[alanjz]

41. Use the given standard reduction potentials to determine
the reduction potential for this half-reaction.
MnO₄^-+e^- --> MnO₄^2- +0.564 V
MnO₄^2-+2e^-+4H⁺ --> MnO₂+2H₂O +2.261 V
(A) 1.695 V (B) 2.825 V
(C) 3.389 V (D) 5.086 V
I just put each in terms of ΔG so I could add them and converted back to E and got choice D (-(1)(96500)(0.564))+(-(2)(96500)(2.261)))/(96500)
However, the given answer is choice A so why is this wrong?

42. How many Faradays are required to reduce all the
chromium in 0.150 L of 0.115 M of Cr2O7^2- to Cr²⁺?
(A) 0.920 F (B) 0.690 F
(C) 0.138 F (D) 0.069 F
I wasn't really sure how to do this since there wasn't a given current and time, which I'm used to seeing in these questions.
There are 0.150*0.115=0.01725 moles in total and I expect to multiply that by two because of the two electrons. I guess you do something with 96500 but I'm not sure what.

[Borek]

41. Use the given standard reduction potentials to determine
the reduction potential for this half-reaction.
MnO₄^-+e^- --> MnO₄^2- +0.564 V
MnO₄^2-+2e^-+4H⁺ --> MnO₂+2H₂O +2.261 V

I don't get the question. Which reaction?

I wasn't really sure how to do this since there wasn't a given current and time, which I'm used to seeing in these questions.

F is given in Coulombs, not without a reason. It is all about charge, and q=it. This is basic physics.

[alanjz]

WhoopsI left that out:MnO ₄+3e^---> MnO₂ +H₂O.

[Dan]

WhoopsI left that out:MnO ₄+3e^---> MnO₂ +H₂O.

Charge?

[alanjz]

sorry about that , should be 2-

[Dan]

WhoopsI left that out:MnO ₄+3e^---> MnO₂ +H₂O.

sorry about that , should be 2-

So you're saying it is:

MnO₄^2- + 3e^- MnO₄ + H₂O

This cannot be correct. That is Mn(VI) to Mn(IV), which requires 2e^-. Hydrogen and oxygen are also not balanced.

Can you write out the question exactly as it was given to you, because this is starting to waste everyone's time.

[alanjz]

Yeah my bad, I was typing this on my phone.
Reduction potential for:MnO₄^-+3e^- + 4H⁺--> MnO₂ +2H₂O.
Given:
MnO₄^-+e^- --> MnO₄^2- (+0.564 V)
MnO₄^2-+2e^-+4H⁺ --> MnO₂+2H₂O (+2.261 V)

[Dan]

Ok, so you are on the right track, you have just missed the final step of the calculation.

Hint: In your calculation, you multiplied the value for E(MnO₄^2-/MnO₂) by a factor of 2 because E is given "per electron" and there is a transfer of two electrons in that process. What you calculated is the voltage associated with a transfer of 3 electrons...

[alanjz]

Aha! So I had to divide by three to get it to be per mole of electrons. Thanks a lot. Sorry I irritated you earlier.