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Why must the charge balance in oxidation-reduction reactions?

Short Answer

Expert verified

The charge should be balanced in oxidation-reduction reactions because the number of electrons lost during oxidation is equal to the number of electrons gained during reduction.

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 halfreactions, net total equation can be obtained. This method of balancing redox reaction is known as half equation method.
02

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

Because number of electrons lost during oxidation is equal to the number of electrons gained during reduction.In an oxidation-reduction reaction, one of the species oxidizes and the other reduces. Here, oxidation means loss of electrons and reduction means gain of electrons.The half- reaction method is used to balance the oxidation reduction reaction. In this method, first the unbalanced chemical reaction splits into two half reactions in which one is oxidation and other is reduction. The solutions in which the electrochemical reactions take place can be acidic, basic or neutral. The number of electrons lost by species undergoing reduction must be equal to the number of electrons gained by the species undergoing oxidation thus, the charge must be balanced.

The charge should be balanced in oxidation-reduction reactions because the number of electrons lost during oxidation is equal to the number of electrons gained during reduction.

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Most popular questions from this chapter

Suppose you have three different metals, \({\rm{A}}\), \(B\), and \({\rm{C}}\). When metals \({\rm{A}}\) and \(B\) come into contact, \(B\) corrodes and \({\rm{A}}\)do not corrode. When metals \({\rm{A}}\)and \({\rm{C}}\) come into contact, \({\rm{A}}\) corrodes and \({\rm{C}}\) do not corrode. Based on this information, which metal corrodes and which metal does not corrode when \(B\) and \({\rm{C}}\) come into contact?

What is the cell potential for the following reaction at room temperature?

\({\bf{Al(s)}}\left| {{\bf{A}}{{\bf{l}}^{{\bf{3 + }}}}{\bf{(aq,0}}{\bf{.15M)}} | | {\bf{C}}{{\bf{u}}^{{\bf{2 + }}}}{\bf{(aq,0}}{\bf{.025M)}}} \right|{\bf{Cu(s)}}\)

What are the values of \(n\) and \(Q\) for the overall reaction? Is the reaction spontaneous under these conditions?

For each reaction listed, determine its standard cell potential at \({25\circ }{\rm{C}}\) and whether the reaction is spontaneous at standard conditions.

(a)\({\mathop{\rm Mn}\nolimits} (s) + {\rm{N}}{{\rm{i}}^{2 + }}(aq) \to {{\mathop{\rm Mn}\nolimits} ^{2 + }}(aq) + {\rm{Ni}}(s)\)

(b)\(3{\rm{C}}{{\rm{u}}^{2 + }}(aq) + 2{\rm{Al}}(s) \to 2{\rm{A}}{{\rm{l}}^{3 + }}(aq) + 3{\rm{Cu}}(s)\)

(c)\({\rm{Na}}(s) + {\rm{LiN}}{{\rm{O}}_3}(aq) \to {\rm{NaN}}{{\rm{O}}_3}(aq) + {\rm{Li}}(s)\)

(d) \({\rm{Ca}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}(aq) + {\rm{Ba}}(s) \to {\rm{Ba}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}(aq) + {\rm{Ca}}(s)\)

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?

What is the standard free energy change and the equilibrium constant for the following reaction at room temperature? Is the reaction spontaneous?

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