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A galvanic cell was constructed as shown in the diagram - HSC - SSCE Chemistry - Question 24 - 2011 - Paper 1
Question 24
A galvanic cell was constructed as shown in the diagram. (a) Calculate the standard cell potential (E°). In your answer, include a net ionic equation for the overal... show full transcript
Worked Solution & Example Answer:A galvanic cell was constructed as shown in the diagram - HSC - SSCE Chemistry - Question 24 - 2011 - Paper 1
Step 1
Calculate the standard cell potential (E°). Include a net ionic equation for the overall cell reaction.
Answer
To calculate the standard cell potential (E°), we first need to determine the half-reactions for nickel and copper:
-
For Nickel reduction:
-
For Copper oxidation:
The overall net ionic equation combining these half-reactions is:
Next, we typically look up the standard reduction potentials:
- E°(Ni²⁺/Ni) = -0.25 V
- E°(Cu²⁺/Cu) = +0.34 V
Now we can calculate the standard cell potential:
Step 2
Calculate the final mass of the nickel electrode.
Answer
To find the final mass of the nickel electrode, we first need to calculate the moles of copper deposited using its molar mass:
- Molar mass of Cu = 63.55 g/mol.
- Moles of Cu deposited = ( \frac{0.395 ext{ g}}{63.55 ext{ g/mol}} = 0.00621 ext{ mol} ).
The reaction indicates that 2 moles of electrons are needed to reduce 1 mole of copper. Thus, 0.00621 moles of Cu will require 0.00621 moles of Ni to be oxidized:
[ 0.00621 ext{ mol Ni} ]
Now, using the molar mass of Ni (58.69 g/mol):
- Mass of Ni lost = ( 0.00621 \text{ mol} \times 58.69 \text{ g/mol} = 0.3641 ext{ g} ).
Finally, we can calculate the final mass of the nickel electrode:
Initial Mass of Ni = 10.27 g. Final Mass of Ni = Initial Mass - Mass lost = 10.27 g - 0.3641 g = 9.9059 g. Thus, the final mass of the nickel electrode is approximately 9.91 g.
Step 3
Calculate the final concentration of the nickel(II) nitrate solution.
Answer
To find the final concentration of the nickel(II) nitrate solution, we first determine the volume of the solution:
Assuming the total volume is unchanged and is 200 mL.
Next, we calculate the moles of Ni lost:
- From the previous calculation, moles of Ni lost = 0.00621 moles.
Now, the initial moles of nickel in the solution: Initial concentration = 0.100 mol/L × 0.200 L = 0.020 mol.
After the loss, the remaining moles of nickel are: Remaining moles of Ni = 0.020 mol - 0.00621 mol = 0.01379 mol.
Finally, to find the final concentration: Final concentration = ( \frac{0.01379 ext{ mol}}{0.200 ext{ L}} = 0.06895 ext{ mol/L} ). Thus, the final concentration of the nickel(II) nitrate solution is approximately 0.069 mol/L.