250 mL of a waste solution obtained from the workshop of a goldsmith contains 0.1 M AgNO$$_3$$ and 0.1 M AuCl. The solution was electrolyzed at 2 V by passing a current of 1 A for 15 minutes. The metal/metals electrodeposited will be: $$\left(E^0_{Ag^+/Ag} = 0.80V, E^0_{Au^+/Au} = 1.69V\right)$$

We are given 250 mL (that is, 0.25 L) of an aqueous solution that contains two cations in the same concentration:

$$[\,Ag^+\,]=0.1\;{\rm M},\qquad [\,Au^+\,]=0.1\;{\rm M}$$

The solution is electrolysed with a constant current of 1 A for 15 min while an external potential difference of 2 V is applied. We have to decide which metal (or metals) will actually get deposited on the cathode.

First we note the standard reduction potentials given in the question:

$$\displaystyle Ag^+ + e^- \longrightarrow Ag,\qquad E^0_{Ag^+/Ag}=+0.80\;{\rm V}$$

$$\displaystyle Au^+ + e^- \longrightarrow Au,\qquad E^0_{Au^+/Au}=+1.69\;{\rm V}$$

In an electrolytic cell the cathode potential is forced to become sufficiently negative (with respect to the standard hydrogen electrode) so that the desired reduction can occur. Among competing reductions, the ion having the higher (more positive) reduction potential is discharged first because it requires a smaller magnitude of cathodic over-potential.

Between silver and gold, gold has the larger reduction potential (1.69 V > 0.80 V). Therefore, as soon as the current starts flowing, the cathode potential will reach a value just negative enough to make the reduction of $$Au^+$$ feasible. At that stage the potential is still not negative enough to reduce $$Ag^+$$. Hence the current supplied will initially deposit only gold.

The second question is: will the entire 15-minute charge be able to exhaust all the gold ions present? If it does, the cathode potential will then be forced to move further negative and silver may start to deposit. We must therefore compare the number of electrons actually supplied with the number required to reduce every $$Au^+$$ present.

Charge passed through the cell:

$$Q = I \times t = 1.0\;{\rm A} \times 15\;{\rm min} \times 60\;\frac{\rm s}{\rm min}=900\;{\rm C}$$

Faraday’s constant is $$F = 96500\;{\rm C\,mol^{-1}}$$, so the moles of electrons supplied are

$$n(e^-) = \frac{Q}{F} = \frac{900}{96500}\;{\rm mol} \approx 9.33\times10^{-3}\;{\rm mol}$$

The electrode reaction for gold requires one electron per gold ion:

$$Au^+ + e^- \rightarrow Au$$

Therefore the maximum moles of gold that can be deposited with the available charge are also $$9.33\times10^{-3}\;{\rm mol}$$.

Initial moles of $$Au^+$$ in the solution are obtained from its concentration and volume:

$$n_0(Au^+) = 0.1\;{\rm mol\,L^{-1}} \times 0.25\;{\rm L}=0.025\;{\rm mol}$$

Because $$0.025\;{\rm mol} > 0.00933\;{\rm mol}$$, the current passed during the experiment is insufficient to remove all the gold ions present. Consequently, when the 15 minutes are over, some $$Au^+$$ will still remain in solution, the cathode potential will never have reached the more negative value necessary for the reduction of $$Ag^+$$, and no silver can have been deposited.

Thus, throughout the electrolysis only gold is electrodeposited.

Hence, the correct answer is Option A.