Predict whether a spontaneous reaction will occur:

Options

A. When a piece of silver wire is dipped in a `ZnSO_4` solution
B. When lodine is added to a NaBr solution
C. When a piece of Zn metal is dipped in a `NiSO_4` solution
D. When a chlorine gas is bubbled through a KI solutiön

Answer

To determine whether a spontaneous reaction will occur, we need to calculate the cell potential (E_cell) for each reaction. A spontaneous reaction is indicated by a positive E_cell value. Here’s a step-by-step solution for the given question:

### Step 1: Understand the Concept of Spontaneity
A spontaneous reaction occurs naturally without external intervention. In electrochemistry, a reaction is spontaneous if the cell potential (E_cell) is positive.

### Step 2: Formula for E_cell
The formula to calculate E_cell is:
\[ E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} \]
Where:
- E_cathode is the reduction potential of the cathode (where reduction occurs).
- E_anode is the reduction potential of the anode (where oxidation occurs).

### Step 3: Gather Reduction Potentials
We need to gather the standard reduction potentials for the relevant half-reactions. The reduction potentials for the elements involved in the reactions are as follows:
- Ag⁺/Ag: +0.80 V
- Zn²⁺/Zn: -0.76 V
- I₂/I⁻: +0.54 V
- Br₂/Br⁻: +1.09 V
- Ni²⁺/Ni: -0.26 V
- Cl₂/Cl⁻: +1.36 V

### Step 4: Analyze Each Reaction

**Reaction 1: Silver wire in Zinc sulfate solution**
- Anode: Ag (oxidized)
- Cathode: Zn²⁺ (reduced)
- E_cell = E_cathode - E_anode = (-0.76 V) - (+0.80 V) = -1.56 V (not spontaneous)

**Reaction 2: Iodine in NaBr solution**
- Anode: Br⁻ (oxidized)
- Cathode: I₂ (reduced)
- E_cell = E_cathode - E_anode = (+0.54 V) - (+1.09 V) = -0.55 V (not spontaneous)

**Reaction 3: Zinc in Nickel sulfate solution**
- Anode: Zn (oxidized)
- Cathode: Ni²⁺ (reduced)
- E_cell = E_cathode - E_anode = (-0.26 V) - (-0.76 V) = +0.50 V (spontaneous)

**Reaction 4: Chlorine gas in KI solution**
- Anode: I⁻ (oxidized)
- Cathode: Cl₂ (reduced)
- E_cell = E_cathode - E_anode = (+1.36 V) - (+0.54 V) = +0.82 V (spontaneous)

### Step 5: Conclusion
From the calculations:
- Reaction 1: Not spontaneous (E_cell < 0)
- Reaction 2: Not spontaneous (E_cell < 0)
- Reaction 3: Spontaneous (E_cell > 0)
- Reaction 4: Spontaneous (E_cell > 0)

Thus, the spontaneous reactions are in options C (Zinc in Nickel sulfate) and D (Chlorine gas in KI solution).

### Final Answer
The spontaneous reactions are in options C and D.

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