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Chapter 17: Q17.9CYL (page 952)

What mass of zinc is required to galvanize the top of a 3.00 m × 5.50 m sheet of iron to a thickness of0.100 mm of zinc? If the zinc comes from a solution of \(Zn{\left( {N{O_3}} \right)_2}\) and the current is 25.5 A, how long will it take to galvanize the top of the iron? The density of zinc is 7.140 g/cm3

Short Answer

Expert verified

The time required to galvanize the top of the iron is 378.8 hours

Step by step solution

01

Define Quantitative Aspects of Electrolysis

The amount of current that is allowed to flow in an electrolytic cell is related to the number of moles of electrons. The number of moles of electrons can be related to the reactants and products using stoichiometry. Recall that the SI unit for current \((I)\) is the ampere \(({\rm{A}})\), which is the equivalent of 1 coulomb per second \(\left( {1\;{\rm{A}} = 1\frac{{\rm{C}}}{{\rm{s}}}} \right)\). The total charge ( \(Q\), in coulombs) is given by

\(Q = I \times t = n \times F\)

Where \(t\) is the time in seconds, \(n\) the number of moles of electrons, and \(F\) is the Faraday constant.

Moles of electrons can be used in stoichiometry problems. The time required to deposit a specified amount of metal might also be requested, as in the second of the following examples.

02

Determine the time how long will it take to galvanize the top of the iron

Given:

Thickness \( = 0.100\;{\rm{mm}} = 0.0001\;{\rm{m}}\)

Area of the sheet \( = 3.00\;{\rm{m}} \times 5.50\;{\rm{m}}\)

Current \( = 2.25\;{\rm{A}}\)

Density of zinc \( = 7.140\;{\rm{g}}/{\rm{c}}{{\rm{m}}^3} = 7140\;{\rm{kg}}/{{\rm{m}}^3}\)

The volume of the zinc can be calculated as follows,

Volume of zinc \( = 3.00m \times 5.50m \times 0.0001m = 0.00165{m^3}\)

\({m_{Zn}} = 0.00165 \times 7140 = 11.781\;{\rm{kg}} = 11781\;{\rm{g}}\)

\({\rm{Z}}{{\rm{n}}^{2 + }} + 2{e^ - } = {\rm{Zn}}\)

\(m({\rm{Zn}}) = 65.39 - \)molar mass of \({\rm{Zn}}\)

\({\rm{n}} = 2\)e or \(2 \times 96500\) coulomb for e each \(65.39\;{\rm{g}}\)

We have \(11781\;{\rm{g}}\)

Hence Q can be calculated as follows,

\(Q = \frac{{2 \times 96500 \times 1178}}{{65.39}} = 347.4C\)

We know that

\(\begin{aligned}{}Q = It\\t = \frac{Q}{I}\\ = \frac{{3477.4}}{{25.5}}\\ = 1363602.9\;{\rm{s}}\\ = 22726.7{\rm{ min}}\\ = 378.8{\rm{ hours }}\end{aligned}\)

Hence the time required to galvanize the top of the iron is 378.8 hours

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Most popular questions from this chapter

From the information provided, use cell notation to describe the following systems:

(a) In one half-cell, a solution of \({\rm{Pt}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}\)forms Pt metal, while in the other half-cell, Cu metal goes into a \({\rm{Cu}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}\)solution with all solute concentrations 1M.

(b) The cathode consists of a gold electrode in a \(0.55{\rm{MAu}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_3}\) solution and the anode is a magnesium electrode in \(0.75{\rm{MMg}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}\) solution.

(c) One half-cell consists of a silver electrode in a \(1{\rm{MAgN}}{{\rm{O}}_3}\) solution, and in the other half-cell, a copper electrode in \(1M{\rm{Cu}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}\) is oxidized.

For each of the following balanced half-reactions, determine whether an oxidation or reduction is occurring.

(a) \({\bf{C}}{{\bf{l}}^{\bf{ - }}}{\bf{ + 3}}{{\bf{e}}^{\bf{ - }}} \to {\bf{C}}{{\bf{l}}_{\bf{2}}}\)

(b) \({\bf{M}}{{\bf{n}}^{{\bf{2 + }}}} \to {\bf{Mn}}{{\bf{O}}_{\bf{2}}}\)

(c) \({{\bf{H}}_{\bf{2}}} \to {{\bf{H}}^{\bf{ + }}}\)

(d) \({\bf{NO}}_{\bf{3}}^{\bf{ - }} \to {\bf{NO}}\)

The mass of three different metal electrodes, each from a different galvanic cell, were determined before andafter the current generated by the oxidation-reduction reaction in each cell was allowed to flow for a few minutes.The first metal electrode, given the label A, was found to have increased in mass; the second metal electrode, giventhe label B, did not change in mass; and the third metal electrode, given the label C, was found to have lost mass.

Make an educated guess as to which electrodes were active and which were inert electrodes, and which wereanode(s) and which were the cathode(s)

Balance the following in basic solution:

(a) \({\bf{S}}{{\bf{O}}_{\bf{3}}}^{{\bf{2 - }}}{\bf{(aq) + Cu(OH}}{{\bf{)}}_{\bf{2}}}{\bf{(s)}} \to {\bf{S}}{{\bf{O}}_{\bf{4}}}^{{\bf{2 - }}}{\bf{(aq) + Cu(OH)(s)}}\)

(b) \({{\bf{O}}_{\bf{2}}}{\bf{(g) + Mn(OH}}{{\bf{)}}_{\bf{2}}}{\bf{(s)}} \to {\bf{Mn}}{{\bf{O}}_{\bf{2}}}{\bf{(s)}}\)

(c) \({\bf{N}}{{\bf{O}}_{\bf{3}}}^{\bf{ - }}{\bf{(aq) + }}{{\bf{H}}_{\bf{2}}}{\bf{(g)}} \to {\bf{NO(g)}}\)

(d) \({\bf{Al(s) + Cr}}{{\bf{O}}_{\bf{4}}}^{{\bf{2 - }}}{\bf{(aq)}} \to {\bf{Al(OH}}{{\bf{)}}_{\bf{3}}}{\bf{(s) + Cr(OH}}{{\bf{)}}_{\bf{4}}}^{\bf{ - }}{\bf{(aq)}}\)

Identify the species that was oxidized, the species that was reduced, the oxidizing agent, and the reducing agentin each of the reactions of the previous problem.

(a) \({\bf{S}}{{\bf{O}}_{\bf{3}}}^{{\bf{2 - }}}{\bf{(aq) + Cu(OH}}{{\bf{)}}_{\bf{2}}}{\bf{(s)}} \to {\bf{S}}{{\bf{O}}_{\bf{4}}}^{{\bf{2 - }}}{\bf{(aq) + Cu(OH)(s)}}\)

(b) \({{\bf{O}}_{\bf{2}}}{\bf{(g) + Mn(OH}}{{\bf{)}}_{\bf{2}}}{\bf{(s)}} \to {\bf{Mn}}{{\bf{O}}_{\bf{2}}}{\bf{(s)}}\)

(c) \({\bf{N}}{{\bf{O}}_{\bf{3}}}^{\bf{ - }}{\bf{(aq) + }}{{\bf{H}}_{\bf{2}}}{\bf{(g)}} \to {\bf{NO(g)}}\)

(d) \({\bf{Al(s) + Cr}}{{\bf{O}}_{\bf{4}}}^{{\bf{2 - }}}{\bf{(aq)}} \to {\bf{Al(OH}}{{\bf{)}}_{\bf{3}}}{\bf{(s) + Cr(OH}}{{\bf{)}}_{\bf{4}}}^{\bf{ - }}{\bf{(aq)}}\)
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