Predict which of the following reactions would proceed spontaneously at `298K?`

Options

A. `Co(s)+Fe^(2+)(aq)rarrCo^(2+)(aq)+Fe(s) `
`E^(c-)._(Co^(2+)//Co)=-0.28V`
B. `Cd^(2+)(aq)+Fe(s)rarrCd(s)+Fe^(2+)(aq) `
`E^(c-)._(Cd^(2+)//Cd)=-0.4V`
C. `Cd(s)+Co^(2+)(aq)rarr Cd^(2+)(aq)+Co(s)`
`E^(c-)._(Fe^(2+)//Fe)=-0.44V`
D. `Zn^(2+)(aq)+H_(2)rarr Zn(s)+2H^(o+)`
`E^(c-)._(Zn^(2+)//Zn)=-0.76V`

Explanation

To determine which reactions proceed spontaneously at 298 K, we need to analyze the standard cell potentials (E°) for each reaction. A reaction is spontaneous if the cell potential (E°) is positive, which corresponds to a negative Gibbs free energy change (ΔG).

### Step-by-Step Solution:

1. **Identify the Reactions**:
We have several reactions to analyze. For each reaction, we need to identify the oxidation and reduction half-reactions and their respective standard reduction potentials (E°).

2. **Calculate E° for Each Reaction**:
- **Reaction A**: Cobalt and Iron
- Cobalt is oxidized (E° = -0.28 V).
- Iron is reduced (E° = -0.44 V).
- E°cell = E°cathode - E°anode = (-0.44) - (-0.28) = -0.16 V (not spontaneous).

- **Reaction B**: Cadmium and Iron
- Cadmium is oxidized (E° = -0.44 V).
- Iron is reduced (E° = -0.40 V).
- E°cell = E°cathode - E°anode = (-0.40) - (-0.44) = 0.04 V (spontaneous).

- **Reaction C**: Cadmium and Cobalt
- Cadmium is oxidized (E° = -0.40 V).
- Cobalt is reduced (E° = -0.28 V).
- E°cell = E°cathode - E°anode = (-0.28) - (-0.40) = 0.12 V (spontaneous).

- **Reaction D**: Zinc and Hydrogen
- Zinc is reduced (E° = -0.76 V).
- Hydrogen is oxidized (E° = 0.00 V).
- E°cell = E°cathode - E°anode = (0.00) - (-0.76) = -0.76 V (not spontaneous).

3. **Determine Spontaneity**:
- From the calculations:
- Reaction A: E°cell = -0.16 V (not spontaneous)
- Reaction B: E°cell = 0.04 V (spontaneous)
- Reaction C: E°cell = 0.12 V (spontaneous)
- Reaction D: E°cell = -0.76 V (not spontaneous)

### Conclusion:
The reactions that proceed spontaneously at 298 K are **Reaction B and Reaction C**.