1. Identify the ions present and their reduction potentials

The solution contains:

• $AgNO_3 \Rightarrow Ag^+$
• $AuCl \Rightarrow Au^+$

Given standard reduction potentials: $Ag^+ + e^- \to Ag, \quad E^\circ = 0.80\,V$ $Au^+ + e^- \to Au, \quad E^\circ = 1.69\,V$

Since $E^\circ$ for $Au^+/Au$ is higher, $Au^+$ is reduced more easily than $Ag^+$.

2. Find initial moles of each ion

Volume of solution: $V = 250\,mL = 0.250\,L$

Concentration of each salt = $0.1\,M$

So moles of each ion initially: $n(Ag^+) = 0.1 \times 0.250 = 0.025\,mol$ $n(Au^+) = 0.1 \times 0.250 = 0.025\,mol$

3. Calculate total charge passed

Current passed: $I = 1\,A$ Time: $t = 15\,min = 900\,s$

Hence charge passed: $Q = It = 1 \times 900 = 900\,C$

Moles of electrons supplied: $n(e^-) = \frac{Q}{F} = \frac{900}{96500} \approx 9.33 \times 10^{-3}\,mol$

4. Compare required electrons with available electrons

Both reductions are one-electron processes: $Ag^+ + e^- \to Ag$ $Au^+ + e^- \to Au$

To reduce all $Au^+$ present, required electrons would be: $0.025\,mol$

But available electrons are only: $0.00933\,mol$

So the total charge is insufficient to reduce all $Au^+$, and since $Au^+$ has the higher reduction potential, it will be discharged preferentially.

Thus only $Au^+$ gets reduced in the given time.

5. Check whether silver can start depositing

As long as $Au^+$ is present, cathode potential required for gold deposition is more favorable than that for silver. Since the available charge cannot even exhaust all the gold ions, silver deposition does not begin.

Hence deposited metal is only gold.

6. Correct option

$\boxed{\text{C: Only gold}}$

7. Comparison with stored answer

Stored correct answer: C

My derived answer: C

So, they agree.