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Judgment of Redox Reactions

CSCA Judgment of Redox Reactions study guide organized around the publicly available CSCA syllabus. Practice Chemistry questions on aicsca.com.

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Judgment of Redox Reactions

1. Core Logic & Definitions

Redox reactions are high-frequency topics in the CSCA exam. We need to identify reaction types and roles quickly by tracking changes in **Oxidation State**.

* **Essence**: Electron Transfer.

* **Feature**: Change in Oxidation State.

* **Mnemonic**:

* **LEO says GER** (Loss of Electrons is Oxidation; Gain of Electrons is Reduction).

* **Rise-Lose-Oxidation, Reducing Agent**.

* **Drop-Gain-Reduction, Oxidizing Agent**.

*Explanation*:

* **Rise**: Oxidation state increases $\rightarrow$ **Lose**: Loses electrons $\rightarrow$ **Oxidation** $\rightarrow$ Acts as **Reducing Agent**.

* **Drop**: Oxidation state decreases $\rightarrow$ **Gain**: Gains electrons $\rightarrow$ **Reduction** $\rightarrow$ Acts as **Oxidizing Agent**.

2. Double-Line Bridge Method

This is the most intuitive method to visualize electron transfer direction and quantity.

**Steps**:

1. **Label**: Mark oxidation states of elements that change.

2. **Connect**: Draw lines from reactant to product for the same element.

3. **Annotate**: Mark "Lose/Gain" electron count (Format: $a \times b\ e^-$).

**Typical Example**: Copper + Dilute Nitric Acid

$$3Cu + 8HNO_3(dilute) \rightarrow 3Cu(NO_3)_2 + 2NO\uparrow + 4H_2O$$

* **Analysis**:

* $Cu$: $0 \rightarrow +2$ (Rise 2, Lose $2e^-$).

* $N$ (in $HNO_3$): $+5 \rightarrow +2$ (in $NO$, Drop 3, Gain $3e^-$).

* *Note*: Some $N$ remains $+5$ in $Cu(NO_3)_2$, acting as acid, not participating in redox.

3. Common Oxidizing & Reducing Agents

| Role | Typical Substance | Product |

| :--- | :--- | :--- |

| **Strong Oxidizing Agents** | $KMnO_4$ ($H^+$) | $Mn^{2+}$ (Colorless) |

| | $HNO_3$ (Conc/Dil) | $NO_2$ (Red-brown gas) / $NO$ (Colorless) |

| | $Cl_2, Br_2$ | $Cl^-, Br^-$ |

| | $Fe^{3+}$ | $Fe^{2+}$ (Pale green) |

| **Strong Reducing Agents** | Active Metals ($Na, Mg, Al$) | Metal Cations ($Na^+, Mg^{2+}...$) |

| | $C, CO, H_2$ | $CO_2, H_2O$ (High temp reduction) |

| | $I^-, S^{2-}$ | $I_2, S$ (Precipitate) |

4. Typical Example Analysis

**Example 1: Identification**

Which of the following is a Redox reaction?

* A. $CaCO_3 \xrightarrow{\Delta} CaO + CO_2\uparrow$

* B. $2NaOH + Cl_2 \rightarrow NaCl + NaClO + H_2O$

* C. $AgNO_3 + NaCl \rightarrow AgCl\downarrow + NaNO_3$

**Analysis**:

* A is Decomposition, valences of Ca(+2), C(+4), O(-2) unchanged $\rightarrow$ Non-Redox.

* B involves Cl changing from 0 to -1 ($NaCl$) and +1 ($NaClO$) $\rightarrow$ **Redox** (Disproportionation).

* C is Double Displacement, no valence change $\rightarrow$ Non-Redox.

**Answer: B**

**Example 2: Electron Transfer Calculation**

In reaction $MnO_2 + 4HCl(conc.) \xrightarrow{\Delta} MnCl_2 + Cl_2\uparrow + 2H_2O$, if 1 mol $Cl_2$ is produced, how many electrons are transferred?

**Analysis**:

1. Check change: $Cl$ goes from -1 $\rightarrow$ 0 (in $Cl_2$).

2. Calculate: To form 1 molecule of $Cl_2$, 2 $Cl^-$ ions each lose 1 electron.

3. Total: $1 \text{ mol} \times 2 = 2 \text{ mol } e^-$ transferred.

* *Trap*: 4 mol HCl are consumed, but only 2 mol are oxidized (to $Cl_2$). The other 2 mol provide acidity (for $MnCl_2$).