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

Short Answer

Expert verified

The standard free energy change value is negative thus the reaction is spontaneous

Step by step solution

01

Define the Standard potential cell

In electrochemistry,a Galvanic cellis a kind of electrochemical cell in which the current is produced using a redox reaction. Redox reactions involve oxidation as well as reduction. A galvanic cell consists of two half cells. In the one-half cell, oxidation occurs. This half-cell acts as the anode. In the other half cell, reduction occurs, This half-cell is termed as the cathode. These two half cells work together and constitute an electrochemical cell.

\({E^\circ }\) cell \( = {E^\circ }\) red ( cathode\() - {E^\circ }\) red ( anode ) were,\({E^\circ }\) cell = standard emf of the cell.

\({E^\circ }\) red = standard reduction potential. And\({E^\circ }\) red (cathode\() > {E^\circ }\) red (anode)

02

Determine the Balance equation

The given cell reaction is,

\({\mathop{\rm Sn}\nolimits} ({\rm{s}}) + 2{\rm{C}}{{\rm{u}}^{2 + }}({\rm{aq}})\rightleftharpoons \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\leftharpoonup\over{\smash{\rightharpoondown}}$}} {\rm{S}}{{\rm{n}}^{2 + }}({\rm{aq}}) + 2{\rm{C}}{{\rm{u}}^ + }({\rm{aq}})\)

\({\mathop{\rm Sn}\nolimits} ({\rm{s}}) + 2{\rm{C}}{{\rm{u}}^{2 + }}({\rm{aq}}) \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\leftharpoonup\over{\smash{\rightharpoondown}}$}} {\rm{S}}{{\rm{n}}^{2 + }}({\rm{aq}}) + 2{\rm{C}}{{\rm{u}}^ + }({\rm{aq}})\)

Here tin(Sn) undergoes oxidation (loss of electrons), acts as an anode and copper undergoes reduction (gain of electrons), and acts as a cathode.

The standard electrode potential of the cell can be calculated using the formula,

\(E_{{\rm{cell }}}^o = E_{{\rm{cathode }}}^o - E_{{\rm{anode }}}^o\)

The relevant half-cell reactions and reduction potentials are:

\(\begin{aligned}{l}2C{u^{2 + }}(aq) + 2{e^ - } \to Cu(s)\\{E^\circ } &= + 0.34Vn\\Sn(s) \to S{n^{2 + }}(aq) + 2{e^ - }\quad \\{E^\circ } &= - 0.14V\end{aligned}\)

Therefore, the standard electrode potential of the cell is,

\(\begin{aligned}{c}E_{{\rm{cell }}}^o = E_{{\rm{cathode }}}^o - E_{{\rm{anode }}}^o\\ = 0.34\;{\rm{V}} - ( - 0.14\;{\rm{V}})\\ = 0.48\;{\rm{V}}\end{aligned}\)

The standard free energy change can be calculated as:

\(\Delta {G^o} = - nFE_{{\rm{cell }}}^o\)

Where,

\(\Delta {G^0}\)is standard free energy change,

\({\rm{n}}\)is the number of electrons transferred in the cell,

\({\rm{F}}\)is the faraday constant \(\left( {96500{\rm{Cmo}}{{\rm{l}}^{ - 1}}} \right)\),

\(E_{{\rm{cell }}}^0\)is the standard electrode potential of the cell.

\(\begin{aligned}{}\Delta {G^\circ } = - nF\\E_{{\rm{cell }}}^o = - 2\;{\rm{mol}} \times 69500{\rm{Cmo}}{{\rm{l}}^{ - 1}} \times 0.48{\rm{J}}{{\rm{C}}^{ - 1}}\\ = - 92640\;{\rm{J}}\\ = - 92.64\;{\rm{kJ}}\end{aligned}\)

The standard free energy change value is negative thus the reaction is spontaneous.

At equilibrium, \(\Delta G = 0\) that is \(Q = K\)

\(\Delta {G^o} = - RT\ln K\)

That is,

\(\begin{aligned}{}K(Q) = {e^{ - \Delta {G^\circ }/RT}}\\ = {e^{ - ( - 92.64kJ/mol)/(0.008314kJ/{\rm{mol}}.K \times 298.15K)}}\\ = 1.70 \times {10^{16}}\end{aligned}\)

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Most popular questions from this chapter

Use the data in Appendix \({\rm{L}}\) to determine the equilibrium constant for the following reactions. Assume 298.15\({\rm{K}}\) if no temperature is given.

(a) \({\bf{AgCl(s)}}\rightleftharpoons {\bf{A}}{{\bf{g}}^{\bf{ + }}}{\bf{(aq) + C}}{{\bf{l}}^{\bf{ - }}}{\bf{(aq)}}\)

(b) \({\bf{CdS(s)}}\rightleftharpoons {\bf{C}}{{\bf{d}}^{{\bf{2 + }}}}{\bf{(aq) + }}{{\bf{S}}^{{\bf{2 - }}}}{\bf{(aq)}}\) at \({\bf{377\;K}}\)

(c) \({\bf{H}}{{\bf{g}}^{{\bf{2 + }}}}{\bf{(aq) + 4B}}{{\bf{r}}^{\bf{ - }}}{\bf{(aq)}}\rightleftharpoons {\left[ {{\bf{HgB}}{{\bf{r}}_{\bf{4}}}} \right]^{{\bf{2 - }}}}{\bf{(aq)}}\)

(d) \({{\bf{H}}_{\bf{2}}}{\bf{O(l)}}\rightleftharpoons {{\bf{H}}^{\bf{ + }}}{\bf{(aq) + O}}{{\bf{H}}^{\bf{ - }}}{\bf{(aq)}}\) at \({\bf{2}}{{\bf{5}}^{\bf{^\circ }}}{\bf{C}}\)

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) \({\bf{Al(s) + Cr}}{{\bf{O}}_{\bf{4}}}^{{\bf{2 - }}}{\bf{(aq)}} \to {\bf{Al(OH}}{{\bf{)}}_{\bf{3}}}{\bf{(s) + Cr(OH}}{{\bf{)}}_{\bf{4}}}^{\bf{ - }}{\bf{(aq)}}\)

Determine the overall reaction and its standard cell potential at \({25\circ }{\rm{C}}\) for these reactions. Is the reaction spontaneous at standard conditions? Assume the standard reduction for \({\rm{B}}{{\rm{r}}_2}(l)\) is the same as for \({\rm{B}}{{\rm{r}}_2}(aq)\).

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?

For each of the following balanced half-reactions, determine whether an oxidation or reduction is occurring.

(a) \({\bf{C}}{{\bf{l}}^{\bf{ - }}}{\bf{ + 3}}{{\bf{e}}^{\bf{ - }}} \to {\bf{C}}{{\bf{l}}_{\bf{2}}}\)

(b) \({\bf{M}}{{\bf{n}}^{{\bf{2 + }}}} \to {\bf{Mn}}{{\bf{O}}_{\bf{2}}}\)

(c) \({{\bf{H}}_{\bf{2}}} \to {{\bf{H}}^{\bf{ + }}}\)

(d) \({\bf{NO}}_{\bf{3}}^{\bf{ - }} \to {\bf{NO}}\)

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