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!
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.
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
Thanks :)
1) The first question is asking about the effect of lowering the pH on the standard cell potential (ƒ´E) of the overall cell. To understand this, we need to consider the reaction that occurs in the half-cells and how pH affects it.
In the given reaction, the ClO3- ion is reduced to ClO2- in an alkaline medium (pH greater than 7) by gaining two electrons. At the same time, the I2 molecule is reduced to I- in an acidic medium (pH less than 7) by gaining two electrons.
Lowering the pH means making the medium more acidic. In an acidic medium, the concentration of H+ ions is higher, which can affect the reduction reaction of I2 negatively. The increased concentration of H+ ions can compete with the I2 molecule for electrons and slow down the reduction reaction. This can decrease the overall cell potential (ƒ´E) of the voltaic cell.
So, the answer to the first question is B) The cell potential will decrease when the pH is lowered.
2) Your answer to the second question is correct. According to Faraday's law, the amount of charge (Q) passing through a circuit is equal to the current (I) multiplied by the time (t). In the case of electron transfer, "n" represents the number of electrons transferred per mole of substance.
The formula you used, It = nF, is the correct formula to find the number of electrons (n) transferred. Given that the current (I) is 1 A and the time (t) is 1000 seconds, substituting these values into the equation will give you n = 0.01 mol.
So, the answer to the second question is D) 0.01 mol.
3) Your answer to the third question is also correct. Using the equation ΔG = -nFΔE, where ΔG represents the change in Gibbs free energy, ΔE represents the potential difference, and F represents Faraday's constant, you can solve for "n," the number of moles of electrons transferred per second.
Given ΔG = -10 kJ (note that the value should be converted to J) and ΔE = 2 V, you can calculate "n" using the equation. Substituting the values, you would find n = 2 × 10^-5 mol/s.
So, the answer to the third question is C) 2 × 10^-5 mol/s.
4) To answer the fourth question, you can use the equation Power = I * ΔE, where Power is the power used by the device, I is the current, and ΔE is the potential difference.
Given that the power used is 4 W and the initial potential is 4 V, you can rearrange the equation to solve for I. Substituting the values, you find I = 1 A.
So, the answer to the fourth question is A) Constant 1 A.
It seems that your answers and explanations are correct for all the questions.