To calculate the mass of calcium deposited, we can use Faraday's Law of Electrolysis, which states:
\[ \text{Mass} = \frac{M \times I \times t}{n \times F} \]
Where:
The time is given as 1 hour 47 minutes 13 seconds.
1 hour = 3600 seconds, 47 minutes = 47 × 60 = 2820 seconds, 13 seconds = 13 seconds
Total time t:
t = 3600 + 2820 + 13 = 6433 seconds
\[ \text{Mass of Ca} = \frac{40 \, \text{g/mol} \times 3 \, \text{A} \times 6433 \, \text{sec}}{2 \times 96500 \, \text{C/mol}} \]
Simplifying the equation:
\[ \text{Mass of Ca} = \frac{40 \times 3 \times 6433}{2 \times 96500} \]
\[ \text{Mass of Ca} = \frac{772960}{193000} \]
\[ \text{Mass of Ca} = 4.0 \, \text{g} \]
The mass of calcium deposited is 4.0 g, making option (D) the correct answer.
For the given cell: \[ {Fe}^{2+}(aq) + {Ag}^+(aq) \to {Fe}^{3+}(aq) + {Ag}(s) \] The standard cell potential of the above reaction is given. The standard reduction potentials are given as: \[ {Ag}^+ + e^- \to {Ag} \quad E^\circ = x \, {V} \] \[ {Fe}^{2+} + 2e^- \to {Fe} \quad E^\circ = y \, {V} \] \[ {Fe}^{3+} + 3e^- \to {Fe} \quad E^\circ = z \, {V} \] The correct answer is: