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Write the following balanced reactions using cell notation. Use platinum as an inert electrode, if needed.

(a) \({\rm{Mg}}(s) + {\rm{N}}{{\rm{i}}^{2 + }}(aq) \to {\rm{M}}{{\rm{g}}^{2 + }}(aq) + {\rm{Ni}}(s)\)

(b) \(2{\rm{A}}{{\rm{g}}^ + }(aq) + {\rm{Cu}}(s) \to {\rm{C}}{{\rm{u}}^{2 + }}(aq) + 2{\rm{Ag}}(s)\)

(c) \({\rm{Mn}}(s) + {\rm{Sn}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}(aq) \to {\rm{Mn}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}(aq) + {\rm{Au}}(s)\)

(d)\(3{\rm{CuN}}{{\rm{O}}_3}(aq) + {\rm{Au}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_3}(aq) \to 3{\rm{Cu}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}(aq) + {\rm{Au}}(s)\)

Short Answer

Expert verified

a. Balanced reaction with cell notation will be: \({\mathop{\rm Mg}\nolimits} (s)\left| {M{g^{2 + }}(aq)||{\rm{N}}{{\rm{i}}^{2 + }}(aq)} \right|{\rm{Ni}}(s)\)

b. Balanced reaction with cell notation will be: \({\rm{Cu}}(s)\left| {{\rm{C}}{{\rm{u}}^{2 + }}(aq) || {\rm{A}}{{\rm{g}}^ + }(aq)} \right|{\rm{Ag}}(s)\)

c. Balanced reaction with cell notation will be: \({\mathop{\rm Mn}\nolimits} (s)\left| {{{{\mathop{\rm Mn}\nolimits} }^{2 + }}(aq) || {{{\mathop{\rm Sn}\nolimits} }^ + }(aq)} \right|{\mathop{\rm Sn}\nolimits} (s)\)

d. Balanced reaction with cell notation will be: \({\mathop{\rm Pt}\nolimits} (s)\left| {{\rm{C}}{{\rm{u}}^ + }(aq),{\rm{C}}{{\rm{u}}^{2 + }}(aq) || A{u^{3 + }}(aq)} \right|Au(s)\)

Step by step solution

01

Define oxidation and reduction

  • The oxidation-reduction reaction is also known as a redox reaction. In this reaction, one reactant is oxidized and other is reduced. In balancing an oxidation-reduction reaction, they must be first divided into two half reactions: one is oxidation reaction and other is reduction reaction.
  • The balancing of redox reaction is complicated as compared to simple balancing. It is necessary to determine the half reactions of reactants undergoing oxidation and reduction. On adding the two half reactions, net total equation can be obtained. This method of balancing redox reaction is known as half equation method.
02

a) Determine balanced reaction for each pair of half reactions in an acidic solution.  

The given balanced reaction is as follows:

\({\rm{Mg}}(s) + {\rm{N}}{{\rm{i}}^{2 + }}(aq) \to {\rm{M}}{{\rm{g}}^{2 + }}(aq) + {\rm{Ni}}(s)\)

The above reaction can be split into two half reactions as follows:

\(\begin{array}{l}{\rm{Mg}}(s) \to + {\rm{M}}{{\rm{g}}^{2 + }}(aq) + 2{e^ - }\\{\rm{N}}{{\rm{i}}^{2 + }}(aq) + 2{e^ - } \to {\rm{Ni}}(s)\end{array}\)

Here, magnesium atom gets oxidized to magnesium ion by release of two electrons and nickel ion gained 2 electrons to form neutral nickel atom which is a reduction reaction. In a cell reaction, first oxidation is shown which is separated from the reduction via a salt bridge represented as two vertical lines.

Thus, balanced reaction with cell notation will be: \({\mathop{\rm Mg}\nolimits} (s)\left| {M{g^{2 + }}(aq)||{\rm{N}}{{\rm{i}}^{2 + }}(aq)} \right|{\rm{Ni}}(s)\)

03

b) Determine balanced reaction for each pair of half reactions in an acidic solution.

The given balanced reaction is as follows:

\(2{\rm{A}}{{\rm{g}}^ + }(aq) + {\rm{Cu}}(s) \to {\rm{C}}{{\rm{u}}^{2 + }}(aq) + 2{\rm{Ag}}(s)\)

The above reaction can be split into two half reactions as follows:

\(\begin{array}{l}{\rm{Cu}}(s) \to {\rm{C}}{{\rm{u}}^{2 + }}(aq) + 2{e^ - }\\2{\rm{A}}{{\rm{g}}^ + }(aq) + 2{e^ - } \to 2{\rm{Ag}}(s)\end{array}\)

Here, copper atom gets oxidized to copper ion by release of two electrons and silver ion gained 2 electrons to form neutral silver atom which is a reduction reaction. In a cell reaction, first oxidation is shown which is separated from the reduction via a salt bridge represented as two vertical lines. Thus, balanced reaction with cell notation will be:\({\rm{Cu}}(s)\left| {{\rm{C}}{{\rm{u}}^{2 + }}(aq) || {\rm{A}}{{\rm{g}}^ + }(aq)} \right|{\rm{Ag}}(s)\)

04

c) Determine balanced reaction for each pair of half reactions in an acidic solution.  

The given balanced reaction is as follows:

\({\rm{Mn}}(s) + {\mathop{\rm Sn}\nolimits} {\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}(aq) \to {\mathop{\rm Mn}\nolimits} {\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}(aq) + {\mathop{\rm Sn}\nolimits} (s)\)

The above reaction can be split into two half reactions as follows:

\(\begin{array}{l}{\mathop{\rm Mn}\nolimits} (s) \to {{\mathop{\rm Mn}\nolimits} ^{2 + }}(aq) + 2{e^ - }\\{{\mathop{\rm Sn}\nolimits} ^{2 + }}(aq) + 2{e^ - } \to {\mathop{\rm Sn}\nolimits} (s)\end{array}\)

Here, manganese atom gets oxidized to manganese ion by release of two electrons and lead ion gained 2 electrons to form neutral lead atom which is a reduction reaction.

In a cell reaction, first oxidation is shown which is separated from the reduction via a salt bridge represented as two vertical lines.

Thus, balanced reaction with cell notation will be: \({\mathop{\rm Mn}\nolimits} (s)\left| {{{{\mathop{\rm Mn}\nolimits} }^{2 + }}(aq) || {{{\mathop{\rm Sn}\nolimits} }^ + }(aq)} \right|{\mathop{\rm Sn}\nolimits} (s)\)

05

d) Determine balanced reaction for each pair of half reactions in an acidic solution.  

The given balanced reaction should be written using the cell notation.

The given balanced reaction is as follows:

\(3{\rm{CuN}}{{\rm{O}}_3}(aq) + {\rm{Au}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_3}(aq) \to 3{\rm{Cu}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}(aq) + {\rm{Au}}(s)\)

The above reaction can be split into two half reactions as follows:

\(\begin{array}{l}{\rm{C}}{{\rm{u}}^ + }(aq) \to {\rm{C}}{{\rm{u}}^{2 + }}(aq) + {e^ - }\\{\rm{A}}{{\rm{u}}^{3 + }}(aq) + 3{e^ - } \to {\rm{Au}}(s)\end{array}\)

Here, \({\rm{Cu}}\) (I) gets oxidized to \({\rm{Cu}}\) (II) ion by release of one electron and gold(III) ion gained 3 electrons to form neutral gold atom which is a reduction reaction. Here, Pt electrode is used as an inert electrode. In a cell reaction, first oxidation is shown which is separated from the reduction via a salt bridge represented as two vertical lines.

Thus, balanced reaction with cell notation will be: \({\mathop{\rm Pt}\nolimits} (s)\left| {{\rm{C}}{{\rm{u}}^ + }(aq),{\rm{C}}{{\rm{u}}^{2 + }}(aq) || A{u^{3 + }}(aq)} \right|Au(s)\)

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Most popular questions from this chapter

List some things that are typically considered when selecting a battery for a new application.

Why would a sacrificial anode made of lithium metal be a bad choice despite its \({\bf{E}}_{{\bf{Li}}}^{\bf{^\circ }}{\bf{ + Li = - 3}}{\bf{.04\;V}}\), which appears to be able to protect all the other metals listed in the standard reduction potential table?

Why is it not possible for hydroxide ion \(\left( {{\bf{O}}{{\bf{H}}^ - }} \right)\)to appear in either of the half-reactions or the overall equation when balancing oxidation-reduction reactions in acidic solution?

Determine the standard cell potential and the cell potential under the stated conditions for the electrochemical reactions described here. State whether each is spontaneous or nonspontaneous under each set of conditions at \({\bf{298}}{\bf{.15\;K}}\).

(a) \({\bf{Hg(l) + }}{{\bf{S}}^{{\bf{2 - }}}}{\bf{(aq,0}}{\bf{.10M) + 2A}}{{\bf{g}}^{\bf{ + }}}{\bf{(aq,0}}{\bf{.25M)}} \to {\bf{2Ag(s) + HgS(s)}}\)

(b) The galvanic cell is made from a half-cell consisting of an aluminium electrode in 0.015M aluminium nitrate solution and a half-cell consisting of a nickel electrode in \({\bf{0}}{\bf{.25M}}\) nickel(II) nitrate solution.

(c) The cell is made of a half-cell in which \({\bf{1}}{\bf{.0M}}\) aqueous bromide is oxidized to \({\bf{0}}{\bf{.11M}}\) bromine ion and a half-cell in which aluminium ion at \({\bf{0}}{\bf{.023M}}\) is reduced to aluminium metal. Assume the standard reduction potential for \({\bf{B}}{{\bf{r}}_{\bf{2}}}{\bf{(l)}}\) is the same as that of \({\bf{B}}{{\bf{r}}_{\bf{2}}}{\bf{(aq)}}\).

Consider the following metals: Ag, Au, \(Mg, Ni,\)\(and\)\(Zn\). Which of these metals could be used as a sacrificial anode in the cathodic protection of an underground steel storage tank? Steel is mostly iron, so use \( - 0.447\;{\rm{V}}\) as the standard reduction potential for steel.

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