Electrochemistry Lesson 7 The Standard Hydrogen Cell

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Electrochemistry Lesson 7 The Standard Hydrogen Cell. The Standard Hydrogen Half Cell The zero point of the reduction chart is the hydrogen half-cell. E is the cell potential or voltage E 0 is the standard cell potential @ 25 o C solutions are 1.0 M gases are 101 KPa. - PowerPoint PPT Presentation

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ElectrochemistryLesson 7

The StandardHydrogen Cell

The Standard Hydrogen Half Cell 

The zero point of the reduction chart is the hydrogen half-cell.

E is the cell potential or voltage

E0 is the standard cell potential @ 25 oCsolutions are 1.0 Mgases are 101 KPa

 

 

   

Ag+ reacts with H2 spontaneously Eo = +0.80 v

Zn2+ is nonspontaneous with H2 Eo = -0.76 v

All half reactions are compared to H2

The voltage for any combination is the difference

Using the Reduction Potential Chart

Look on your reduction chart to find the voltage of each agent. Oxidizing agents are on the left and reducing agents are on the right. Formula Agent Type E0

 HClO4 oxidizing

 

Using the Reduction Potential Chart

Look on your reduction chart to find the voltage of each agent. Oxidizing agents are on the left and reducing agents are on the right. Formula Agent Type E0

 ClO4

- Acid oxidizing 1.39 v

HNO3 oxidizing

Take higher oneStronger Oxidizing Agent

HNO3 is both H+ and NO3

-

Using the Reduction Potential Chart

Look on your reduction chart to find the voltage of each agent. Oxidizing agents are on the left and reducing agents are on the right. Formula Agent Type E0

 ClO4

- Acid oxidizing 1.39 v

HNO3 oxidizing 0.96 v

HCl oxidizing

Using the Reduction Potential Chart

Look on your reduction chart to find the voltage of each agent. Oxidizing agents are on the left and reducing agents are on the right. Formula Agent Type E0

 ClO4

- Acid oxidizing 1.39 v

HNO3 oxidizing 0.96 v

HCl oxidizing 0.00 vSO4

2- reducing

The reaction is reversed so the E0 = -2.01 v

Oxidation

Using the Reduction Potential Chart

Look on your reduction chart to find the voltage of each agent. Oxidizing agents are on the left and reducing agents are on the right. Formula Agent Type E0

 ClO4

- Acid oxidizing 1.39 v

HNO3 oxidizing 0.96 v

HCl oxidizing 0.00 vSO4

2- reducing -2.01 v

MnO4- Acid oxidizing

Using the Reduction Potential Chart

Look on your reduction chart to find the voltage of each agent. Oxidizing agents are on the left and reducing agents are on the right.  Formula Agent Type E0

 ClO4

- Acid oxidizing 1.39 v

HNO3 oxidizing 0.96 v

HCl oxidizing 0.00 v SO4

2- reducing -2.01 v

MnO4- Acid oxidizing 1.51 v

MnO4- Alkaline oxidizing

Using the Reduction Potential Chart

Look on your reduction chart to find the voltage of each agent. Oxidizing agents are on the left and reducing agents are on the right. Formula Agent Type E0

 ClO4

- Acid oxidizing 1.39 v

HNO3 oxidizing 0.96 v

HCl oxidizing 0.00 vSO4

2- reducing -2.01 v

MnO4- Acid oxidizing 1.51 v

MnO4- Alkaline oxidizing 0.60 v

Formula Agent Type E0

H2SO4 oxidizing

H2SO4 → HSO4- + H+

Don’t have SO42-

Formula Agent Type E0

H2SO4 oxidizing 0.00 v

 H2O Neutral oxidizing

 

Formula Agent Type E0

H2SO4 oxidizing 0.00 v

 H2O Neutral oxidizing -0.41 v

 H2O Neutral reducing

Formula Agent Type E0

H2SO4 oxidizing 0.00 v

 H2O Neutral oxidizing -0.41 v

 H2O Neutral reducing -0.82 v

Fe2+ reducing

Formula Agent Type E0

H2SO4 oxidizing 0.00 v

 H2O Neutral oxidizing -0.41 v

 H2O Neutral reducing -0.82 v

Fe2+ reducing -0.77 v

Fe2+ oxidizing

Formula Agent Type E0

H2SO4 oxidizing 0.00 v

 H2O Neutral oxidizing -0.41 v

 H2O Neutral reducing -0.82 v

Fe2+ reducing -0.77 v

Fe2+ oxidizing -0.45 v 

Determine if the reaction below is spontaneous. Write a balanced equation for the reaction and calculate the E0.  1. MnO4

- & Mn2+ Acid

Reduction

Oxidation

The top reaction is written forward

The bottom reaction is reversed and the voltage is changed to negative

Determine if the reaction below is spontaneous. Write a balanced equation for the reaction and calculate the E0.  1. MnO4

- & Mn2+ Acid

2(MnO4- + 8H+ + 5e-→ Mn2+ + 4H2O) 1.51 v

5(Mn2+ + 2H2O → MnO2(s)+ 4H+ + 2e-) -1.22 v

2MnO4-+16H++5Mn2++10H2O → 2Mn2++8H2O+5MnO2(s)+20H+

2MnO4- + 3Mn2+ + 2H2O → 5MnO2(s) + 4H+ 0.29 v

add potentials

40.29 v

3 2

simplify

positive voltage- spontaneous

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