Question

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.

Answer 1

Part:A

The
`E_(cell)` of the reaction is +0.50V

Part-B:

a)
`E_(cell)^(o)` of the reaction is -1.61 V.

b)
`E_(cell)^(o)` of the reaction is +0.27 V.

Part-C:

"K" of the oxidation reaction is
`5.43*10^(4)`.

Step-by-step explanation:

Part:A

The given chemical reaction is as follows.

`3Ni^(2+)+2Cr\rightarrow 3Ni+2Cr^(3+)`

Reduction half reaction:

At cathode:

`Ni^(2+)+2e^(-)\rightarrow Ni`

Oxidation half reaction;

At anode:

`Cr \rightarrow Cr^(3+)+3e^(-)`

`E_(cell)=E_(cathode)-E_(anode)`

`=-0.23-(-0.73)=+0.50V`

Part-B:

a)

The given chemical reaction is as follows.

`Zn+Mg^(2+) \rightarrow Zn^(2+)+Mg`

`E_(cell)=E_(cathode)-E_(anode)`

`=-2.37-(-0.76)=-1.61V(Non\,spontaneous)`

b)

The given chemical reaction is as follows.

`Cd+Pb^(2+) \rightarrow Cd^(2+)+Pb`

`E_(cell)=E_(cathode)-E_(anode)`

`=-0.13-(-0.40)=+0.27V(spontaneous)`

Part-C;

The given chemical reaction is as follows.

`Sn+H^(+) \rightarrow Sn^(2+)+H_(2)`

Oxidation half reaction;

At anode:

`Sn \rightarrow Sn^(2+)+2e^(-)`

Reduction half reaction;

At cathode:

`2H^(+)+2e^(-)\rightarrow H_(2)`

`E_(cell)=E_(cathode)-E_(anode)`

`=0-(-0.14)=+0.14V`

`-RTlnK= -nFE_(cell)^(o)`

`-2.303* 8.314*298\,logK= -2* 96487* 0.14`

logK = 4.737

`K=5.43* 10^(4)`