Question

A)2Cu(s)+Mn2+(ag)--> 2Cu+(aq)+ Mn(s)
Ecell=??V

B)Determine whether the reaction is spontaneous as written.
Spontaneous or nonspontaneous?

C) MNO2(s)+4H+(aq)+Zn(s)--> Mn2+(aq)+2H2O(l)+Zn2+(aq)
Ecell=??V

D)Determine whether the reaction is spontaneous as written.
Spontaneous or nonspontaneous?

E)Cl2(l)+2F-(aq)-->F2(aq)+2Cl-(aq)
Ecell=??V

F)Determine whether the reaction is spontaneous as written.
Spontaneous or nonspontaneous?

Answer 1

A) Overallcellreaction: \(2 \mathrm{Cu}(\mathrm{s})+\mathrm{Mn}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Mn}(\mathrm{s})+2 \mathrm{Cu}^{+}(\mathrm{aq})\)

Oxidation half reaction: \(2 \mathrm{Cu}(\mathrm{s}) \rightarrow 2 \mathrm{Cu}^{+}(a q)+2 e^{-} ; \mathrm{E}_{\mathrm{Cu}^{+} \mathrm{KCu}}^{0}=0.52 \mathrm{~V}\)

Reduction half reaction: \(\mathrm{Mn}^{2+}(a q)+2 e^{-} \rightarrow \mathrm{Mn}(s) ; \mathrm{E}_{\mathrm{Mn}^{2+} / \mathrm{Ma}}^{0}=-1.18 \mathrm{~V}\)

\(\begin{aligned} \text { Cell potential, } \mathrm{E}_{\text {cell }}^{\circ} &=\mathrm{E}_{\text {cathode }}^{0}-\mathrm{E}_{\text {anode }}^{0} \\ &=\mathrm{E}_{\mathrm{Mn}^{2+}}^{0}-\mathrm{E}_{\mathrm{Cu}^{+} / \mathrm{C}_{4}}^{0} \\ &=-1.18 \mathrm{~V}-0.52 \mathrm{~V} \\ &=-1.70 \mathrm{~V} \end{aligned}\)

B) since cell potential comes out to be negative, this reaction is a nonspontaneous one.

\(\begin{aligned} \text { C) Overallcellreaction: } \mathrm{MnO}_{2}(\mathrm{~s})+4 \mathrm{H}^{+}(\mathrm{aq})+\mathrm{Zn}(\mathrm{s}) \rightarrow & \mathrm{Mn}^{2+}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(l) \\ &+\mathrm{Zn}^{2+}(\mathrm{aq}) \end{aligned}\)

Oxidation half reaction: \(\mathrm{Zn}(\mathrm{s}) \rightarrow \mathrm{Zn}^{2+}(a q)+2 e^{-} ; \mathrm{E}_{2 n^{2}+/ 2 \mathrm{~A}}^{0}=-0.76 \mathrm{~V}\)

Reduction half reaction: \(\mathrm{MnO}_{2}(a q)+4 \mathrm{H}^{+}(\mathrm{aq})+2 e^{-} \rightarrow \mathrm{Mn}^{2+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

\(\mathrm{E}_{\mathrm{MAO}_{2} \mathrm{Mh}^{2}}^{0}=1.23 \mathrm{~V}\)

\(\begin{aligned} \text { Cellpotential, } \mathrm{E}_{\text {cell }}^{\circ} &=\mathrm{E}_{\text {eathode }}^{\circ}-\mathrm{E}_{\text {anode }}^{\circ} \\ &=\mathrm{E}_{\mathrm{MrO}_{2} \mathrm{Ma}^{2+}}^{0}-\mathrm{E}_{2 n^{2}+\mathrm{Zn}}^{0} \\ &=1.23 \mathrm{~V}-(-0.76 \mathrm{~V}) \\ &=1.99 \mathrm{~V} \end{aligned}\)

D) since cell potential comes out to be positive, this reaction is a spontaneous one.

E) Overall cellreaction: \(\mathrm{Cl}_{2}(\mathrm{I})+2 \mathrm{~F}^{-}(\mathrm{aq}) \rightarrow \mathrm{F}_{2}(\mathrm{~g})+2 \mathrm{Cl}^{-}(\mathrm{aq})\)

Oxidation half reaction: \(2 \mathrm{~F}^{-}(\mathrm{aq}) \rightarrow \mathrm{F}_{2}(\mathrm{~g})+2 e^{-} ; \mathrm{E}_{\mathrm{R} / \mathrm{F}}^{0}=2.87 \mathrm{~V}\)

Reduction half reaction: \(\mathrm{Cl}_{2}(1)++2 e^{-} \rightarrow 2 \mathrm{Cl}^{-}(\mathrm{aq}) ; \mathrm{E}_{\alpha_{2} \mathrm{Cr}}^{0}=1.36 \mathrm{~V}\)

\(\begin{aligned} \text { Cell potential, } E_{\text {cell }}^{\circ} &=E_{\text {cathoods }}^{\circ}-E_{\text {anode }}^{0} \\ &=\mathrm{E}_{\alpha_{2} \mathrm{Cr}}^{0}-\mathrm{E}_{\mathrm{F}_{2} / \mathrm{F}}^{0} \\ &=1.36 \mathrm{~V}-2.87 \mathrm{~V} \\ &=-1.51 \mathrm{~V} \end{aligned}\)

F) since cell potential comes out to be negative, this reaction is a nonspontaneous one.