A galvanic cell is set up from a zinc bar weighing `100g` and `1.0L` of `1.0M CuSO_(4)` solution. How long would the cell run if it is assumed to deliver a steady current of `1.0A. (` Atomic mass of `Zn=65)`.

Options

A. `1.1` hours
B. 46 hours
C. `53.6` hourse
D. 24 hours.

Explanation

To solve the problem step-by-step, we will follow these calculations:

### Step 1: Determine the number of moles of Zinc
Given:
- Weight of Zinc (Zn) = 100 g
- Atomic mass of Zinc = 65 g/mol

**Calculation:**
\[ \text{Number of moles of Zn} = \frac{\text{Weight of Zn}}{\text{Atomic mass of Zn}} = \frac{100 \, \text{g}}{65 \, \text{g/mol}} \approx 1.54 \, \text{mol} \]

### Step 2: Determine the number of moles of Copper ions
Given:
- Concentration of CuSO₄ = 1.0 M
- Volume of CuSO₄ solution = 1.0 L

**Calculation:**
\[ \text{Number of moles of Cu}^{2+} = \text{Concentration} \times \text{Volume} = 1.0 \, \text{mol/L} \times 1.0 \, \text{L} = 1.0 \, \text{mol} \]

### Step 3: Identify the limiting reagent
- Moles of Zn = 1.54 mol
- Moles of Cu²⁺ = 1.0 mol

Since the reaction involves the reduction of Cu²⁺ to Cu and the oxidation of Zn to Zn²⁺, and each mole of Cu²⁺ requires 2 moles of electrons (from Zn), we can see that Cu²⁺ is the limiting reagent because it has fewer moles.

### Step 4: Calculate the charge (Q) using Faraday's law
Each mole of Cu²⁺ accepts 2 electrons. Therefore, the total charge can be calculated as follows:
- Charge (Q) = Number of moles of Cu²⁺ × Number of electrons × Faraday's constant (F)

Given:
- Faraday's constant, F = 96500 C/mol

**Calculation:**
\[ Q = 1.0 \, \text{mol} \times 2 \, \text{mol e}^- \times 96500 \, \text{C/mol} = 193000 \, \text{C} \]

### Step 5: Calculate the time (T) the cell will run
Using the formula \( T = \frac{Q}{I} \), where I is the current in amperes.

Given:
- Current (I) = 1.0 A

**Calculation:**
\[ T = \frac{193000 \, \text{C}}{1.0 \, \text{A}} = 193000 \, \text{s} \]

### Step 6: Convert time from seconds to hours
To convert seconds to hours:
\[ T \, \text{(in hours)} = \frac{193000 \, \text{s}}{3600 \, \text{s/h}} \approx 53.6 \, \text{hours} \]

### Final Answer
The galvanic cell will run for approximately **53.6 hours**.

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