Electrolysis of Water And Generation of Hydrogen Gas over Water | CHEM 237, Lab Reports of Chemistry

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Electrolysis of Water

And

Generation of Hydrogen Gas over Water

Jessica McPhail

Jonathan Gawinski

Molly Doyle

Chem 237

Professor Wang

Performed 3/19/

Due 3/26/

Part 1 of 2: Electrolysis of Water

Purpose

The purpose of this experiment is to be able to determine the amount of hydrogen and

oxygen gases by determining the volume of the gas collected and calculating the partial

pressure for the hydrogen gas. From this data and the recorded time and current we can

find an experimental value for Faraday’s constant.

Procedure

For this experiment, the Hoffman Electrolysis apparatus was set up where the solution

was poured into the center and main container. A certain amount of voltage was applied

to the solution added (in our case solution was 1.5 M H 2 SO 4 ). As a result of the applied

voltage the solution dissociated into hydrogen and oxygen molecules; hydrogen gas

towards the right tube and oxygen gas to the left tube. After 20 minutes of recording the

current every 30 seconds, we were able to determine the amount of moles of hydrogen

and oxygen gas and from there determine Faraday’s constant using:

H 2 O! 1 / 2 O 2 + H+^ (2 moles of electrons/1mole H+^ )

Pertinent Data and Calculations

Trial 1: at 6 volts

P H2 = P atm + P water column – P water vapor

P = 0.966 atm V = 0.0175 L R = 0.0821 (L atm)/(mol K) T = 295 K

n = PV/RT = 6.98 x 10 -4^ mol H 2

Faraday’s constant: (∑ t x I) / [(2 mol e-^ x 6.98 x 10-4^ mol H 2 )]

F = 98993 coulombs

Trial 2: at 5 volts

P = 0.964 atm V = 0.0156 L R = 0.0821 (L atm)/(mol K) T = 295 K

n = PV/RT = 6.21 x 10 -4^ mol H 2

Faraday’s constant: (∑ t x I) / [2 mol e-^ x (6.21 x 10-4^ mol H 2 )]

F = 97326 coulombs

Part 2 of 2: Generation of Hydrogen Gas Over Water

Purpose

The purpose of this experiment is to apply a method for collecting gases. By this method,

we can determine the volume, pressure, number of moles, and temperature. In

performing this experiment, we determine how many moles of hydrogen gas are

produced from a 1 mL of solution consisting 1% NaBH 4 in 1 M NaOH.

Procedure

For this experiment, we set up the apparatus to collect gases over water. We determined

the volume and pressure of the gas from the reading of the amount of water that was

displaced by the gas. We then could perform the calculations to determine moles of gas.

In order to initiate the reaction, a catalyst was added so that hydrogen gas was generated.

The reaction taking place was:

NaBH 4 + 2 H 2 O! NaBO 2 + 4H 2

Pertinent Data and Calculations

Trial 1:

P H2 = P atm - P water column – P water vapor

P H2 = 0.932 atm V = 0.0104 L R = 0.0821 (L atm)/ (mol/K) T = 295 K

n = PV/RT = 4.0 x 10 -4^ mol H 2

Trial 2:

P H2 = P atm - P water column – P water vapor

P H2 = 0.930 atm V = 0.012864 L R = 0.0821 (L atm)/ (mol/K) T = 295 K

n = PV/RT = 4.94 x 10 -4^ mol H 2

Error Analysis

A quantitative error may occur in the accuracy of the pressure since we did not measure

the atmospheric pressure. Also, some fluctuation may have occurred in the conversion

from mL of the water column to mmHg for the pressure. Rounding is also a factor in

quantitative error. Qualitative error may occur in the measurement of the gas after it has

been trapped in the burette such as water collecting at the top of the burette. Also, it is

possible that not all of the gas was collected due to leaks in the tubing or connections.

Error may have also occurred in the placement of the catalyst. For the first trial we had

placed catalyst B into the 1 mL of solution, the reaction did not occur so we placed

catalyst A with catalyst B. Therefore, Trial 2 should be more accurate in this sense

because both catalyst A and B were present for all the reaction taking place. The size of

the catalyst compared to the 1 mL of solution may have effected the outcome of the

amount of substance reacting.

 - Trial - Trial - y = 4E-09x 2 - 9E-06x + 0. - 0. - 0. - 0. - 0. - 0. - 0. - 0. - 0. - 0. - Series Current (A) - 0 0.1204 120.4 630 0.115 115 36. time (s) Current (A) mA t(s) * I(A) time (s) Current (A) mA t(s) * I(A) 

• 60 0.1183 118.3 7.161 660 0.1149 114.9 3.
• 90 0.1179 117.9 3.543 690 0.1148 114.8 3.
• 120 0.1175 117.5 3.531 720 0.1146 114.6 3.
• 150 0.117 117 3.5175 750 0.1145 114.5 3.
• 180 0.1167 116.7 3.5055 780 0.1145 114.5 3.
• 210 0.1166 116.6 3.4995 810 0.1144 114.4 3.
• 240 0.1163 116.3 3.4935 840 0.1144 114.4 3.
• 270 0.1161 116.1 3.486 870 0.1143 114.3 3.
• 300 0.1159 115.9 3.48 900 0.1138 113.8 3.
• 330 0.1158 115.8 3.4755 930 0.1138 113.8 3.
• 360 0.1157 115.7 3.4725 960 0.1138 113.8 3.
• 390 0.1155 115.5 3.468 990 0.1138 113.8 3.
• 420 0.1155 115.5 3.465 1020 0.1137 113.7 3.
• 450 0.1154 115.4 3.4635 1050 0.1137 113.7 3.
• 480 0.1153 115.3 3.4605 1080 0.1137 113.7 3.
• 510 0.115 115 3.4545 1110 0.1137 113.7 3.
• 540 0.1153 115.3 3.4545 1140 0.1136 113.6 3.
• 570 0.1151 115.1 3.456 1170 0.1136 113.6 3. - 101. 600 0.115 115 3.4515 1200 0.1136 113.6 3.408Sum(t*I) - Trial - y = 6E-09x 2 - 1E-05x + 0. at 5 volts - 0. - 0. - 0. - 0. - 0. - 0. - 0. - 0. - 0 0.1085 108.5 630 0.1003 100.3 3. time (s) Current (A) mA t(s) * I(A) time (s) Current (A) mA t(s) * I(A)
  • 30 0.1061 106.1 3.219 660 0.1002 100.2 3.
  • 60 0.1052 105.2 3.1695 690 0.1 100 3.
  • 90 0.1046 104.6 3.147 720 0.0998 99.8 2.
• 120 0.1042 104.2 3.132 750 0.0997 99.7 2.
• 150 0.1038 103.8 3.12 780 0.0996 99.6 2.
• 180 0.1034 103.4 3.108 810 0.0995 99.5 2.
• 210 0.1031 103.1 3.0975 840 0.0993 99.3 2.
• 240 0.1028 102.8 3.0885 870 0.0992 99.2 2.
• 270 0.1026 102.6 3.081 900 0.0991 99.1 2.
• 300 0.1023 102.3 3.0735 930 0.0989 98.9 2.
• 330 0.1021 102.1 3.066 960 0.0988 98.8 2.
• 360 0.1019 101.9 3.06 990 0.0987 98.7 2.
• 390 0.1017 101.7 3.054 1020 0.0986 98.6 2.
• 420 0.0997 99.7 3.021 1050 0.0984 98.4 2.
• 450 0.1013 101.3 3.015 1080 0.0983 98.3 2.
• 480 0.1011 101.1 3.036 1110 0.0982 98.2 2.
• 510 0.1009 100.9 3.03 1140 0.0981 98.1 2.
• 540 0.1007 100.7 3.024 1170 0.0978 97.8 2.
• 600 0.1005 100.5 3.0165 121. 570 0.1006 100.6 3.0195 1200 0.0977 97.7 2.9325Sum(t*I)