Determine the overall reaction and its standard cell potential at 25 °C 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?

Short Answer

Expert verified

The overall reaction is \(3Cu(s) + 2A{u^{3 + }}(aq) \to 3C{u^{2 + }}(aq) + 2Au(s)\)

The standard cell potential at \({25{\circ }}{\rm{C}}\) is \( + 1.16\;{\rm{V}}\)

The standard cell potential is positive, so the reaction is spontaneous.

Step by step solution

01

Define standard cell potential

  • The shorthand notation of a cell reaction indicates phase boundary by a single vertical line (|) and salt bridge is denotes by a double vertical line . In this notation on the left side of salt bridge oxidation reaction which occurred at anode and on the right side of salt bridge; reduction reaction which occurred at cathode.
  • The electrodes are written on the extreme corners; anode on extreme left and cathode on extreme right. The reactants in each half-cell are always first, followed by the products.
  • If the standard cell potential is positive, then the reaction isspontaneousand if the standard cell potential is negative, then the reaction isnon-spontaneous
02

Determine the standard cell potential

The cell reaction that involves silver electrode in \(1{\rm{M}}\) silver nitrate and zinc electrode in \(1{\rm{M}}\) zinc nitrate.

Therefore, the cell notation is as follows:

\(Zn(s)\left| {{Z^{2 + }} || (1M)A{g^ + }(1M)} \right|Ag(s)\)

In the given reaction zinc is oxidized at anode and silver is reduced at cathode.

Anode reaction:

\({\rm{Zn}}(s) \to {\rm{Z}}{{\rm{n}}^{2 + }}(aq) + 2{e^ - }\quad {E^o} = - 0.7618\;{\rm{V}}\)

Cathode reaction:

\(A{g^ + }(aq) + {e^ - } \to Ag(s)\quad {E^o} = + 0.7996\;{\rm{V}}\)

Because the reduction potential of Silver is more than that of Zinc so Ag half-cell acts as cathode Zinc half-cell acts as anode.

The overall or net cell reaction is as follows:

\(2\left( {A{g^ + }(aq) + {e^ - } \to Ag(s)} \right)\)

\(\frac{{1\left( {{\rm{Zn}}(s) \to {\rm{Z}}{{\rm{n}}^{2 + }}(aq) + 2{e^ - }} \right)}}{{{\rm{Zn}}(s) + 2A{g^ + }(aq) \to 2Ag(s) + {\rm{Z}}{{\rm{n}}^{2 + }}(aq)}}\)

Now calculate the standard cell potential at \({25\circ }{\rm{C}}\), using the following expression:

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Most popular questions from this chapter

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?

A galvanic cell consists of a Mg electrode in \({\bf{1M}}\)\({\bf{Mg}}{\left( {{\bf{N}}{{\bf{O}}_{\bf{3}}}} \right)_{\bf{2}}}\)solution and a Ag electrode in 1M AgNO solution. Calculate the standard cell potential at \({25^\circ }{\rm{C}}\).

An active (metal) electrode was found to gain mass as the oxidation-reduction reaction was allowed to proceed. Was the electrode part of the anode or cathode? Explain.

Given the following pairs of balanced half-reactions, determine the balanced reaction for each pair of half reactions in an acidic solution.

(a) \({\bf{Ca}} \to {\bf{C}}{{\bf{a}}^{{\bf{2 + }}}}{\bf{ + 2}}{{\bf{e}}^{\bf{ - }}}{\bf{,}}\quad {{\bf{F}}_{\bf{2}}}{\bf{ + 2}}{{\bf{e}}^{\bf{ - }}} \to {\bf{2\;}}{{\bf{F}}^{\bf{ - }}}\)

(b) \({\bf{Li}} \to {\bf{L}}{{\bf{i}}^{\bf{ + }}}{\bf{ + }}{{\bf{e}}^{\bf{ - }}}{\bf{,}}\quad {\bf{C}}{{\bf{l}}_{\bf{2}}}{\bf{ + 2}}{{\bf{e}}^{\bf{ - }}} \to {\bf{2C}}{{\bf{l}}^{\bf{ - }}}\)

(c) \({\bf{Fe}} \to {\bf{F}}{{\bf{e}}^{{\bf{3 + }}}}{\bf{ + 3}}{{\bf{e}}^{\bf{ - }}}{\bf{,}}\quad {\bf{B}}{{\bf{r}}_{\bf{2}}}{\bf{ + 2}}{{\bf{e}}^{\bf{ - }}} \to {\bf{2B}}{{\bf{r}}^{\bf{ - }}}\)

(d) \({\bf{Ag}} \to {\bf{A}}{{\bf{g}}^{\bf{ + }}}{\bf{ + }}{{\bf{e}}^{\bf{ - }}}{\bf{,}}\quad {\bf{MnO}}_{\bf{4}}^{\bf{ - }}{\bf{ + 4}}{{\bf{H}}^{\bf{ + }}}{\bf{ + 3}}{{\bf{e}}^{\bf{ - }}} \to {\bf{Mn}}{{\bf{O}}_{\bf{2}}}{\bf{ + 2}}{{\bf{H}}_{\bf{2}}}{\bf{O}}\)

For the \(\Delta {G\circ }\) values given here, determine the standard cell potential for the cell.

(a) \(12\;{\rm{kJ}}/{\rm{mol}},{\rm{n}} = 3\)

(b) \( - 45\;{\rm{kJ}}/{\rm{mol}},{\rm{n}} = 1\)

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