Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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

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

b.\({\rm{Ni}}(s)\left| {{\rm{N}}{{\rm{i}}^{2 + }}(aq) || {\rm{A}}{{\rm{g}}^ + }(aq)} \right|{\rm{Ag}}(s)\)

Short Answer

Expert verified

a. Mg is oxidized and it acts as reducing agent and \({\rm{Cu}}\) is reduced and it acts an oxidizing agent.

b. Ni is oxidized and it acts as reducing agent and \({\rm{Ag}}\) is reduced and it acts an oxidizing agent.

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.  

Consider the following cell notation.

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

Since half-cell \({\rm{Mg}}({\rm{s}})\mid {\rm{M}}{{\rm{g}}^{2 + }}({\rm{aq}})\) left side of the cell notation, it represents anode or oxidation half-cell. As oxidation state of \({\rm{Mg}}\) increases from 0 to \( + 2\), so it is getting oxidized and acts as reducing agent.

Since Half-cell \({\rm{C}}{{\rm{u}}^{2 + }}({\rm{aq}})\mid {\rm{Cu}}({\rm{s}})\) lies on the right side of cell notation. Hence it represents cathode or reduction half-cell. As oxidation state of \({\rm{C}}{{\rm{u}}^2} + \) decreases from \( + 2\) to 0, so it is getting reduced and acts as oxidizing agent. Mg is oxidized and it acts as reducing agent and \({\rm{Cu}}\) is reduced and it acts an oxidizing agent.

03

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

Consider the following cell notation

\(Ni(s)\left| {N{i^{2 + }}(aq)} \right|\left| {A{g^ + }(aq)} \right|Ag(s)\)

Since half-cell Ni(s) \(\mid {\rm{N}}{{\rm{i}}^{2 + }}\) (aq) left side of the cell notation, it represents anode or oxidation half-cell. As oxidation state of \({\rm{N}}{{\rm{i}}^2} + \) increases from 0 to \( + 2\), so it is getting oxidized and acts as reducing agent. Since Half-cell \({\rm{A}}{{\rm{g}}^ + }({\rm{aq}})\mid {\rm{Ag}}({\rm{s}})\) lies on the right side of cell notation. Hence it represents cathode or reduction half cell. As oxidation state of \({\rm{Ag}} + \) decreases from \( + 1\) to 0 , so it is getting reduced and acts as oxidizing agent. Ni is oxidized and it acts as reducing agent and \({\rm{Ag}}\) is reduced and it acts an oxidizing agent.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Determine the overall reaction and its standard cell potential at 25 ยฐC for the reaction involving the galvanic cell made from a half-cell consisting of a silver electrode in 1 M silver nitrate solution and a half-cell consisting of azinc electrode in 1 M zinc nitrate. Is the reaction spontaneous at standard conditions?

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 \({\bf{B}}{{\bf{r}}_{\bf{2}}}{\bf{(aq)}}\).

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

Determine \({\bf{\Delta G}}\) and \({\bf{\Delta G}}^\circ \) for each of the reactions in the previous problem.

Balance the following equation in acidic solution:

\({\mathop {{\rm{Hg}}}\nolimits_2 ^{2 + }}{\rm{ + Ag}} \to {\rm{Hg}} + {\rm{A}}{{\rm{g}}^ + }\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free