Calculate the cell potential, E, for the following reactions at 26.29 °C using t

Question

Calculate the cell potential, E, for the following reactions at 26.29 °C using the ion concentrations provided. Then, determine if the cells are spontaneous or nonspontaneous as written. Standard reduction potentials (E°red) may be found here.'

[Fe2+]=0.0066M [Pt2+]=0.057M

E= ? V

[Cu2+]=0.013M [Ag+]=0.013M

E= ?V

[Co2+]=0.050M [Co3+]=0.030M

[Ti3+]=0.0055M [Ti2+]=0.0110M

E=?V

Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Mg2 ] = 0.796 M and [Sn2 ] = 0.0170 M. Standard reduction potentials can be found here.

E=?V

Explanation / Answer

Answer – We are given redox reacts and the concentration of ions. We need to calculate the cell potential of the reaction.

1) Pt(s) + Fe2+(aq) <----> Pt2+(aq) + Fe(s)

[Fe2+] = 0.0066M [Pt2+] = 0.057M

We need to calculate the standard cell potential for this reaction. In this one Pt gets oxidized and Fe2+ is gets reduced as follow –

Pt(s) ------> Pt2+(aq) + 2e- Eo = -1.2 V

Fe2+(aq) + 2e- ----> Fe(s) Eo = -0.44 V

Pt(s) + Fe2+(aq) <----> Pt2+(aq) + Fe(s) , Eo = -1.64 V

We know the Nernst equation –

Ecell = Eocell -0.0592/n * ln K

          = Eocell – 0.0592 /2 * ln [Pt2+(aq)] / [Fe2+(aq)]

Ecell = -1.64 – 0.0592/2 *ln (0.057) / (0.0066)

          = -1.70 V

2) Cu(s) + 2Ag+(aq) <----> Cu2+(aq) +2Ag(s)

[Cu2+] = 0.013M , [Ag+] = 0.013M

We need to calculate the standard cell potential for this reaction. In this one Cu gets oxidized and Ag+ is gets reduced as follow –

Cu(s) ------> Cu2+(aq) + 2e- Eo = - 0.337 V

2Ag+(aq) + 2e- ----> 2Ag(s) Eo = 0.80 V

Cu(s) + 2Ag+(aq) <----> Cu2+(aq) +2Ag(s) , Eo = 0.463 V

We know the Nernst equation –

Ecell = Eocell -0.0592/n * ln K

          = Eocell – 0.0592 /2 * ln [Cu2+(aq)] / [Ag+(aq)]2

Ecell = 0.463 V – 0.0592/2 *ln (0.013) / (0.013)2

          = 0.334 V

3) Co2+(aq) + Ti3+(aq) <----> Co3+(aq) + Ti2+(aq)

[Co2+] = 0.050M, [Co3+] = 0.030M , [Ti3+] = 0.0055M ,[Ti2+] = 0.0110M

We need to calculate the standard cell potential for this reaction. In this one Co2+(aq) gets oxidized and Ti2+(aq) is gets reduced as follow –

Co2+(aq) ------> Co3+(aq) + e- Eo = - 1.82 V

Ti3+(aq) + e- ----> Ti2+(aq) Eo = - 0.37 V

Co2+(aq) + Ti3+(aq) <----> Co3+(aq) + Ti2+(aq) , Eo = -2.19 V

We know the Nernst equation –

Ecell = Eocell -0.0592/n * ln K

          = Eocell – 0.0592 /1 * ln [Co3+] [Ti2+] / [Co2+] [Ti3+]

Ecell = -2.19 V – 0.0592/1 *ln (0.030)(0.0110) / (0.050)(0.0055)

          = -2.20 V

4) Mg(s) + Sn2+(aq) <----> Mg2+(aq) + Sn(s)

[Mg2+] = 0.796M [Sn2+] = 0.0170M

We need to calculate the standard cell potential for this reaction. In this one Mg gets oxidized and Sn2+ is gets reduced as follow –

Mg(s) ------> Mg2+(aq) + 2e- Eo = +2.37 V

Sn2+(aq) + 2e- ----> Sn(s) Eo = -0.14 V

Mg(s) + Sn2+(aq) <----> Mg2+(aq) + Sn(s), Eo = 2.23 V

We know the Nernst equation –

Ecell = Eocell -0.0592/n * ln K

          = Eocell – 0.0592 /2 * ln [Mg2+] / [Sn2+]

Ecell = 2.23 V – 0.0592/2 *ln (0.796) / (0.0170)

          = 2.12 V

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