Current strength of 0.2A carry out electrolysis of aqueous potassium chloride for 10 minutes.The volume of solution is 125 ml and it is assumed that during the electrolysis does not change.What is the plural concentration of OH- ions at the end of the...?

Question

...end of the electrolysis?Calculate the volume of hydrogen and chlorine which develop on the electrodes at 1 bar and 20 degrees Celsius.

Noah Aldrich · Accepted Answer

The quantity of electricity passed #Ixxt=0.2xx10xx60=120C=120/96500faraday=12/9650faraday#
We know  1 faraday electricity produces 1 gm equivalent of different substance on electrolysis So strength of #(OH^- )=12/9650# gm equivalent in 125 ml
so 1000ml contains# 12/9650xx1000/125# g eqiv=#96/9650#i.e strength#~~.00995 N# hydrogen  produced #=12/9650# gm equivalent =0.00062mole
Chlorine  produced #=12/9650# gm equivalent =0.00062mole
Hence same volume of each gas will be) produced. under same condition of temperature and pressure
It canbe calculated as follows
#v=(nRT)/P=(0.00062xx0.082xx293)/1~~0.015L#

Owen Ambrose · Answer

#"[OH"^"-"] = "0.009 95 mol/L"#.

The volumes of #"H"_2# and #"Cl"_2# are #"15.2 mL"#.

The halved responses are: Oxidation: #"2Cl"^"-" → "Cl"_2 + "2e"^"-"# Reduction: #"2H"_2"O" + "2e"^"-" → "H"_2 + "2OH"^"-"# The amount that was transferred in charge The quantity of charge (#Q#) transferred is #color(blue)(|bar(ul(color(white)(a/a) Q =It color(white)(a/a)|)))" "# where #I# is the current in amperes (#"1 A = 1 C·s"^"-1"#) and #t# is the time in seconds. #Q = "0.2 C"·color(red)(cancel(color(black)(s^"-1"))) × 10 color(red)(cancel(color(black)("min"))) × (60 color(red)(cancel(color(black)("s"))))/(1 color(red)(cancel(color(black)("min")))) = "120 C"# One mol of electrons is produced by one faraday of electricity. #color(blue)(|bar(ul(color(white)(a/a)"1 F" = "96 485 C" = "1 mol electrons"color(white)(a/a)|)))" "# ∴ #Q = 120 color(red)(cancel(color(black)("C"))) ×"1 mol electrons"/("96 485" color(red)(cancel(color(black)("C")))) = 1.24 × 10^"-3"color(white)(l) "mol electrons"# Volume of #"H"_2# Given the halved responses, #1.24 × 10^"-3"color(red)(cancel(color(black)("mol electrons"))) × ("1 mol H"_2)/(2 color(red)(cancel(color(black)("mol electrons")))) = 6.22 × 10^"-4"color(white)(l) "mol H"_2# Drawing on the Ideal Gas Law, #color(blue)(|bar(ul(color(white)(a/a)V = (nRT)/Pcolor(white)(a/a)|)))" "# #V_"H₂" = (6.22 × 10^"-4"color(white)(l) color(red)(cancel(color(black)("mol"))) × "0.083 14" color(red)(cancel(color(black)("bar")))·"L"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1")))× 293.15 color(red)(cancel(color(black)("K"))))/(1 color(red)(cancel(color(black)("bar")))) = "0.0152 L" = "15.2 mL"# Volume of #"Cl"_2# Given the halved responses, #1.24 × 10^"-3"color(red)(cancel(color(black)("mol electrons"))) × ("1 mol Cl"_2)/(2 color(red)(cancel(color(black)("mol electrons")))) = 6.22 × 10^"-4"color(white)(l) "mol Cl"_2# This value is equivalent to that of hydrogen, so #V_"Cl₂" = "15.2 mL"# Concentration of #"OH"^"-"# #1.24 × 10^"-3"color(red)(cancel(color(black)("mol electrons"))) × ("2 mol OH"^"-")/(2 color(red)(cancel(color(black)("mol electrons")))) = 1.24 × 10^"-3"color(white)(l)"mol OH"^"-"# #color(blue)(|bar(ul(color(white)(a/a)"Molarity" = "moles"/"litres" color(white)(a/a)|)))" "# ∴ #["OH"^"-"] = (1.24 × 10^"-3"color(white)(l)"mol")/("0.125 L") = "0.009 95 mol/L"#

Wyatt Atkins · Answer

undefined To find the final concentration of OH⁻ ions at the end of the electrolysis of aqueous potassium chloride (KCl), we need to consider the reduction of water at the cathode.

The overall reaction at the cathode during the electrolysis of water is:

$$ 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq) $$

Given that the current strength is 0.2 A and the electrolysis lasts for 10 minutes, we can calculate the total charge passed through the solution using the formula:

$$ Q = I \times t $$

where:
- Q is the total charge passed in coulombs (C),
- I is the current strength in amperes (A), and
- t is the time in seconds (s).

Convert the time to seconds:

$$ t = 10 \text{ minutes} \times 60 \text{ seconds/minute} = 600 \text{ seconds} $$

Now, calculate the total charge passed:

$$ Q = 0.2 \text{ A} \times 600 \text{ s} = 120 \text{ C} $$

Since 1 mole of electrons (e⁻) is equal to 1 faraday (F) of charge (96,485 C), we can calculate the moles of electrons passed through the solution:

$$ \text{Moles of electrons} = \frac{120 \text{ C}}{96,485 \text{ C/mol}} $$

Next, we can use the stoichiometry of the reaction to find the moles of OH⁻ ions produced:

- For every mole of electrons, 2 moles of OH⁻ ions are produced.
- The volume of the solution is 125 ml, which is 0.125 L.

$$ \text{Moles of OH⁻ ions} = \text{Moles of electrons} \times 2 $$

Finally, we can calculate the concentration of OH⁻ ions in the solution:

$$ \text{Concentration of OH⁻ ions} = \frac{\text{Moles of OH⁻ ions}}{\text{Volume of solution}} $$

Plug in the values to find the final concentration of OH⁻ ions at the end of the electrolysis.