is the site of an oxidizing agent, which means that the element at the cathode is being reduced. as a result, the voltage there becomes -0.638.

-0.638 - 0.762 = -1.4

so anything with a reducing potential greater than 0.638 is good. the question attempts to trick you if you have your minus signs backwards.

 

[bullsfan1986]

To clear up some of the confusion: the reduction potential is 0.638, but the cathode is the site of an oxidizing agent, which means that the element at the cathode is being reduced. as a result, the voltage there becomes -0.638.

-0.638 - 0.762 = -1.4

so anything with a reducing potential greater than 0.638 is good. the question attempts to trick you if you have your minus signs backwards.

I think we might be saying the same thing, but I need to make sure my reasoning is correct. So if the question says the anode is -0.762V, wouldn't the negative sign become positive if you're using the formula: E Zn-HgO = Ecat-Eanode, where EZn-HgO would have to be greater than 1.4 V. So, Something > 1.4V = Ecat - (-0.762V) = Ecat + 0.762V. So the actual total voltage would be positive? I don't know if that made any sense.

 

[IamSuperDoctor]

Traditionally, the anode voltage in a galvanic cell is taken as the negative of the reduction potential (oxidation here not reduction). so yes, what you said is correct, the -0.762 would become positive and the cathode voltage (reduction here) is simply the reduction potential = 0.638. Add them up and you get a positive cell voltage, meaning that the reaction is spontaneous, so what i said before was backwards