2.7 Redox

2.7 Redox

For the assessment you need to:

•  Write both oxidation and reduction half equations and full equation

•  Identify the species that is reduced and the species that is oxidized

  
  

•  Identify the colour changes

  
  

• Explain reduction and oxidation in terms of the TRANSFER OF ELECTRONS or the CHANGE IN OXIDATION STATE

A piece of zinc placed in a solution of CuSO_4.

Cu²⁺ _(aq)    +      2e^-        --> Cu_(s)

 

Zn_(s)         --> Zn²⁺ _(aq)      +    2 e^-

 

Cu²⁺_(aq)      + Zn _(s)      --> Cu_(s)    +  Zn²⁺_(aq)    

 

The blue CuSO₄ reduced to pink Cu.  The grey Zn is oxidized to colourless Zn²⁺. 

The Copper in CuSO₄  has an oxidation number of +2, which is reduced to an oxidation number of 0 in Cu. This is a decrease in oxidation number as electrons are gained, so this is reduction. 

The zinc Zn has an oxidation number of 0, which is increased to an oxidation number of +2 in  Zn²⁺. This is an increase in oxidation number, as electrons are lost, so this is oxidation

A piece of magnesium placed in a solution of HCl_.

•  Write both oxidation and reduction half equations and full equation

•  Identify the species that is reduced and the species that is oxidized

•  Identify the colour changes

• Explain reduction and oxidation in terms of the TRANSFER OF ELECTRONS or the CHANGE IN OXIDATION STATE

A piece of magnesium placed in a solution of HCl_.

2H⁺_(aq)           +      2e^{-          -->} H_2(g)

 

Mg _(s)      --> Mg²⁺ _(aq)          +     2 e^-

Mg _(s)   +      2H⁺ _(aq)           -->   H_2(g)   +  Mg²⁺_(aq)    

 The colourless H⁺ is reduced to colourless H_2­ gas, which would be seen as bubbles. The grey Mg is oxidized to colourless Mg²⁺. 

The hydrogen in HCl  has an oxidation number of + 1 which is reduced to an oxidation number of 0 in H₂. This is a decrease in oxidation number as electrons are gained, so reduction.  

 

The magnesium in grey Mg has an oxidation number of 0, which is increased to an oxidation number of +2 in colourless Mg²⁺. This is an increase in oxidation number, as electrons are lost, so oxidation. 

Potassium iodide reacting with Cl₂ gas.

•  Write both oxidation and reduction half equations and full equation

•  Identify the species that is reduced and the species that is oxidized

•  Identify the colour changes

• Explain reduction and oxidation in terms of the TRANSFER OF ELECTRONS or the CHANGE IN OXIDATION STATE

Potassium iodide reacting with Cl₂ gas.

2I^-_(aq)        -->      l_{2 (l)}    +    2e^- (oxidation)

 

Cl_{2 (g)}       + 2e^- -->      2Cl^-_(aq)    (Reducation)            

 

2l^-_(aq)       +      Cl_{2 (g)}  -->  2Cl^-_(aq) +   I_2(l)

The pale green Cl₂ is being reduced into colourless Cl^-_. This is reduction as the oxidation state of the Cl in Cl_2­ is 0 and it decreases to -1 in Cl^-. For the oxidation state to decrease the Cl has gained one electron.  

The colourless I^- is being oxidized into yellow I₂. This is oxidation as the oxidation state of the I in I^- is -1 and it increases to 0 in I₂. For the oxidation state to increase the I^- has lost one electron. 

Sulfur Dioxide (SO₂) is added to a solution of potassium permanganate (KMnO₄).

•  Write both oxidation and reduction half equations and full equation

•  Identify the species that is reduced and the species that is oxidized

•  Identify the colour changes

• Explain reduction and oxidation in terms of the TRANSFER OF ELECTRONS or the CHANGE IN OXIDATION STATE

Sulfur Dioxide (SO₂) is added to a solution of potassium permanganate (KMnO₄).

MnO₄^-    + 8H⁺    +   5e^-     --> Mn²⁺    +  4H₂O          (reduction)  

SO₂  +  2H₂O   --> SO₄^2-   +  4H⁺       +  2e^-         (oxidation)

2MnO₄^-    + 16H⁺    +   10e^-    --> 2Mn²⁺    +  8H₂O     (reduction)  

5SO₂  +  10H₂O        -->   5SO₄^2-   +  20H⁺   +  10e^-   (oxidation)

2MnO₄^- + 5SO₂      +  2H₂O    --> 2Mn²⁺     + 5SO₄^2-        + 4H⁺           

The purple MnO₄^- is being reduced into colourless Mn²⁺_.  The colourless SO₂ is being oxidized into colourless SO₄^2-.

The MnO₄^- is reduced as the oxidation state of the Mn in MnO₄^- _­ is +7 and it decreases to +2 in Mn ²⁺. For the oxidation state to decrease the Mn has gained 5 electrons.

The SO₂ is oxidation as the oxidation state of the S in SO₂ is +4 and it increases to +6 in SO₄^2-. For the oxidation state to increase the SO₂ has lost 2 electrons.

Acidified potassium permanganate solution is added to some hydrogen peroxide.

•  Write both oxidation and reduction half equations and full equation

•  Identify the species that is reduced and the species that is oxidized

•  Identify the colour changes

• Explain reduction and oxidation in terms of the TRANSFER OF ELECTRONS or the CHANGE IN OXIDATION STATE

Acidified potassium permanganate solution and hydrogen peroxide.

MnO₄^-_(aq)   + 8H⁺_(aq) + 5e^{-     -->}  Mn²⁺_(aq) + 4H₂O_(l)

 

H₂O_2(aq)    ^--> O_2(g)    + 2H⁺_(aq)    -     2 e^-         

2MnO₄^- _(aq)   + 16H+_(aq) + 10e-     -->  2Mn²⁺_(aq) + 8H₂O_(l)

 

5H₂O₂(aq)    --> 5O_2(g)    + 10H+_(aq)    -     10 e-     

 

2MnO₄^-_(aq) + 6H⁺_(aq) ⁺ 5H₂O_2(aq)  ^-->  2Mn²⁺_(aq) + 8H₂O_(l)  +  5 O_2 (g)  

 

The purple MnO₄^- is reduced to colourless Mn²⁺. The colourless H₂O₂ is oxidized to colourless O₂ gas_, which can be seen as bubbles. 

The Mn MnO₄^- has an oxidation number of +7, which is reduced to an oxidation state of +2 in Mn²⁺. This is a decrease in oxidation number as electrons are gained, so reduction. 

 

The oxygen in H₂O₂ has an oxidation number of -1, which is increased to an oxidation state of 0 in O₂. This is an increase in oxidation number, as electrons are lost, so oxidation. 

An iron nail is placed in a solution of hydrogen peroxide.

•  Write both oxidation and reduction half equations and full equation

•  Identify the species that is reduced and the species that is oxidized

•  Identify the colour changes

• Explain reduction and oxidation in terms of the TRANSFER OF ELECTRONS or the CHANGE IN OXIDATION STATE

An iron nail is placed in a solution of hydrogen peroxide.

Fe_{(s)                 -->}        Fe²⁺_{(aq)    +}   2e^-

 

H₂O_{2 (aq)}    +     2H⁺_(aq)      + 2e^-   -->     2H₂O_(l)  

 

H₂O_{2 (aq)}    +     2H⁺_(aq)     + Fe_(s)   ^-->     2H₂O_(l)    + Fe²⁺_(aq

The grey Fe is oxidised to green Fe²⁺.  The colourless H₂O₂ is reduced to colourless H₂O. 

 

The iron in Fe has an oxidation number of 0, which is oxidies to an oxidation state of +2 in Fe²⁺. This is a increase in oxidation number as electrons are lost, so oxidation. 

 

he O in colourless H₂O₂ has an oxidation number of -1, which is decreased to an oxidation state of -2 in colourless H₂O. This is a decrease in oxidation number, as electrons are gained, so reduction. 

Hydrogen peroxide (H₂O₂) is added a solution of potassium iodide (KI). 

•  Write both oxidation and reduction half equations and full equation

•  Identify the species that is reduced and the species that is oxidized

•  Identify the colour changes

• Explain reduction and oxidation in terms of the TRANSFER OF ELECTRONS or the CHANGE IN OXIDATION STATE

Hydrogen peroxide (H₂O₂) is added a solution of potassium iodide (KI). 

2I^-_{(aq)            -->}        l_{2 (g)    +}    2e^-

 

H₂O_{2 (aq)}      +     2H⁺_(aq)      + 2e^-   -->     2H₂O_(l)              

 

2l^-_(aq)       +      H₂O_2(aq) +        2H⁺_(aq)        -->  2H₂O_(l)     +   I_2(g)

The colourless l^- is oxidized to brown l₂ .  The colourless H₂O₂ is reduced to colourless H₂O. 

 

The iodine in colourless l^- has an oxidation number of -1, which is increased to an oxidation state of 0 in brown l₂ gas. This is an increase in oxidation makes the Cl^- the reductant.

 

he O in colourless H₂O₂ has an oxidation number of -1, which is decreased to an oxidation state of -2 in colourless H₂O. This is a decrease in oxidation number, as electrons are gained, so reduction. 

Potassium dichromate (K₂Cr₂O₇) is added to a solution of hydrogen peroxide. 

•  Write both oxidation and reduction half equations and full equation

•  Identify the species that is reduced and the species that is oxidized

•  Identify the colour changes

• Explain reduction and oxidation in terms of the TRANSFER OF ELECTRONS or the CHANGE IN OXIDATION STATE

Potassium dichromate (K₂Cr₂O₇) is added to a solution of hydrogen peroxide. 

Cr₂O₇^2-_(aq)+  14H⁺_(aq)+ 6e^-  -->  2Cr³⁺_(aq) + 7H₂O_(l)   (Reduction)            

H₂O_2(aq)        --> O_2(g)  + 2H⁺_(aq) + 2e                              (Oxidation)

Cr₂O₇^2-_(aq) +14H⁺_(aq) + 6e^-  --> 2Cr³⁺_(aq) + 7H₂O_(l)                                    

3H₂O_2(aq)  -->  3O_2(g)  + 6H⁺_(aq) + 6e                                 

Cr₂O₇^2-_(aq)  +   3 H₂O_2(aq) + 8H⁺ _(aq) --> 2Cr³⁺ _(aq) +   3 O_2(g)   + 7H₂O _(l)

The orange Cr₂O₇^2- is being reduced into green Cr³⁺_. This is reduction as the oxidation state of the Cr in Cr₂O₇^2-_­ is +6 and it decreases to +3 in Cr³⁺. For the oxidation state to decrease the Cr in Cr₂O₇^2-/dichromate has gained 3 electrons each (6 in total).

 

The colourless H₂O₂ is being oxidized into colourless O₂ (bubbles seen). This is oxidation as the oxidation state of the O in H₂O₂ is -1 and it increases to zero in O₂. For the oxidation state to increase each oxygen atom in H₂O₂ has lost 1 electron (2 in total).