6.5 Redox Reactions | The Transition Metals | Chemistry 12

Predicting Feasibility Using $E^{\circ}$ Values

The feasibility (spontaneity) of a redox reaction can be predicted using standard electrode potentials ( $E^{\circ}$ ).

Standard Cell Potential

The standard cell potential is calculated as:

$E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ}$

where the cathode is the reduction half-reaction and the anode is the oxidation half-reaction.

Rule: A redox reaction is feasible (spontaneous) if $E_{ce l l}^{\circ} > 0$ .

This corresponds to a negative Gibbs free energy change: $Δ G = - n F E_{ce l l}^{\circ} < 0$ .

Example: Predicting Feasibility

Given:

$E^{\circ} (F e^{3 +} / F e^{2 +}) = + 0.77$ V
$E^{\circ} (I_{2} / I^{-}) = + 0.54$ V

Question: Is the reaction $2 F e^{3 +} + 2 I^{-} \to 2 F e^{2 +} + I_{2}$ feasible?

Solution:

$F e^{3 +}$ is reduced (cathode): $E^{\circ} = + 0.77$ V
$I^{-}$ is oxidized (anode): $E^{\circ} = + 0.54$ V

$E_{ce l l}^{\circ} = + 0.77 - 0.54 = + 0.23 V$

Since $E_{ce l l}^{\circ} = + 0.23$ V $> 0$ , the reaction is feasible.

Oxidation State Changes in Transition Metal Redox Reactions

Transition metals can exist in multiple oxidation states, making them versatile participants in redox reactions.

Example — Chromium oxidation:

When $C r^{3 +}$ is oxidized to $C r O_{4}^{2 -}$ in alkaline medium using $H_{2} O_{2}$ :

$2 C r^{3 +} + 3 H_{2} O_{2} + 10 O H^{-} \to 2 C r O_{4}^{2 -} + 8 H_{2} O$

Chromium's oxidation state increases from +3 to +6 (oxidation).

The $C u^{2 +} / I^{-}$ Redox System

This is an important redox system in analytical chemistry.

Reaction

When excess potassium iodide ( $K I$ ) is added to a copper(II) sulfate solution:

$2 C u_{(a q)}^{2 +} + 4 I_{(a q)}^{-} \to 2 C u I_{(s)} + I_{2 (a q)}$

Observations:

A white precipitate of copper(I) iodide ( $C u I$ ) forms
The solution turns brown due to the formation of iodine ( $I_{2}$ )

Roles:

$C u^{2 +}$ is the oxidizing agent — reduced from +2 to +1
$I^{-}$ is the reducing agent — oxidized from -1 to 0

Half-Equations

Reduction (cathode): $C u^{2 +} + e^{-} \to C u^{+}$

Oxidation (anode): $2 I^{-} \to I_{2} + 2 e^{-}$

Overall (multiply reduction by 2 and add): $2 C u^{2 +} + 4 I^{-} \to 2 C u I + I_{2}$

Feasibility Check

Using $E^{\circ}$ values:

$E^{\circ} (C u^{2 +} / C u^{+}) = + 0.15$ V
$E^{\circ} (I_{2} / I^{-}) = + 0.54$ V

$E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ} = + 0.15 - 0.54 = - 0.39 V$

Note: Although $E_{ce l l}^{\circ}$ appears negative using standard values, the reaction proceeds because the formation of the insoluble $C u I$ precipitate removes $C u^{+}$ from solution, shifting the equilibrium to the right (Le Chatelier's principle). The actual cell potential under these conditions becomes positive.

Iodometric Titration (Back-Titration)

The $I_{2}$ produced in the $C u^{2 +} / I^{-}$ reaction can be quantified by titration with standard sodium thiosulfate solution ( $N a_{2} S_{2} O_{3}$ ):

$I_{2} + 2 S_{2 O_{3 (a q)}}^{2 -} \to 2 I_{(a q)}^{-} + S_{4 O_{6 (a q)}}^{2 -}$

Indicator: Starch solution — forms a blue-black complex with $I_{2}$ . The endpoint is when the blue-black color disappears (all $I_{2}$ consumed).

Calculation Example

Given: 25.0 cm³ of $C u^{2 +}$ solution reacts with excess $K I$ . The $I_{2}$ produced requires 20.0 cm³ of 0.100 mol dm⁻³ $N a_{2} S_{2} O_{3}$ for complete reaction.

Find: Concentration of $C u^{2 +}$

Step 1: Moles of $N a_{2} S_{2} O_{3}$ : $n (N a_{2} S_{2} O_{3}) = 0.100 \times \frac{20.0}{1000} = 2.00 \times 1 0^{- 3} mol$

Step 2: Moles of $I_{2}$ (ratio 1:2 from equation): $n (I_{2}) = \frac{2.00 \times 1 0 ^{- 3}}{2} = 1.00 \times 1 0^{- 3} mol$

Step 3: Moles of $C u^{2 +}$ (ratio 2:1 from $2 C u^{2 +} \to I_{2}$ ): $n (C u^{2 +}) = 2 \times 1.00 \times 1 0^{- 3} = 2.00 \times 1 0^{- 3} mol$

Step 4: Concentration of $C u^{2 +}$ : $[C u^{2 +}] = \frac{2.00 \times 1 0 ^{- 3}}{0.0250} = 0.0800 mol dm^{- 3}$

$K M n O_{4}$ Reactions in Acidic Medium

Potassium manganate(VII) ( $K M n O_{4}$ ) is a powerful oxidizing agent in acidic solution.

Half-equation for reduction of $M n O_{4}^{-}$ in acid: $M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$

Manganese is reduced from +7 to +2.

Reaction with $F e^{2 +}$ (Mohr's Salt)

$M n O_{4}^{-} + 8 H^{+} + 5 F e^{2 +} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$

Balanced equation (multiply and combine half-equations): $M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O (\times 1)$ $F e^{2 +} \to F e^{3 +} + e^{-} (\times 5)$

$M n O_{4}^{-} + 8 H^{+} + 5 F e^{2 +} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$

Color change: Deep purple $M n O_{4}^{-}$ → colorless $M n^{2 +}$ . $K M n O_{4}$ is a self-indicator — endpoint is the first permanent pale pink color.

Reaction with $F e S O_{4}$ vs Mohr's Salt

$F e S O_{4}$ (iron(II) sulfate): Contains $F e^{2 +}$ ions — reacts identically to above
Mohr's Salt $(N H_{4})_{2} F e (S O_{4})_{2} \cdot 6 H_{2} O$ : Also contains $F e^{2 +}$ ions — same reaction. Mohr's salt is preferred in titrations as it is more stable (less prone to oxidation by air)

Predicting Feasibility Using $E^{\circ}$ Values

The feasibility (spontaneity) of a redox reaction can be predicted using standard electrode potentials ( $E^{\circ}$ ).

Standard Cell Potential

The standard cell potential is calculated as:

$E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ}$

where the cathode is the reduction half-reaction and the anode is the oxidation half-reaction.

Rule: A redox reaction is feasible (spontaneous) if $E_{ce l l}^{\circ} > 0$ .

This corresponds to a negative Gibbs free energy change: $Δ G = - n F E_{ce l l}^{\circ} < 0$ .

Example: Predicting Feasibility

Given:

$E^{\circ} (F e^{3 +} / F e^{2 +}) = + 0.77$ V
$E^{\circ} (I_{2} / I^{-}) = + 0.54$ V

Question: Is the reaction $2 F e^{3 +} + 2 I^{-} \to 2 F e^{2 +} + I_{2}$ feasible?

Solution:

$F e^{3 +}$ is reduced (cathode): $E^{\circ} = + 0.77$ V
$I^{-}$ is oxidized (anode): $E^{\circ} = + 0.54$ V

$E_{ce l l}^{\circ} = + 0.77 - 0.54 = + 0.23 V$

Since $E_{ce l l}^{\circ} = + 0.23$ V $> 0$ , the reaction is feasible.

Oxidation State Changes in Transition Metal Redox Reactions

Transition metals can exist in multiple oxidation states, making them versatile participants in redox reactions.

Example — Chromium oxidation:

When $C r^{3 +}$ is oxidized to $C r O_{4}^{2 -}$ in alkaline medium using $H_{2} O_{2}$ :

$2 C r^{3 +} + 3 H_{2} O_{2} + 10 O H^{-} \to 2 C r O_{4}^{2 -} + 8 H_{2} O$

Chromium's oxidation state increases from +3 to +6 (oxidation).

The $C u^{2 +} / I^{-}$ Redox System

This is an important redox system in analytical chemistry.

Reaction

When excess potassium iodide ( $K I$ ) is added to a copper(II) sulfate solution:

$2 C u_{(a q)}^{2 +} + 4 I_{(a q)}^{-} \to 2 C u I_{(s)} + I_{2 (a q)}$

Observations:

A white precipitate of copper(I) iodide ( $C u I$ ) forms
The solution turns brown due to the formation of iodine ( $I_{2}$ )

Roles:

$C u^{2 +}$ is the oxidizing agent — reduced from +2 to +1
$I^{-}$ is the reducing agent — oxidized from -1 to 0

Half-Equations

Reduction (cathode): $C u^{2 +} + e^{-} \to C u^{+}$

Oxidation (anode): $2 I^{-} \to I_{2} + 2 e^{-}$

Overall (multiply reduction by 2 and add): $2 C u^{2 +} + 4 I^{-} \to 2 C u I + I_{2}$

Feasibility Check

Using $E^{\circ}$ values:

$E^{\circ} (C u^{2 +} / C u^{+}) = + 0.15$ V
$E^{\circ} (I_{2} / I^{-}) = + 0.54$ V

$E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ} = + 0.15 - 0.54 = - 0.39 V$

Iodometric Titration (Back-Titration)

The $I_{2}$ produced in the $C u^{2 +} / I^{-}$ reaction can be quantified by titration with standard sodium thiosulfate solution ( $N a_{2} S_{2} O_{3}$ ):

$I_{2} + 2 S_{2 O_{3 (a q)}}^{2 -} \to 2 I_{(a q)}^{-} + S_{4 O_{6 (a q)}}^{2 -}$

Indicator: Starch solution — forms a blue-black complex with $I_{2}$ . The endpoint is when the blue-black color disappears (all $I_{2}$ consumed).

Calculation Example

Given: 25.0 cm³ of $C u^{2 +}$ solution reacts with excess $K I$ . The $I_{2}$ produced requires 20.0 cm³ of 0.100 mol dm⁻³ $N a_{2} S_{2} O_{3}$ for complete reaction.

Find: Concentration of $C u^{2 +}$

Step 1: Moles of $N a_{2} S_{2} O_{3}$ : $n (N a_{2} S_{2} O_{3}) = 0.100 \times \frac{20.0}{1000} = 2.00 \times 1 0^{- 3} mol$

Step 2: Moles of $I_{2}$ (ratio 1:2 from equation): $n (I_{2}) = \frac{2.00 \times 1 0 ^{- 3}}{2} = 1.00 \times 1 0^{- 3} mol$

Step 3: Moles of $C u^{2 +}$ (ratio 2:1 from $2 C u^{2 +} \to I_{2}$ ): $n (C u^{2 +}) = 2 \times 1.00 \times 1 0^{- 3} = 2.00 \times 1 0^{- 3} mol$

Step 4: Concentration of $C u^{2 +}$ : $[C u^{2 +}] = \frac{2.00 \times 1 0 ^{- 3}}{0.0250} = 0.0800 mol dm^{- 3}$

$K M n O_{4}$ Reactions in Acidic Medium

Potassium manganate(VII) ( $K M n O_{4}$ ) is a powerful oxidizing agent in acidic solution.

Half-equation for reduction of $M n O_{4}^{-}$ in acid: $M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$

Manganese is reduced from +7 to +2.

Reaction with $F e^{2 +}$ (Mohr's Salt)

$M n O_{4}^{-} + 8 H^{+} + 5 F e^{2 +} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$

Balanced equation (multiply and combine half-equations): $M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O (\times 1)$ $F e^{2 +} \to F e^{3 +} + e^{-} (\times 5)$

$M n O_{4}^{-} + 8 H^{+} + 5 F e^{2 +} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$

Color change: Deep purple $M n O_{4}^{-}$ → colorless $M n^{2 +}$ . $K M n O_{4}$ is a self-indicator — endpoint is the first permanent pale pink color.

Reaction with $F e S O_{4}$ vs Mohr's Salt

$F e S O_{4}$ (iron(II) sulfate): Contains $F e^{2 +}$ ions — reacts identically to above
Mohr's Salt $(N H_{4})_{2} F e (S O_{4})_{2} \cdot 6 H_{2} O$ : Also contains $F e^{2 +}$ ions — same reaction. Mohr's salt is preferred in titrations as it is more stable (less prone to oxidation by air)