To find the final molarity of the
CuSO4 solution, we need to calculate how much copper sulfate is reduced during the electrolysis process. We'll follow these steps:
Step 1: Calculate the total charge passed through the solution
Given:
Current,
I=3.0ATime,
t=2 hours
=2×3600 s=7200 sThe total charge
Q can be calculated using the formula:
‌Q=I×t‌Q=3.0A×7200 s=21600c Step 2: Calculate the amount of copper deposited
The electrochemical equivalent
(z) of copper ( Cu
) can be derived from Faraday's laws, where the valency
n of copper is 2 , and Faraday's constant
F=96485c∕molFor copper,
M=63.5g∕mol(molar mass of Cu ).
The number of moles of copper deposited can be found using the formula:
n(Cu)‌=‌| Q×‌ current efficiency ‌ |
| n×F |
n(Cu)‌=‌n(Cu)‌=‌n(Cu)‌=0.1008‌ moles ‌ Step 3: Calculate the change in the moles of
CuSO4One mole of Cu is deposited from one mole of
CuSO4 solution:
n(CuSO4)‌ depleted ‌=n(Cu)=0.1008‌ moles ‌ Step 4: Calculate the initial moles of
CuSO4Initial molarity of
CuSO4 solution is 0.45 M and the volume is 0.750 L :
‌n(CuSO4)‌ initial ‌=‌ molarity ‌×‌ volume ‌‌n(CuSO4)‌ initial ‌=0.45×0.750‌n(CuSO4)‌ initial ‌=0.3375‌ moles ‌ Step 5: Calculate the final moles of
CuSO4‌n(CuSO4)‌ final ‌=n(CuSO4)‌ initial ‌−n(CuSO4)‌ depleted ‌‌n(CuSO4)‌ final ‌=0.3375−0.1008‌n(CuSO4)‌ final ‌=0.2367‌ moles ‌ Step 6: Calculate the final molarity of the
CuSO4 solution
The final molarity is given by the number of moles divided by the volume:
‌‌ Molarity ‌‌final ‌=‌| ‌ moles of CusO ‌‌4‌remaining ‌ |
| ‌ volume ‌ |
‌‌ Molarity ‌‌final ‌=‌‌‌ Molarity ‌‌final ‌=0.3156MRounding to three significant figures, the answer is:
Option D: 0.316 M