/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 Two electrodes are placed in a c... [FREE SOLUTION] | 91影视

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Chapter 18: Problem 7

Two electrodes are placed in a calcium chloride solution and a potential difference is maintained between them. If $3.8 \times 10^{16} \mathrm{Ca}^{2+}\( ions and \)6.2 \times 10^{16} \mathrm{Cl}^{-}$ ions per second move in opposite directions through an imaginary area between the electrodes, what is the current in the solution?

Short Answer

Expert verified
Answer: The current in the calcium chloride solution is approximately 3.584 x 10鈦宦� A.

Step by step solution

01

Calculate the charge each type of ion carries

To find the charge carried by each ion, we will use the elementary charge. The calcium ion \((\mathrm{Ca}^{2+})\) has a charge of \(2e\) (where \(e\) is the elementary charge) and the chloride ion \((\mathrm{Cl}^{-})\) has a charge of \(-e\). The elementary charge is \(e=1.6 \times 10^{-19}\,\mathrm{C}\).
02

Multiply the charge by the number of ions for each ion type

Now we need to find the amount of charge that moves in one second for each type of ion. For \(\mathrm{Ca}^{2+}\) ions, we have: Charge due to Ca\(^{2+}\) ions = Number of Ca\(^{2+}\) ions \(\times\) charge per Ca\(^{2+}\) ion Charge due to Ca\(^{2+}\) ions = \((3.8 \times 10^{16}\,\mathrm{ions}) \times (2 \times 1.6 \times 10^{-19}\,\mathrm{C})\) Similarly, for \(\mathrm{Cl}^-\) ions, we have: Charge due to Cl\(^-\) ions = Number of Cl\(^-\) ions \(\times\) charge per Cl\(^-\) ion Charge due to Cl\(^-\) ions = \((6.2 \times 10^{16}\,\mathrm{ions}) \times (-1.6 \times 10^{-19}\,\mathrm{C})\)
03

Add the contributions from both ions and find the total current

The total charge that moves in one second can be represented as the total current. Therefore, the total current \(I\) is given by: \(I\) = Charge due to Ca\(^{2+}\) ions + Charge due to Cl\(^-\) ions Substituting the values calculated in step 2: \(I\) = \((3.8 \times 10^{16}\,\mathrm{ions}) \times (2 \times 1.6 \times 10^{-19}\,\mathrm{C}) + (6.2 \times 10^{16}\,\mathrm{ions}) \times (-1.6 \times 10^{-19}\,\mathrm{C})\) Performing the calculations, we get: \(I \approx 3.584 \times 10^{-2}\,\mathrm{A}\) The current in the calcium chloride solution is approximately \(3.584 \times 10^{-2}\,\mathrm{A}\).

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Most popular questions from this chapter

A source of emf \(\&\) has internal resistance \(r\), (a) What is the terminal voltage when the source supplies a current \(I ?\) (b) The net power supplied is the terminal voltage times the current. Starting with \(P=I \Delta V,\) derive Eq. \((18-22)\) for the net power supplied by the source. Interpret each of the two terms. (c) Suppose that a battery of emf 8 and internal resistance \(r\) is being recharged: another emf sends a current \(I\) through the battery in the reverse direction (from positive terminal to negative). At what rate is electric energy converted to chemical energy in the recharging battery? (d) What is the power supplied by the recharging circuit to the battery?
A solar cell provides an emf of \(0.45 \mathrm{V}\). (a) If the cell supplies a constant current of \(18.0 \mathrm{mA}\) for \(9.0 \mathrm{h},\) how much electric energy does it supply? (b) What is the power-the rate at which it supplies electric energy?
Suppose a collection of five batteries is connected as shown. (a) What is the equivalent emf of the collection? Treat them as ideal sources of emf. (b) What is the current through the resistor if its value is \(3.2 \Omega ?\)
An ammeter with a full scale deflection for \(I=10.0 \mathrm{A}\) has an internal resistance of \(24 \Omega .\) We need to use this ammeter to measure currents up to 12.0 A. The lab instructor advises that we get a resistor and use it to protect the ammeter. (a) What size resistor do we need and how should it be connected to the ammeter, in series or in parallel? (b) How do we interpret the ammeter readings?
During a "brownout," which occurs when the power companies cannot keep up with high demand, the voltage of the household circuits drops below its normal $120 \mathrm{V} .\( (a) If the voltage drops to \)108 \mathrm{V},$ what would be the power consumed by a "100-W" light-bulb (that is, a light-bulb that consumes \(100.0 \mathrm{W}\) when connected to \(120 \mathrm{V}\) )? Ignore (for now) changes in the resistance of the light-bulb filament. (b) More realistically, the light-bulb filament will not be as hot as usual during the brownout. Does this make the power drop more or less than that you calculated in part (a)? Explain.
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