To solve the problem, we need to determine which metal will be deposited when a 100 ml solution of each 0.1 M AuCl and 0.1 M AgCl is electrolyzed by passing a current of 1 ampere for 15 minutes. We will use the standard reduction potentials provided for Au and Ag to help us.
### Step-by-Step Solution:
1. **Calculate the Total Charge (Q) Passed:**
\[
Q = I \times t
\]
where \( I = 1 \, \text{A} \) and \( t = 15 \, \text{minutes} = 15 \times 60 \, \text{seconds} = 900 \, \text{seconds} \).
\[
Q = 1 \, \text{A} \times 900 \, \text{s} = 900 \, \text{C}
\]
**Hint:** Remember to convert time from minutes to seconds when calculating charge.
2. **Convert Charge to Faradays:**
The Faraday constant \( F \) is approximately \( 96500 \, \text{C/mol} \).
\[
\text{Number of Faradays} = \frac{Q}{F} = \frac{900 \, \text{C}}{96500 \, \text{C/mol}} \approx 0.00932 \, \text{mol}
\]
**Hint:** Use the Faraday constant to convert charge in coulombs to moles of electrons.
3. **Determine Moles of Each Metal Ion Available:**
For both AuCl and AgCl, we have:
\[
\text{Moles of Au} = 0.1 \, \text{M} \times 0.1 \, \text{L} = 0.01 \, \text{mol}
\]
\[
\text{Moles of Ag} = 0.1 \, \text{M} \times 0.1 \, \text{L} = 0.01 \, \text{mol}
\]
**Hint:** Use the molarity and volume to find the total moles of each metal ion.
4. **Identify the Reduction Potentials:**
The standard reduction potentials are given as:
- \( E^\circ \) for \( Au^+ + e^- \rightarrow Au \) is \( 1.69 \, \text{V} \)
- \( E^\circ \) for \( Ag^+ + e^- \rightarrow Ag \) is \( 0.80 \, \text{V} \)
**Hint:** Higher reduction potential means a greater tendency to be reduced.
5. **Determine Which Metal Will Be Deposited First:**
Since \( E^\circ \) for Au is greater than that for Ag, Au will be reduced first. However, we need to check if there is enough charge to deposit Au.
6. **Calculate the Charge Required to Deposit Each Metal:**
The charge required to deposit 1 mole of metal is equal to the Faraday constant (96500 C).
- For Au: \( 1 \, \text{mol} \) requires \( 96500 \, \text{C} \)
- For Ag: \( 1 \, \text{mol} \) requires \( 96500 \, \text{C} \)
Since we have \( 0.00932 \, \text{mol} \) of charge available, we can calculate:
\[
\text{Charge required for Au} = 0.01 \, \text{mol} \times 96500 \, \text{C/mol} = 965 \, \text{C}
\]
\[
\text{Charge required for Ag} = 0.01 \, \text{mol} \times 96500 \, \text{C/mol} = 965 \, \text{C}
\]
**Hint:** Calculate the charge needed for each metal to determine if it can be deposited.
7. **Conclusion:**
Since the total charge available (900 C) is less than the charge required to deposit either metal (965 C), the current will stop before either metal can be deposited. Therefore, no metal will be deposited.
**Final Answer:** None of the metals (Au or Ag) will be deposited.
