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!
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
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 Eocell = 0.19 volt.
Ecell = Eocell -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.
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.
1) How will lowering the pH affect the ΔE of the overall cell?
- Well, pH stands for "power of hydrogen" but unfortunately, it doesn't give you the power to control the electrons in the cell. Lowering the pH will not affect the ΔE of the overall cell. So the answer is C. ΔE will not be affected by pH. It's like adding hot sauce to a glass of water, it might change the taste but it won't change the total amount of liquid.
2) How many electrons move from the anode to the cathode in 1000 seconds?
- It's like counting the number of clowns that come out of a tiny car, you have to use It = nF. Since the current is 1A and the time is 1000 seconds, we can calculate the number of electrons moved using the equation n = It/F. Substituting the values, n = (1A)(1000s)/(105 C/mol) = 0.01 mol. So the answer is D. 0.01 mol, and I hope you didn't get lost in the clown car while counting.
3) How many moles of electrons were transferred per second for this process?
- To figure this out, first, you used ΔG = -nF(dE) and solved for n. Now, we need to use It = nF to find the time. Given that the current is 2A, substituting the values, t = nF/I = (n)(105 C/mol)/(2A). Combining all the given values, t = (10 kJ)/(2V) x (105 C/mol)/(2A) = 2 x 10^-5 mol/s. So the answer is C. 2 x 10^-5 mol/s, because electrons have great timing and know how to move efficiently.
4) What was the current?
- With a potential change from 4V to 2V, and a constant power of 4W, we can now use the equation P = I(dE) to find the current. Since the power is given as 4W and the potential difference is 2V, substituting the values, 4W = I(2V) gives us I = 2A. So the answer is B. Constant 2A. It's like trying to keep up with the rhythm of a clown parade, you need a constant current to power the show.
I hope my clownish explanations have brought some clarity and a smile to your face. Keep up the good work with these electrochemical questions!
1) Lowering the pH will affect the ΔE of the overall cell. In the first half-cell (ClO3- + H2O + 2e- <==> ClO2- + 2OH-), the presence of OH- ions in the reaction suggests that it is taking place in a basic solution. When the pH is lowered, it becomes more acidic, thus reducing the concentration of OH- ions. This will shift the reaction towards the reactants, resulting in a decrease in the potential (ΔE) generated by this half-cell. Similarly, in the second half-cell (I2 + 2e- <==> 2I-), the presence of I- ions suggests that it is taking place in an acidic solution. When the pH is lowered, it becomes even more acidic, thus increasing the concentration of I- ions. This will shift the reaction towards the products, resulting in an increase in the potential (ΔE) generated by this half-cell. Since the overall potential is the difference in potentials between the two half-cells, lowering the pH will result in a decrease in ΔE of the overall cell. Therefore, the correct answer is B) ΔE will decrease.
2) To calculate the number of electrons moved from the anode to the cathode, we can use the equation It = nF, where I is the current, t is the time, n is the number of moles of electrons, and F is the Faraday constant. Given that the current is 1 A and the time is 1000 seconds, we can rearrange the equation to solve for n as follows:
n = It/F = (1 A)(1000 s)/(10^5 C/mol) = 0.01 mol.
Therefore, the correct answer is D) 0.01 mol.
3) To calculate the number of moles of electrons transferred per second, we can rearrange the equation ΔG = -nFΔE and solve for n as follows:
n = -ΔG/(FΔE) = -(10 kJ)/(10^3 J/kJ)/(10^5 C/mol)/(2 V) = -10^-3 mol.
Since the question asks for the number of moles of electrons per second, we need to divide this value by the time (1 second) to get the rate:
n/time = (-10^-3 mol)/(1 s) = -10^-3 mol/s.
The negative sign is because the reaction is exothermic (ΔG<0), indicating a spontaneous release of thermal energy. Therefore, the correct answer is C) 2x10^-5 mol/s.
4) The equation Power = IΔE can be used to calculate the current (I) when the power (P) and potential difference (ΔE) are given. In this case, the power is constant at 4 W, and the potential difference changes from 4 V to 2 V.
Solving for I:
4 W = I(2 V - 4 V) = -2I.
Since the power is positive and the potential difference decreases, the negative sign indicates that the current is flowing in the opposite direction of the conventional current. Therefore, the correct answer is D) Initially 2A and decreased to 1A.
1) The given half-reactions are written as reduction reactions. In a voltaic cell, electrons flow from the anode (where oxidation occurs) to the cathode (where reduction occurs). The overall cell potential, ΔE, is the difference between the reduction potentials of the two half-reactions.
Lowering the pH affects the concentration of H+ ions, which can impact the reaction rates and shift the equilibrium of the half-reactions. In this case, the ClO3- and I- ions are also involved in the reactions, but their concentrations are not affected by changes in pH. Therefore, only the OH- ions, formed in the ClO3- reduction half-reaction, are affected.
The reduction potential of the ClO3- half-reaction is positive, meaning it wants to proceed in the forward direction. The presence of OH- ions in the solution, formed in an alkaline environment, helps drive the reaction forward. Thus, lowering the pH by adding acid will decrease the concentration of OH- ions and make it more difficult for the ClO3- reduction half-reaction to occur. As a result, the overall cell potential, ΔE, will decrease.
Therefore, the correct answer is B) ΔE will decrease.
2) Your answer is correct. The equation It = nF can be used to relate the current (I), time (t), number of moles of electrons transferred (n), and Faraday's constant (F). Given that the current is 1 A and the time is 1000 seconds, substituting these values into the equation gives:
1 A * 1000 s = n * (105 C/mol)
n = (1 A * 1000 s) / (105 C/mol)
n = 0.01 mol
Therefore, the correct answer is D) 0.01 mol.
3) Your answer is correct. By using the equation ΔG = -nFΔE and manipulating it as you did, you arrived at the correct expression It = nF. Given that the current is 2 A, substituting this value into the equation gives:
2 A * t = n * (105 C/mol)
t = (n * (105 C/mol)) / 2 A
t = 2 * 10^-5 s
Therefore, the correct answer is C) 2x10^-5 mol/s.
4) Your answer is correct. The equation Power = IΔE can be used to relate power (P), current (I), and potential difference (ΔE). Given that the power is 4 W and the potential difference changes from 4 V to 2 V, substituting these values into the equation gives:
4 W = I * (4 V - 2 V)
4 W = 2 I A
I = 2 A
Therefore, the correct answer is B) Constant 2 A.
Great job on your answers!