# 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...?

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

Apr 2, 2016

The quantity of electricity passed $I \times t = 0.2 \times 10 \times 60 = 120 C = \frac{120}{96500} f a r a \mathrm{da} y = \frac{12}{9650} f a r a \mathrm{da} y$
We know 1 faraday electricity produces 1 gm equivalent of different substance on electrolysis

So strength of $\left(O {H}^{-}\right) = \frac{12}{9650}$ gm equivalent in 125 ml
so 1000ml contains$\frac{12}{9650} \times \frac{1000}{125}$ g eqiv=$\frac{96}{9650}$i.e strength$\approx .00995 N$

hydrogen produced $= \frac{12}{9650}$ gm equivalent =0.00062mole
Chlorine produced $= \frac{12}{9650}$ 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
$v = \frac{n R T}{P} = \frac{0.00062 \times 0.082 \times 293}{1} \approx 0.015 L$

May 1, 2016

$\text{[OH"^"-"] = "0.009 95 mol/L}$.

The volumes of ${\text{H}}_{2}$ and ${\text{Cl}}_{2}$ are $\text{15.2 mL}$.

#### Explanation:

The half-reactions are:

Oxidation: $\text{2Cl"^"-" → "Cl"_2 + "2e"^"-}$

Reduction: $\text{2H"_2"O" + "2e"^"-" → "H"_2 + "2OH"^"-}$

The quantity of charge transferred

The quantity of charge ($Q$) transferred is

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} Q = I t \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

where $I$ is the current in amperes ($\text{1 A = 1 C·s"^"-1}$) and $t$ is the time in seconds.

$Q = \text{0.2 C"·color(red)(cancel(color(black)(s^"-1"))) × 10 color(red)(cancel(color(black)("min"))) × (60 color(red)(cancel(color(black)("s"))))/(1 color(red)(cancel(color(black)("min")))) = "120 C}$

One faraday of electricity produces 1 mol of electrons.

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 ${\text{H}}_{2}$

From the half-reactions,

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

From the Ideal Gas Law,

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} V = \frac{n R T}{P} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

${V}_{\text{H₂" = (6.22 × 10^"-4"color(white)(l) color(red)(cancel(color(black)("mol"))) × "0.083 14" color(red)(cancel(color(black)("bar")))·"L"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1")))× 293.15 color(red)(cancel(color(black)("K"))))/(1 color(red)(cancel(color(black)("bar")))) = "0.0152 L" = "15.2 mL}}$

Volume of ${\text{Cl}}_{2}$

From the half-reactions,

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 is the same value as for hydrogen, so

${V}_{\text{Cl₂" = "15.2 mL}}$

Concentration of $\text{OH"^"-}$

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"