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!

1) In a voltaic cell, the potential difference (ΔE) is related to the cell potential (Eo) by the Nernst equation:

ΔE = Eo - (0.0592/n) * log(Q)

where n is the number of moles of electrons involved in the reaction, and Q is the reaction quotient. The reaction quotient depends on the concentrations of the reactants and products.

In this case, lowering the pH will increase the concentration of H+ ions in the solution. Since the reaction involves H2O and OH- ions, changing the concentration of H+ ions will have an effect on the concentration of OH- ions. This will change the value of Q, which in turn will affect the value of ΔE.

To determine the specific effect of lowering the pH on ΔE, we need to know the concentrations of ClO3-, ClO2-, H2O, and OH-. Without that information, we can still make a general assumption: since lowering the pH increases the concentration of H+ ions, which are a reactant in the first half-reaction, we can expect the reaction quotient for the first half-reaction to increase. This would shift the equilibrium towards the products and increase the value of ΔE. Therefore, the answer is:

A) ΔE will increase

2) To determine the number of electrons (n) transferred in a redox reaction, we can use Faraday's law of electrolysis, which states that the amount of substance (n) produced or consumed during an electrochemical reaction is directly proportional to the total charge (Q) passed through the system. The equation is:

n = Q/F

where F is the Faraday constant and is equal to 96,485 C/mol. In this case, we are given the current (I = 1 A) and time (t = 1000 s), and we need to calculate the number of electrons (n):

n = It/F = (1 A)(1000 s)/(96,485 C/mol) = 0.01036 mol ≈ 0.01 mol

Therefore, the answer is:

D) 0.01 mol

3) To determine the number of moles of electrons transferred per second, we can use the equation:

n = It/F

where n is the number of moles of electrons, I is the current (2 A), t is the time, and F is the Faraday constant. In this case, we are given the potential (2 V) and the change in Gibbs free energy (ΔG = -10 kJ), and we need to calculate the number of moles of electrons transferred per second:

First, we can use the equation:

ΔG = -nFE

to solve for n:

n = -ΔG/(FE)

where E is the potential difference (2 V) and F is the Faraday constant (105 C/mol):

n = -(-10 kJ)/(10^3 J/kJ)/(2 V)(105 C/mol) ≈ 0.0952 mol

Then, we can divide this value by the time (1 s):

n/sec = 0.0952 mol/1 s ≈ 0.0952 mol/s ≈ 9.52 x 10^-2 mol/s ≈ 2 x 10^-5 mol/s

Therefore, the answer is:

C) 2x10^-5 mol/s

4) To determine the current (I) in the electrochemical cell, we can use the equation:

Power = IV

where Power is the power used by the device (4 W) and V is the potential difference between the anode and the cathode. We are given that the potential difference changes from 4 V to 2 V. Since the power is constant, we can assume that the current remains constant. Therefore, the answer is:

A) Constant 1 A