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In an ionic solution, a current consists of \(\mathrm{Ca}^{2+}\) ions (of charge \(+2 e )\) and \(\mathrm{Cl}^{-}\) ions (of charge \(-e )\) traveling in opposite directions. If \(5.11 \times 10^{18} \mathrm{Cl}^{-}\) ions go from \(A\) to \(B\) every 0.50 min, while \(3.24 \times 10^{18} \mathrm{Ca}^{2+}\) ions move from \(B\) to \(A\), what is the current (in mA) through this solution, and in which direction \((\) from \(A\) to \(B\) or from \(B\) to \(A)\) is it going?

Short Answer

Expert verified
7.313 mA, from B to A.

Step by step solution

01

Understanding Ions and Current Flow

In this step, identify the charges associated with each ion. Chloride ions, \(\mathrm{Cl}^-\), have a charge of \(-e\), and calcium ions, \(\mathrm{Ca}^{2+}\), have a charge of \(+2e\). The electron charge \(e\) is approximately \(1.60 \times 10^{-19}\, \text{Coulombs}\). In a current, positive ion flow contributes to current in its direction, whereas negative ion flow contributes to current in the opposite direction.
02

Calculate Ionic Charge Flow

Convert ion numbers to charge per minute. First calculate for \(\mathrm{Cl}^-\) ions:\[\text{Charge of } \mathrm{Cl}^- = 5.11 \times 10^{18} \times (-1.60 \times 10^{-19})\, \text{C} = -8.176 \times 10^{-1}\, \text{C per 0.5 min}\]For \(\mathrm{Ca}^{2+}\) ions:\[\text{Charge of } \mathrm{Ca}^{2+} = 3.24 \times 10^{18} \times 2 \times 1.60 \times 10^{-19}\, \text{C} = 1.037 \times 10^{0}\, \text{C per 0.5 min}\]
03

Calculate Net Charge Flow per Second

Calculate the net charge difference and divide by time to find charge per second (current in Amperes). Net charge is the sum of the individual charges:\[\text{Net charge } = -8.176 \times 10^{-1}\, \text{C} + 1.037 \times 10^{0}\, \text{C} = 2.194 \times 10^{-1}\, \text{C per 0.5 min}\]Convert to seconds (1 min = 60 s):\[\text{Current (A)} = \frac{2.194 \times 10^{-1}}{30}\, \text{C/s} = 7.313 \times 10^{-3}\, \text{A}\]
04

Convert Amps to Milliamps (mA)

Convert current from Amperes to milliamperes:\[\text{Current (mA)} = 7.313 \times 10^{-3}\, \text{A} \times 1000\, \text{mA/A} = 7.313\, \text{mA}\]
05

Determine Direction of Current Flow

A positive net charge moved in the direction from \(B\) to \(A\) as \(\mathrm{Ca}^{2+}\) ions outnumber the \(\mathrm{Cl}^-\) ions when inverted for current flow direction assignment of positive ions, which produce a current flowing in their own moving direction, thus current flows from \(B\) to \(A\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ionic Solutions
When discussing ionic solutions, it's crucial to understand that they involve ionic compounds, where ions are the carriers of electric charge. Ionic solutions are primarily composed of cations (positively charged ions) and anions (negatively charged ions). These ions are dissolved in a solvent, giving the solution the ability to conduct electricity.

An example is a saline solution, where sodium ions (\( ext{Na}^+\)) and chloride ions (\( ext{Cl}^-\)) are freely moving in water. The ability of these ions to move and transfer charge is what allows ionic solutions to carry a current. In the given exercise, \( ext{Ca}^{2+}\) and \( ext{Cl}^-\) ions are responsible for this. Understanding how ions move in a solution provides foundational knowledge for calculating ionic currents and interpreting the direction of charge flow.
Charge Flow
Charge flow in an ionic solution is analogous to the movement of water in a pipe. It's the movement of charged particles that contributes to the total current. In our exercise context, the movement of \( ext{Cl}^-\) ions contributes negatively to charge flow, whereas \( ext{Ca}^{2+}\) ions contribute positively. Each ion's charge and its quantity determines its total contribution to the charge flow.

To calculate the total charge flow, you need to account for both ion types separately. Multiplying the number of ions by their respective charge gives the total charge moved per unit time. This reveals not just quantity but direction of the charge flow, which ultimately influences current orientation.
Current Direction
Current direction in ionic solutions is determined by the predominant direction of positive charge flow. Because current is defined as the flow of positive charge, in a solution where ions move opposite to each other, the direction of current will align with the positive ions.

In the problem provided, calcium ions (\( ext{Ca}^{2+}\)), which carry a positive charge, dominate over the chloride ions (\( ext{Cl}^-\)), leading to a net flow from \(B\) to \(A\). Even though chloride ions move in the reverse direction, their negative contribution doesn't change current direction when \( ext{Ca}^{2+}\) ions have a stronger effect. Recognizing this principle clarifies why current direction adheres to the flow of positive ions.
Physics Problem Solving
Engaging with physics problems like ionic current calculation develops problem-solving skills by applying mathematical principles to physical phenomena. First, understand the problem: recognize the charges involved and their movements. This insight aids in setting up equations accurately, reflecting the true nature of ion interactions.

Then, break down the problem into manageable steps, as illustrated by calculating individual charge contributions and converting them to current. Finally, evaluate the result: Does the calculation align with known principles, and does the current direction make physical sense?

Using a systematic approach ensures you consider all elements of a problem, leading to a comprehensive understanding and solution. This approach not only solves the current issue but builds a methodology applicable to other complex physics problems.

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Most popular questions from this chapter

Copper has \(8.5 \times 10^{28}\) electrons per cubic meter. (a) How many electrons are there in a 25.0 \(\mathrm{cm}\) length of 12 -gauge copper wire (diameter 2.05 \(\mathrm{mm} ) ?\) (b) If a current of 1.55 \(\mathrm{A}\) is flowing in the wire, what is the average drift speed of the electrons along the wire? (There are \(6.24 \times 10^{18}\) electrons in 1 coulomb of charge.)

Charging and discharging a capacitor. A 1.50\(\mu \mathrm{F}\) capacitor is charged through a 125\(\Omega\) resistor and then discharged through the same resistor by short-circuiting the battery. While the capacitor is being charged, find (a) the time for the charge on its plates to reach \(1-1 / e\) of its maximum value and (b) the current in the circuit at that time. (c) During the discharge of the capacitor, find the time for the charge on its plates to decrease to 1/e of its initial value. Also, find the time for the current in the circuit to decrease to 1\(/ e\) of its initial value.

Navigation of electric fish. Certain fish, such as the Nile fish (Gnathonemus), concentrate charges in their head and tail, thereby producing an electric field in the water around them. (See Figure \(19.67 . )\) This field creates a potential difference of a few volts between the head and tail, which in turn causes current to flow in the conducting seawater. As the fish swims, it passes near objects that have resistivities different from that of seawater, which in turn causes the current to vary. Cells in the skin of the fish are sensitive to this current and can detect changes in it. The changes in the current allow the fish to navigate. (In the next chapter, we shall investigate how the fish might detect this current.) Since the electric field is weak far from the fish, we shall consider only the field running directly from the head to the tail. We can model the seawater through which that field passes as a conducting tube of area 1.0 \(\mathrm{cm}^{2}\) and having a potential difference of 3.0 \(\mathrm{V}\) across its ends. The length of a Nile fish is about \(20 \mathrm{cm},\) and the resistivity of seawater is 0.13\(\Omega \cdot \mathrm{m} .\) (a) How large is the current through the tube of seawater? (b) Suppose the fish swims next to an object that is 10 \(\mathrm{cm}\) long and 1.0 \(\mathrm{cm}^{2}\) in cross-sectional area and has half the resistivity of seawater. This object replaces the seawater for half the length of the tube. What is the current through the tube now? How large is the change in the current that the fish must detect? (Hint: How are this object and the remaining water in the tube connected, in series or in parallel?)

The battery for a certain cell phone is rated at 3.70 \(\mathrm{V}\) . According to the manufacturer it can produce \(3.15 \times 10^{4} \mathrm{J}\) of electrical energy, enough for 5.25 \(\mathrm{h}\) of operation, before needing to be recharged. Find the average current that this cell phone draws when turned on.

If you triple the length of a cable and at the same time double its diameter, what will be its resistance if its original resistance was \(R\) ?

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