Answer:
Part A. The half-cell B is the cathode and the half-cell A is the anode
Part B. 0.017V
Explanation:
Part A
The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).
So, the half-cell B is the cathode and the half-cell A is the anode.
Part B
By the Nersnt equation:
E°cell = E° - (0.0592/n)*log[anode]/[cathode]
Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.
E°cell = 0 - (0.0592/2)*log(0.23/0.87)
E°cell = 0.017V
The correct answer to the problem is 0.193 which is three significant figures.
<h3>What are significant figures?</h3>
The term significant figures has to do with the figures that have a mathematical meaning. We know that the result has to correspond to the highest number of significant figures.
Hence, If we multiply 0.34 x 0.568 the result ought to be recorded as 0.193 which is three significant figures.
Learn more about significant figures:brainly.com/question/14804345
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Answer:
Calcium sulphate is the right answer for this question.
It is because calcium sulphate helps to obtain the hardness in water. hardness.
The presence of this resource is a main reason for permanent hardness.
Hard drinking water has moderate health benefits, but can pose serious problems for the industrial settings.
Explanation:
Answer:
1.69.
Explanation:
- The solution = 12.0 / 7.11 = 1.687 = 1.69.
- The rule of significant figures for division states that: the results are reported to the fewest significant figures.
- 12.0 contains 3 significant figures.
- 7.11 contains 3 significant figures.
So, the solution should contain 3 significant figures.
- Now, the issue id of rounding; In a series of calculations, carry the extra digits through to the final result, then round.
- If the digit to be removed is equal to or greater than 5, the preceding digit is increased by 1.
- The digit that should be removed is 7 that is larger than 5 so increase the preceding digit by 1.
