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Chapter 17: Q17.2E (page 957)

For the scenario in the previous question, how many electrons moved through the circuit.

Short Answer

Expert verified

Number of electrons is \({\rm{3}}{\rm{.28 \times 1}}{{\rm{0}}^{{\rm{22}}}}\)

Step by step solution

01

Formula of number of electron

The charge of 1 electron is equal to\({\bf{1}}{\bf{.6022 \times 1}}{{\bf{0}}^{{\bf{ - 19}}}}{\bf{C}}\). Thus, in 1C, the number of electrons will be:

\(\begin{array}{l}{\bf{1C = }}\frac{{\bf{1}}}{{{\bf{1}}{\bf{.6022 \times 1}}{{\bf{0}}^{{\bf{ - 19}}}}}}{\bf{ }}\\{\bf{electrons = 6}}{\bf{.2414 \times 1}}{{\bf{0}}^{{\bf{18}}}}{\bf{ electrons}}\end{array}\)

02

Determine the number of electrons

For 5250 C of charge, the number of electrons can be calculated as follows:

\({\rm{1C}} \to {\rm{6}}{\rm{.2414 \times 1}}{{\rm{0}}^{{\rm{18}}}}{\rm{ electrons }}\)

Thus, \(5250{\rm{C}} \to (5250)\left( {6.2414 \times {{10}^{18}}{\rm{ electrons }}} \right) = 3.28 \times {10^{22}}{\rm{ electrons }}\)

Therefore, the number of electrons are \(3.28 \times 1{0^{22}}\).

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Most popular questions from this chapter

Why do batteries go dead, but fuel cells do not?

Given the following cell notations, determine the species oxidized, species reduced, and the oxidizing agent and reducing agent, without writing the balanced reactions

a. \({\rm{Mg}}(s)\left| {{\rm{M}}{{\rm{g}}^{2 + }}(aq) || {\rm{C}}{{\rm{u}}^{2 + }}(aq)} \right|{\rm{Cu}}(s)\)

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Use cell notation to describe the galvanic cell where copper(II) ions are reduced to copper metal and zinc metal is oxidized to zinc ions.

Identify the species oxidized, species reduced, and the oxidizing agent and reducing agent for all the reactions in the previous problem.

(a) \({\rm{Al}}(s) + {\rm{Z}}{{\rm{r}}^{4 + }}(aq) \to {\rm{A}}{{\rm{l}}^{3 + }}(aq) + {\rm{Zr}}(s)\)

(b) \({\rm{A}}{{\rm{g}}^ + }(aq) + {\rm{NO}}(g) \to {\rm{Ag}}(s) + {\rm{N}}{{\rm{O}}_3}^ - (aq)\)(acidic solution)

(c) \({\rm{Si}}{{\rm{O}}_3}^{2 - }(aq) + {\rm{Mg}}(s) \to {\rm{Si}}(s) + {\rm{Mg}}{({\rm{OH}})_2}(s)\)(basic solution)

(d) \({\rm{Cl}}{{\rm{O}}_3}^ - (aq) + {\rm{Mn}}{{\rm{O}}_2}(s) \to {\rm{C}}{{\rm{l}}^ - }(aq) + {\rm{Mn}}{{\rm{O}}_4}^ - (aq)\)(basic solution)

What mass of each product is produced in each of the electrolytic cells of the previous problem if a total charge of \({\bf{3}}.{\bf{33}} \times {\bf{1}}{{\bf{0}}^5}{\rm{ }}{\bf{C}}\) passes through each cell? Assume the voltage is sufficient to perform the reduction
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