9. The reaction 2 H2O(l)>>>O2(g) + 2 H2(g) requires 2.06 V, but the reaction
2 H2O(l) + 2 Cl –(aq)>>> H2(g) + 2 OH –(aq) + Cl2(g) requires 2.19 V. If a 1.0 mol/L solution of aqueous NaCl is electrolyzed, what voltage should be applied to get O2(g), but not Cl2(g)?
How would I do this?
10. The electrolysis of aqueous NaCl produces the reaction that follows.
2 H2O(l) + 2 Cl –(aq)>>> H2(g) + 2 OH –(aq) + Cl2(g)
Which other aqueous solution could be electrolyzed to produce the same reaction?
How would I do this?
12. A student wanted to produce lithium using the reaction 2 Li +(l) + 2 Cl –(l)>>> 2 Li(l) + Cl2(g) at 650°C. Give three reasons why the voltages in the reduction half-reaction table in your data booklet would not help him estimate the required voltage.
My answer:
1. It does not contain Li liquid
2. Neither does it contain Cl liquid
3. The Voltages of the equations we do have do not add up to give enough power to produce lithium.
Can u check it please?
13. Why is it sometimes possible for a weaker oxidizing agent or a weaker reducing agent to be involved in an electrolytic reaction rather than the strongest agents?
How is this possible?
20. A pot is to be electroplated with 48.6 g of nickel from a Ni(NO3)2(aq) solution. The cathode compartment of an electrolytic cell can hold 2.50 L of solution. What concentration of Ni(NO3)2(aq) solution is necessary to supply the necessary nickel?
Can u help on these questions...they r the only ones I don't get...Thanks in advance!
9. The reaction 2 H2O(l)>>>O2(g) + 2 H2(g) requires 2.06 V, but the reaction
2 H2O(l) + 2 Cl –(aq)>>> H2(g) + 2 OH –(aq) + Cl2(g) requires 2.19 V. If a 1.0 mol/L solution of aqueous NaCl is electrolyzed, what voltage should be applied to get O2(g), but not Cl2(g)?
How would I do this?
Shouldn't it be above 2.06 but below 2.19?
10. The electrolysis of aqueous NaCl produces the reaction that follows.
2 H2O(l) + 2 Cl –(aq)>>> H2(g) + 2 OH –(aq) + Cl2(g)
Which other aqueous solution could be electrolyzed to produce the same reaction? What about KCl or LiCl or any chloride whose metal salt (Na, K, Li, etc) is above hydrogen in the activity series?
How would I do this?
12. A student wanted to produce lithium using the reaction 2 Li +(l) + 2 Cl –(l)>>> 2 Li(l) + Cl2(g) at 650°C. Give three reasons why the voltages in the reduction half-reaction table in your data booklet would not help him estimate the required voltage.
My answer:
1. It does not contain Li liquid
2. Neither does it contain Cl liquid
3. The Voltages of the equations we do have do not add up to give enough power to produce lithium.
I think both 1 and 2 of your answers are another way of saying that the voltages found in your booklet are those of a 1 M solution of the ion and the solid metal (in the case of Li) or gas at 1 atm pressure vs 1 M Cl^-(in the case of Cl). In molten LiCl this isn't so. Actually this is anactivity of 1 as opposed to a concentration of 1 M. I don't understand your #3 answer.
Can u check it please?
13. Why is it sometimes possible for a weaker oxidizing agent or a weaker reducing agent to be involved in an electrolytic reaction rather than the strongest agents?
Check out how the voltage changes with concentratin; i.e., the Nernst equation.
How is this possible?
20. A pot is to be electroplated with 48.6 g of nickel from a Ni(NO3)2(aq) solution. The cathode compartment of an electrolytic cell can hold 2.50 L of solution. What concentration of Ni(NO3)2(aq) solution is necessary to supply the necessary nickel?
Can't you determine how much Ni(NO3)2 must be in 2.5 L to have a minimum of 48.6 g Ni? Actually, you would need more Ni than this.
48.6 g Ni x (1 mol Ni(NO3)2/1 mol Ni) I hope this helps.
9. To determine the voltage required to get oxygen gas (O2) without chlorine gas (Cl2) during the electrolysis of a 1.0 mol/L solution of NaCl, we need to compare the voltages of the two reactions. The reaction that produces O2 and H2 requires 2.06 V, while the reaction that produces Cl2, H2, and OH requires 2.19 V.
Since we want to produce O2 but not Cl2, we need to apply a voltage higher than 2.06 V but lower than 2.19 V. In other words, the voltage should be within this range (2.06 V < Voltage < 2.19 V).
To achieve this, you can adjust the current passing through the electrolytic cell or use electrodes with different properties to control the voltages during the electrolysis process.
10. The given reaction during the electrolysis of aqueous NaCl is:
2 H2O(l) + 2 Cl –(aq) >>> H2(g) + 2 OH –(aq) + Cl2(g)
To produce the same reaction, any other aqueous solution containing a chloride ion (Cl –) can be electrolyzed. For example, potassium chloride (KCl) or lithium chloride (LiCl) can undergo the same reaction.
The key is to have a solution with a chloride ion and a metal that is above hydrogen in the activity series, as these metals can displace hydrogen from water to produce the desired reaction.
12. Your answer regarding the reduction half-reaction table is partially correct. The voltages listed in the reduction half-reaction table in the data booklet are standard reduction potentials at 1 M concentration for the given species.
However, in the case of producing lithium (Li) from 2 Li+(l) and 2 Cl –(l) at 650°C, the reduction half-reaction table's voltages cannot be directly applied. This is because the reaction is taking place at a different temperature and with different concentrations than what is listed in the table.
Additionally, the reduction half-reaction table may not contain specific values for the half-reactions involving Li+ and Cl – at 650°C. So, the required voltage for the reaction cannot be estimated from the values in the table.
13. In electrolytic reactions, it is sometimes possible for weaker oxidizing agents or weaker reducing agents to be involved instead of the strongest agents due to factors such as concentration and conditions.
This is because the rate of a redox reaction depends not only on the strength of the oxidizing or reducing agent but also on the concentrations and conditions of the reaction. The Nernst equation can be used to calculate the electrode potential at non-standard conditions.
If the concentration or conditions of the strongest agent are not optimal, a weaker oxidizing agent or reducing agent may be involved in the reaction instead. This allows for better control and flexibility in the electrochemical process.
20. To determine the concentration of Ni(NO3)2(aq) solution necessary to supply the necessary nickel for electroplating, you need to use the molar mass of nickel (Ni) to calculate the number of moles of nickel required.
First, calculate the number of moles of nickel in 48.6 g of nickel:
48.6 g Ni x (1 mol Ni/atomic mass of Ni)
Then, if the cathode compartment of the electrolytic cell can hold 2.50 L of solution, you can calculate the concentration of Ni(NO3)2(aq) using the moles of nickel and the volume of the solution:
Concentration = moles of Ni / volume of solution (in L)
This will give you the concentration of Ni(NO3)2(aq) solution necessary to supply the required nickel for electroplating.