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Current passes through a solution of sodium chloride. In \(1.00 \mathrm{~s}, 2.68 \times 10^{16} \mathrm{Na}^{+}\) ions arrive at the negative electrode and \(3.92 \times 10^{16} \mathrm{Cl}^{-}\) ions arrive at the positive electrode. (a) What is the current passing between the electrodes? (b) What is the direction of the current?

Short Answer

Expert verified
a) The current passing between the electrodes is \(6.27A\). b) The direction of the current is from the negative electrode to the positive electrode.

Step by step solution

01

Identify given values

The number of ions from the problem are given as \(2.68 \times 10^{16}\) \(Na^+\) ions at the negative electrode and \(3.92 \times 10^{16}\) \(Cl^-\) ions at the positive electrode. The time given is 1 second.
02

Calculate the charge

The charge transferred by an ion is given by its charge. For \(Na^+\) and \(Cl^-\), this is the charge of a single electron or proton, \(1.60 \times 10^{-19}\) Coulombs (C). Multiply the number of ions arriving at each electrode per second by this charge. So for the \(Na^+\) ions to the negative electrode it's \(2.68 \times 10^{16} \times 1.60 \times 10^{-19} = 4.29 C\). For the \(Cl^-\) ions to the positive electrode it's \(3.92 \times 10^{16} \times 1.60 \times 10^{-19} = 6.27 C\). Since the amount of \(Cl^-\) arriving is bigger, this will be the determining factor for current.
03

Determine the current

Use the formula \(I = Q/t\) to calculate the current. Substituting the values \(Q = 6.27C\) and \(t = 1s\), the current \(I\) is then \(6.27A\).
04

Determine the current direction

The direction of the current is the direction in which positive charges move. Because the sodium ions, which are positive, are moving to the negative electrode, the current will be in the opposite direction to their movement, from the negative electrode to the positive electrode.

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Key Concepts

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

Charge Transfer
Charge transfer in an electrolytic process, such as the one described in the exercise where sodium chloride is involved, refers to the movement of electric charge through a medium due to the migration of ions. The medium in this scenario is the sodium chloride solution. To comprehend the fundamentals of charge transfer, it's crucial to recognize that ions are charged particles. An ion becomes positively charged, denoted as a cation (like Na+), when it loses electrons; conversely, it gains electrons and becomes negatively charged, expressed as an anion (like Cl-), when it gains electrons.

During electrolysis, direct current (DC) forces these ions to move towards electrodes with an opposite charge. Positive sodium ions (Na+) are attracted to the negative electrode (cathode) while negative chloride ions (Cl-) move towards the positive electrode (anode). When these ions reach the electrodes, they either gain or lose electrons; this electron exchange represents the fundamental mechanism behind charge transfer. The calculated charge in our example uses the quantity of migrating ions and their electronic charge to determine the total amount of charge transferred in a given time.
Electron Charge
Electron charge, quantified as approximately \(1.60 \times 10^{-19}\) Coulombs (C), plays a vital part in calculations pertaining to electrochemistry. This minuscule charge is considered a fundamental constant and is one of nature's elementary charges. Electrons carrying this charge move in the electrolyte and contribute to the overall current in an electrolytic cell.

The calculation in the exercise requires the understanding that each sodium (Na+) and chloride (Cl-) ion contributes a single charge—equivalent to the charge of one electron—when it arrives at the electrode. Thus, to determine the total charge transferred by all ions arriving within one second, you multiply the number of ions by the electron's charge. Despite the seeming smallness of an individual electron's charge, the massive quantity of ions involved results in a significant cumulative charge, hence considerably influencing the total current flow in the system.
Current Direction
Understanding the current direction is essential when dealing with electrical circuits, and in an electrolytic cell, it's equally significant. The convention in electricity states that current flows from positive to negative. However, this might seem a bit counterintuitive in the case of electrolysis because the current flow within the electrolyte is driven by the movement of ions, which can be either positive or negative.

In the example we're discussing, the exercise seeks to ascertain the current's direction based on ion movement. During electrolysis, despite the negative chloride ions (Cl-) moving toward the positive electrode, and positively-charged sodium ions (Na+) heading to the negative electrode, the current direction is defined from the negative (cathode) towards the positive (anode) electrode. This is due to the historic definition of current as the flow of positive charge. Thus, the direction of the conventional current is opposite to the flow of negatively charged ions (anions) and in the same direction as the flow of positively charged ions (cations). This understanding aligns with the solution presented in Step 4 explaining the direction of current based on positive charge movement.

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Most popular questions from this chapter

A dielectric of permittivity \(3.5 \times 10^{-11} \mathrm{~F} / \mathrm{m}\) completely fills the volume between two capacitor plates. For \(t>0\) the electric flux through the dielectric is \(\left(8.0 \times 10^{3} \mathrm{~V} \cdot \mathrm{m} / \mathrm{s}^{3}\right) t^{3} .\) The dielectric is ideal and nonmagnetic; the conduction current in the dielectric is zero. At what time does the displacement current in the dielectric equal \(21 \mu \mathrm{A} ?\)

A resistor with resistance \(R\) is connected to a battery that has emf \(12.0 \mathrm{~V}\) and internal resistance \(r=0.40 \Omega .\) For what two values of \(R\) will the power dissipated in the resistor be \(80.0 \mathrm{~W} ?\)

A typical small flashlight contains two batteries, each having an emf of \(1.5 \mathrm{~V},\) connected in series with a bulb having resistance \(17 \Omega .\) (a) If the internal resistance of the batteries is negligible, what power is delivered to the bulb? (b) If the batteries last for \(5.0 \mathrm{~h}\), what is the total energy delivered to the bulb? (c) The resistance of real batteries increases as they run down. If the initial internal resistance is negligible, what is the combined internal resistance of both batteries when the power to the bulb has decreased to half its initial value? (Assume that the resistance of the bulb is constant. Actually, it will change somewhat when the current through the filament changes, because this changes the temperature of the filament and hence the resistivity of the filament wire.

A metal wire has a circular cross section with radius \(0.800 \mathrm{~mm}\) You measure the resistivity of the wire in the following way: You connect one end of the wire to one terminal of a battery that has emf \(12.0 \mathrm{~V}\) and negligible internal resistance. To the other terminal of the battery you connect a point along the wire so that the length of wire between the battery terminals is \(d\). You measure the current in the wire as a function of \(d\). The currents are small, so the temperature change of the wire is very small. You plot your results as \(I\) versus \(1 / d\) and find that the data lie close to a straight line that has slope \(600 \mathrm{~A} \cdot \mathrm{m} .\) What is the resistivity of the material of which the wire is made?

A closely wound rectangular coil of 80 turns has dimensions of \(25.0 \mathrm{~cm}\) by \(40.0 \mathrm{~cm} .\) The plane of the coil is rotated from a position where it makes an angle of \(37.0^{\circ}\) with a magnetic field of \(1.70 \mathrm{~T}\) to a position perpendicular to the field. The rotation takes \(0.0600 \mathrm{~s}\). What is the average emf induced in the coil?

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