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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 10⁻² 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

Show that \(A^{2} \times \Omega=W\) (amperes squared times ohms \(=\) watts).
A person can be killed if a current as small as \(50 \mathrm{mA}\) passes near the heart. An electrician is working on a humid day with hands damp from perspiration. Suppose his resistance from one hand to the other is $1 \mathrm{k} \Omega$ and he is touching two wires, one with each hand. (a) What potential difference between the two wires would cause a 50 -mA current from one hand to the other? (b) An electrician working on a "live" circuit keeps one hand behind his or her back. Why?
What is the current in a \(60.0-\mathrm{W}\) bulb when connected to a \(120-\mathrm{V}\) emf?
What is the energy stored in a small battery if it can move \(675 \mathrm{C}\) through a potential difference of \(1.20 \mathrm{V} ?\)
A portable radio requires an emf of \(4.5 \mathrm{V}\). Olivia has only two non rechargeable \(1.5-\mathrm{V}\) batteries, but she finds a larger 6.0 -V battery. (a) How can she arrange the batteries to produce an emf of $4.5 \mathrm{V} ?$ Draw a circuit diagram. (b) Is it advisable to use this combination with her radio? Explain.
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