Raj Bose
IB Chemistry HL
Period 4
15/10/2014
Research Question: How does increasing the concentration of copper (II) nitrate solution in a voltaic cell affect the voltage produced?
Variables:
Variables Description of the variable Method of controlling variable
Independent Concentration of copper (II) nitrate solution used in each trial of the experiment.
N/A
Dependent The voltage being produced by the two electrodes which is measured by the voltmeter
N/A
Controlled Concentration of the zinc nitrate solution. The concentration of zinc nitrate was kept at the same concentration of 0.1M. The solution was made by measuring out 3.788g±0.001g of zinc nitrate powder and a weighing balance was used to measure it. Then it was placed in 500 cm3 beaker. …show more content…
This was done by measuring out 2.022g±0.001g of potassium crystals onto a weighing balance and adding 200 cm3 of distilled water using a 250 cm3 graduated cylinder. Then the solution was mixed and dissolved in the 500 cm3 beaker. Three paper towels were then twisted and soaked in the potassium nitrate solution. It was then placed as the salt bridge across the two beakers which constituted the voltaic cell.
Controlled Material of cathode and anode Copper is to be used as cathode and zinc is to be used as …show more content…
Then using a spatula, measure out 2.022g ±0.001g of KNO3. Add this into the 500〖cm〗^3beaker. Using the 250〖cm〗^3 ±5〖cm〗^3 graduated cylinder measure out 200〖cm〗^3 of distilled water. Dissolve the contents of the beaker using the stirring rod.
Zn(NO3)2 Solution:
0.1M was chosen as the starting concentration for the Zn(NO3)2 solution, the same starting concentration as the copper (II) nitrate solution. In order to calculate the mass required to make a 200〖cm〗^3of 0.1M solution of Zn(NO3)2, the number of moles of the compound required must first be calculated. To calculate the number of moles required, the following equation was used:
C=n/V
(1 mol)/(dm^3 )=n/0.20 n=0.02 mol
n=m/Mr
0.02mol=m/(65.37+(2(14.01+(3*16.00)) g/mol)
0.02mol=m/(189.39 g/mol)
3.788g=m
Procedure: Place a plastic weigh boat on the 3 decimal scales, and then tare the scale. Then using a spatula, measure out 3.788g ±0.001g of Zn(NO3)2. Add this into the 500〖cm〗^3beaker. Using the 250〖cm〗^3 ±5〖cm〗^3 graduated cylinder measure out 200〖cm〗^3 of distilled
IB Chemistry HL
Period 4
15/10/2014
Research Question: How does increasing the concentration of copper (II) nitrate solution in a voltaic cell affect the voltage produced?
Variables:
Variables Description of the variable Method of controlling variable
Independent Concentration of copper (II) nitrate solution used in each trial of the experiment.
N/A
Dependent The voltage being produced by the two electrodes which is measured by the voltmeter
N/A
Controlled Concentration of the zinc nitrate solution. The concentration of zinc nitrate was kept at the same concentration of 0.1M. The solution was made by measuring out 3.788g±0.001g of zinc nitrate powder and a weighing balance was used to measure it. Then it was placed in 500 cm3 beaker. …show more content…
This was done by measuring out 2.022g±0.001g of potassium crystals onto a weighing balance and adding 200 cm3 of distilled water using a 250 cm3 graduated cylinder. Then the solution was mixed and dissolved in the 500 cm3 beaker. Three paper towels were then twisted and soaked in the potassium nitrate solution. It was then placed as the salt bridge across the two beakers which constituted the voltaic cell.
Controlled Material of cathode and anode Copper is to be used as cathode and zinc is to be used as …show more content…
Then using a spatula, measure out 2.022g ±0.001g of KNO3. Add this into the 500〖cm〗^3beaker. Using the 250〖cm〗^3 ±5〖cm〗^3 graduated cylinder measure out 200〖cm〗^3 of distilled water. Dissolve the contents of the beaker using the stirring rod.
Zn(NO3)2 Solution:
0.1M was chosen as the starting concentration for the Zn(NO3)2 solution, the same starting concentration as the copper (II) nitrate solution. In order to calculate the mass required to make a 200〖cm〗^3of 0.1M solution of Zn(NO3)2, the number of moles of the compound required must first be calculated. To calculate the number of moles required, the following equation was used:
C=n/V
(1 mol)/(dm^3 )=n/0.20 n=0.02 mol
n=m/Mr
0.02mol=m/(65.37+(2(14.01+(3*16.00)) g/mol)
0.02mol=m/(189.39 g/mol)
3.788g=m
Procedure: Place a plastic weigh boat on the 3 decimal scales, and then tare the scale. Then using a spatula, measure out 3.788g ±0.001g of Zn(NO3)2. Add this into the 500〖cm〗^3beaker. Using the 250〖cm〗^3 ±5〖cm〗^3 graduated cylinder measure out 200〖cm〗^3 of distilled