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You may want to reference (Pages 906 - 909) Section 19.5 while completing this problem. Standard reduction half-cell potentials at 25∘C Half-reaction E∘ (V) Half-reaction E∘ (OR) Au3+(aq)+3e−→Au(s) 1.50 Fe2+(aq)+2e−→Fe(s) −0.45 Ag+(aq)+e−→Ag(s) 0.80 Cr3+(aq)+e−→Cr2+(aq) −0.50 Fe3+(aq)+3e−→Fe2+(aq) 0.77 Cr3+(aq)+3e−→Cr(s) −0.73 Cu+(aq)+e−→Cu(s) 0.52 Zn2+(aq)+2e−→Zn(s) −0.76 Cu2+(aq)+2e−→Cu(s) 0.34 Mn2+(aq)+2e−→Mn(s) −1.18 2H+(aq)+2e−→H2(g) 0.00 Al3+(aq)+3e−→Al(s) −1.66 Fe3+(aq)+3e−→Fe(s) −0.036 Mg2+(aq)+2e−→Mg(s) −2.37 Pb2+(aq)+2e−→Pb(s) −0.13 Na+(aq)+e−→Na(s) −2.71 Sn2+(aq)+2e−→Sn(s) −0.14 Ca2+(aq)+2e−→Ca(s) −2.76 Ni2+(aq)+2e−→Ni(s) −0.23 Ba2+(aq)+2e−→Ba(s) −2.90 Co2+(aq)+2e−→Co(s) −0.28 K+(aq)+e−→K(s) −2.92 Cd2+(aq)+2e−→Cd(s) −0.40 Li+(aq)+e−→Li(s) −3.04 Part A Use the tabulated electrode potentials to calculate K for the oxidation of tin by H+ (at 25 ∘C): Sn(s)+2H+(aq)→Sn2+(aq)+H2(g) Express your answer using two significant figures. KK = Previous Answer Request Answer Incorrect; Try Again; 5 attempts remaining Provide Feedback Incorrect. Incorrect; Try Again; 5 attempts remaining. No additional feedback.

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Answer:

Part:A

The [tex]E_{cell}[/tex] of the reaction is +0.50V

Part-B:

a) [tex]E_{cell}^{o}[/tex] of the reaction is -1.61 V.

b) [tex]E_{cell}^{o}[/tex] of the reaction is +0.27 V.

Part-C:

"K" of the oxidation reaction is [tex]5.43\times10^{4}[/tex].

Explanation:

Part:A

The given chemical reaction is as follows.

[tex]3Ni^{2+}+2Cr\rightarrow 3Ni+2Cr^{3+}[/tex]

Reduction half reaction:

At cathode:

[tex]Ni^{2+}+2e^{-}\rightarrow Ni[/tex]

Oxidation half reaction;

At anode:

[tex]Cr \rightarrow Cr^{3+}+3e^{-}[/tex]

[tex]E_{cell}=E_{cathode}-E_{anode}[/tex]

[tex]=-0.23-(-0.73)=+0.50V[/tex]

Part-B:

a)

The given chemical reaction is as follows.

[tex]Zn+Mg^{2+} \rightarrow Zn^{2+}+Mg[/tex]

[tex]E_{cell}=E_{cathode}-E_{anode}[/tex]

[tex]=-2.37-(-0.76)=-1.61V(Non\,spontaneous)[/tex]

b)

The given chemical reaction is as follows.

[tex]Cd+Pb^{2+} \rightarrow Cd^{2+}+Pb[/tex]

[tex]E_{cell}=E_{cathode}-E_{anode}[/tex]

[tex]=-0.13-(-0.40)=+0.27V(spontaneous)[/tex]

Part-C;

The given chemical reaction is as follows.

[tex]Sn+H^{+} \rightarrow Sn^{2+}+H_{2}[/tex]

Oxidation half reaction;

At anode:

[tex]Sn \rightarrow Sn^{2+}+2e^{-}[/tex]

Reduction half reaction;

At cathode:

[tex]2H^{+}+2e^{-}\rightarrow H_{2}[/tex]

[tex]E_{cell}=E_{cathode}-E_{anode}[/tex]

[tex]=0-(-0.14)=+0.14V[/tex]

[tex]-RTlnK= -nFE_{cell}^{o}[/tex]

[tex]-2.303\times 8.314\times298\,logK= -2\times 96487\times 0.14[/tex]

logK = 4.737

[tex]K=5.43\times 10^{4}[/tex]

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