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Design
Research Question: –
What affect does the potential difference (V) across the circuit has on the copper coating on an iron metal?
Hypothesis: –
My hypothesis is based on the Ohm’s Law which states V = IR, where V is the potential difference across the circuit and ‘I’ is the current present in the circuit and ‘R’ is the resistance.
(When ‘R’ is constant)
The above equation states that as the current increases, the electron flow will also increase. Thus increasing the potential difference across the circuit which in turn increases the copper coating on iron metal.
Variables: –
Independent Variable: – Potential difference
Dependent Variable: – Change in mass
Control Variable: –
Materials Required: –
Diagram of the experiment: –
Procedure: –
How to control the Control Variables: –
Data Control and Processing
Raw Data Table:-
The Raw Data Table consists of the data obtained from two trials and is shown below in a tabular form
Table 1
Trial 1 |
||
Potential Difference (V) |
Mass (g) |
|
Initial Mass (g) |
Final mass (g) |
|
1 |
1.560 |
1.596 |
1.5 |
1.596 |
1.641 |
2 |
1.641 |
1.706 |
2.5 |
1.706 |
1.791 |
3 |
1.791 |
1.881 |
Table 2
Trial 2 |
||
Potential Difference (V) |
Mass (g) |
|
Initial Mass (g) |
Final mass (g) |
|
1 |
1.580 |
1.616 |
1.5 |
1.616 |
1.664 |
2 |
1.664 |
1.711 |
2.5 |
1.711 |
1.797 |
3 |
1.797 |
1.889 |
Qualitative Observation: –
After taking down the readings and noting the associated qualitative data, the raw data table is now processed and the data is represented graphically so that a better analysis could be done.
Processed Data Table: –
The processed Data table consists of two tables showing the change in mass after the data is obtained. A graphical representation of the data is also done.
Processed Table 1: –
Trial 1 |
|||
Potential difference (V±0.1) |
Mass (g±0.005) |
||
initial mass (g) |
final mass (g) |
change in mass (g) |
|
1.0 |
1.560 |
1.596 |
0.036 |
1.5 |
1.596 |
1.641 |
0.045 |
2.0 |
1.641 |
1.706 |
0.065 |
2.5 |
1.706 |
1.791 |
0.085 |
3.0 |
1.791 |
1.881 |
0.090 |
Change in Mass was calculated by subtracting the initial mass from the final mass,
For V = 1.0 ± 0.1,
Change in mass (ΔM) = Final Mass – Initial Mass
= 1.596g – 1.560g
= 0.036g
Similarly, Change in Mass was calculated for different Potential difference shown in the table.
Processed Table 2: –
Trial 2 |
|||
Potential difference (V±0.1) |
Mass (g±0.005) |
||
initial mass (g) |
final mass (g) |
change in mass (g) |
|
1.0 |
1.580 |
1.616 |
0.036 |
1.5 |
1.616 |
1.664 |
0.048 |
2.0 |
1.664 |
1.711 |
0.067 |
2.5 |
1.711 |
1.797 |
0.086 |
3.0 |
1.797 |
1.889 |
0.092 |
Similarly for this table,
For V = 3.0 ± 0.1
Change in mass (ΔM) = Final Mass – Initial Mass
= 1.889 – 1.797
= 0.092g
Graphical Representation: –
Trial 1
The above graph shows the relationship between Change in mass and Potential Difference. The obtained graph resembles a linear graph and the error bars are plotted. A trend line is also made to pass through all obtained values.
From the analysis of the graph of Trial 1, it can be observed that as the Potential difference increases the change in mass also increases which means that the amount of copper deposited increases with increasing potential difference.
Similarly, graph for trial 2 is shown below,
Graph 2
The graph obtained for Trial 2 also resembles a linear graph but there are some differences between the graph obtained for trial 1 and trial 2.
As Potential difference across the circuit increases, the Change in mass also increases which means that the copper coating thickens with increasing Potential difference.
Further Calculations are done to find the exact difference between both the trials,
Calculations: –
For trial 1,
The equation for the graph is y = 0.0296x + 0.005,
Finding the slope of the above graph,
m = 0.0296
Slope = 0.0296 gV-1
The slop obtained from differentiating the equation states that when 1V is increased, 0.0296 of Copper deposition takes place. This cycle continues for 1V increase.
Calculating the moles of Copper,
Moles =
Moles =
Moles =
In Copper Sulphate, a +2 charge is present on Copper hence the required electrons will be as follows,
Calculating the transfer of charge,
Charge transferred =
= 89.60 C
Increase in the current =
= 0.149 A
For Trial 2,
The equation of this graph is y = 0.03x + 0.0058
By differentiating the above equation,
m = 0.03 gV-1
Calculating the moles of Copper,
No. of moles = Mass ÷ Molar Mass
Required amount of moles =
= 9.44×10-4 mol at 1 p.d
Calculating the transfer of Charge,
Charge transfer =
= 90.95 C
Increase in the Current = 90.95 C ÷ 600s
= 0.151 A
Conclusion and Evaluation
Conclusion
For Trial 1,
The Charge transfer obtained is 89.60 C
For Trial 2,
The Charge transfer is 90.95 C
The difference between both the charges = 90.95 – 89.60
= 1.35 C
Uncertainty of the experiment cannot be found out because the slope of both the graphs are different.
From the graph, it is seen that,
Evaluation
The major reason for this deviation from the trend line plotted in the graph indicates that the resistance was not controlled properly. The graph is plotted on the basis of the Ohm’s Law stating, when the Resistance ‘R’ is constant but in this experiment, both trial one and trial 2 have different value for Charge transfer thus it can be concluded that the experiment was not accurate because of the presence of fluctuating Resistance.
Errors that may have occurred
How to improve the experiment
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