# Chemistry --Need Help!!

Ive posted twice, but no one's answered yet...I could really use any kind of guidance--particularly for the 1st question :)

1)) The following two half-cells are paired up in a voltaic cell. How will lowering the pH affect the ƒ´E of the overall cell?

ClO3-(aq) + H2O(l) + 2e- ==> ClO2-(aq) + 2OH-(aq) Eo= 0.35 V
I2(S) + 2e-==> 2I-(aq) Eo = 0.54 V

A)dE will increase
B)dE will decrease
C)dE will not be affected by pH

I really have no idea...about any of this...so an explanation of any concepts involved would be great

2)) An electrochemical cell has a constant potential of 2 V and provides a steady current of 1 A. How many electrons move from the anode to the cathode in 1000 seconds? (F ~ 105 C mol-1)

A. 10 mol
B. 1 mol
C. 0.1 mol
D. 0.01 mol

My answer is D, using It=nF

3)) An electrochemical cell has a constant potential of 2 V and provides a steady current of 2 A. After supplying power to a device for a particular amount of time, £GG = -10 kJ. How many mole of electrons were transferred per second for this process? (F ~ 105 C/mol)

A. 1x105 mol/s
B. 2x105 mol/s
C. 2x10-5 mol/s
D. 1x10-5 mol/s

My answer is C. I first used dG=-nF(dE), solving for n, then used that for It=nF, solving for t, finally dividing n by t to get 2E-5 mol/s

4)) An electrochemical cell starts with a potential of 4 V, powers a device for a while, and ends with a potential of 2 V. If a constant 4 W of power was used by the device, what was the current?

A. Constant 1 A
B. Constant 2 A
C. Initially 1 A and increased to 2 A
D. Initially 2 A and decreased to 1 A

My answer is C..I went by the equation Power=I(dE)

Just looking for confirmation of my answers to make sure I'm doing these correctly, and for an explanation for the first question. Thanks!

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1. I answered #1 at a post below. I looked at #2 and it is ok. I didn't look at the others. Let me find that post and Ill give you a link here.

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2. Here is a link to the other post. I think it is detailed enough that you can follow it.
http://www.jiskha.com/display.cgi?id=1269384338

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3. Thanks :)

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posted by Val

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