Predict which of the following reactions would proceed spontaneously at 298 K? [Hint : Refer to Electrochemical series]

Options

A. `Co(s)+ Fe^(2+)(aq) rightarrow Co^(2+) (aq) + Fe(s)`
B. `Cd^(2+) (aq) + Fe(s)rightarrow Cd(s)+ Fe^(2+) (aq)`
C. `Cd(s)+Co^(2+)(aq) rightarrow Cd^(2+) (aq) + Co(s))`
D. `Zn^(2+) ((aq))+^H_2rightarrow Zn((s)) + 2H^(+)`

Explanation

To determine which of the given reactions would proceed spontaneously at 298 K, we need to calculate the standard cell potential (E°) for each reaction. A reaction is considered spontaneous if the E° value is positive.

### Step-by-Step Solution:

1. **Identify the Reactions**: We have four reactions to analyze. For each reaction, we need to identify the half-reactions involved.

2. **Gather Standard Reduction Potentials**: We will use the standard reduction potentials (E°) for the relevant half-reactions:
- Co²⁺ + 2e⁻ → Co; E° = -0.28 V
- Cd²⁺ + 2e⁻ → Cd; E° = -0.40 V
- Fe²⁺ + 2e⁻ → Fe; E° = -0.44 V
- Zn²⁺ + 2e⁻ → Zn; E° = -0.76 V
- H⁺ + 2e⁻ → H₂; E° = 0.00 V (standard hydrogen electrode)

3. **Calculate E° for Each Reaction**:
- **Reaction 1**: Co + Fe²⁺ → Co²⁺ + Fe
- E°cell = E°(cathode) - E°(anode)
- E°cell = (-0.44 V) - (-0.28 V) = -0.44 + 0.28 = -0.16 V (not spontaneous)

- **Reaction 2**: Cd²⁺ + Fe → Cd + Fe²⁺
- E°cell = E°(cathode) - E°(anode)
- E°cell = (-0.40 V) - (-0.44 V) = -0.40 + 0.44 = +0.04 V (spontaneous)

- **Reaction 3**: Cd + Co²⁺ → Cd²⁺ + Co
- E°cell = E°(cathode) - E°(anode)
- E°cell = (-0.28 V) - (-0.40 V) = -0.28 + 0.40 = +0.12 V (spontaneous)

- **Reaction 4**: Zn + 2H⁺ → Zn²⁺ + H₂
- E°cell = E°(cathode) - E°(anode)
- E°cell = (0.00 V) - (-0.76 V) = 0.00 + 0.76 = +0.76 V (spontaneous)

4. **Conclusion**: The reactions that proceed spontaneously at 298 K are:
- Reaction 2: Cd²⁺ + Fe → Cd + Fe²⁺
- Reaction 3: Cd + Co²⁺ → Cd²⁺ + Co
- Reaction 4: Zn + 2H⁺ → Zn²⁺ + H₂

### Final Answer:
The spontaneous reactions are Reaction 2, Reaction 3, and Reaction 4.

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