Why is a salt bridge necessary in galvanic cells like the one in Figure 17.4?

Short Answer

Expert verified

Salt bridge is necessary to functionalize the galvanic cell.

Step by step solution

01

Define oxidation and reduction

  • A voltaic or a galvanic cellis an electrochemical cell which derives the electrical energy from the spontaneous redox chemical reactions which occur inside the cell.
  • A galvanic cell usually contains two different metals which are dipped in an electrolytic solution or of an individual half-cells having different metals and their ions present in the solution, joined by a salt-bridge or divided by a porous membrane.
02

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

Salt bridge is U shaped glass tube that contains concentrated solution of inert electrolyte like \({\rm{KC}}{{\rm{l}}_,}{\rm{KN}}{{\rm{O}}_3}\) in agar-agar or gelatin. The electrolyte present in salt bridge do not involve in oxidation reduction taken place in cell.

It is necessary in galvanic cell because of the following reasons:

  1. It helps to maintain the electrical neutrality of two half cells.
  2. It completes the circuit in a cell and allows the cell to function. The flow of ions from one solution to other solution takes place in inner circuit and mediated via salt bridge.

Salt bridge is necessary to functionalize the galvanic cell.

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Most popular questions from this chapter

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 equationwhen balancing oxidation-reduction reactions in basic solution?

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Anode:\({\bf{CO(g) + }}{{\bf{H}}_{\bf{2}}}{\bf{O(l)}} \to {\bf{C}}{{\bf{O}}_{\bf{2}}}{\bf{(g) + 2}}{{\bf{H}}^{\bf{ + }}}{\bf{(aq) + 2}}{{\bf{e}}^{\bf{ - }}}\;\;\;{\bf{E}}_{{\bf{anode }}}^{\bf{^\circ }}{\bf{ = - 0}}{\bf{.53\;V}}\)

Cathode:\({\bf{2}}{{\bf{H}}^{\bf{ + }}}{\bf{(aq) + 2}}{{\bf{e}}^{\bf{ - }}} \to {{\bf{H}}_{\bf{2}}}{\bf{(g)}}\;\;\;{\bf{E}}_{{\bf{cathode }}}^{\bf{^\circ }}{\bf{ = 0\;V}}\)

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Overall:\({\bf{CO(g) + }}{{\bf{H}}_{\bf{2}}}{\bf{O(l)}} \to {\bf{C}}{{\bf{O}}_{\bf{2}}}{\bf{(g) + }}{{\bf{H}}_{\bf{2}}}{\bf{(g)}}\;\;\;{\bf{E}}_{{\bf{cell }}}^{\bf{^\circ }}{\bf{ = + 0}}{\bf{.53\;V}}\)

Would this make a good battery for smartphones? Why or why not?

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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