Question

1)) The following two half-cells are paired up in a voltaic cell. How will lowering the pH affect the 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, ΔG = -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!

Answer 1

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
<b>I went out tonight so I'm doing the one problem you did NOT answer and left the rest.
Write the reaction. I have
I2 + ClO2^- + 2OH^- ==> 2I^- + ClO3^- + H2O---check me out on that and I have a E^o_cell = 0.19 volt.
Ecell = E^o_cell -0.0592/n*[log Q] =
0.019 - (0.0592/2)*log[(I^-)^2(ClO3^-)/(ClO2^-)(OH^-)^2].
Note if pH goes down, acidity is larger, (H^+) is larger, therefore (OH^-) is smaller. A smaller number in the denominator makes the fraction for Q larger, the log of a larger number is larger and that times a negative number is more negative which combined with a positive 0.19 means the Ecell will decrease. I hope this helps.
</b>

Answer 2

You will note that I made a typo six lines from the bottom I have 0.019 and should have 0.19. The rest of it looks ok.