In the given electrochemical cell, $$Ag(s)|AgCl(s)|FeCl_{2}(aq),FeCl_{3}(aq)|Pt(s)$$ at298 K, theceU potential $$E_{cell}$$ will increase when:
A. Concentration of $$Fe^{2+}$$ is increased.
B. Concentration of $$Fe^{3+}$$ is decreased.
C. Concentration of $$Fe^{2+}$$ is decreased.
D. Concentration of $$Fe^{3+}$$ is increased.
E. Concentration of $$Cl^{-}$$ is increased.
Choose the correct answer from the options given below :

Let us write the two half‐reactions with their standard reduction potentials:

AgCl(s) + e- → Ag(s) + Cl- ; E° = +0.222 V

Fe3+ + e- → Fe2+ ; E° = +0.771 V

Since the Fe3+/Fe2+ couple has the higher E°, it acts as the cathode and AgCl/Ag as the anode. The overall cell reaction is:

$$Ag(s) + Cl^{-}(aq) + Fe^{3+}(aq) \to AgCl(s) + Fe^{2+}(aq)$$ (1)

The standard cell potential is:

$$E^\circ_{cell} \;=\; E^\circ_{cathode} \;-\; E^\circ_{anode} \;=\; 0.771 \;-\; 0.222 \;=\; 0.549\text{ V}$$

The Nernst equation for a one‐electron transfer is:

$$E_{cell} = E^\circ_{cell} \;-\; \frac{RT}{nF}\ln Q$$ (2)

Here $$n = 1$$, and the reaction quotient is:

$$Q = \frac{[Fe^{2+}]}{[Fe^{3+}]\,[Cl^{-}]}$$ (3)

Substituting (3) into (2) gives:

$$E_{cell} = 0.549 \;-\; \frac{RT}{F}\ln\!\Bigl(\frac{[Fe^{2+}]}{[Fe^{3+}]\,[Cl^{-}]}\Bigr)$$

We now examine how changes in concentrations affect $$E_{cell}$$:

Case 1: Increase in $$[Fe^{2+}]$$. This raises the numerator of $$Q$$, so $$Q$$ increases and $$\ln Q$$ increases. Hence $$E_{cell}$$ decreases.

Case 2: Decrease in $$[Fe^{3+}]$$. This lowers the denominator of $$Q$$, so $$Q$$ increases and $$\ln Q$$ increases. Hence $$E_{cell}$$ decreases.

Case 3: Decrease in $$[Fe^{2+}]$$. This lowers the numerator of $$Q$$, so $$Q$$ decreases and $$\ln Q$$ decreases. Hence $$E_{cell}$$ increases.

Case 4: Increase in $$[Fe^{3+}]$$. This raises the denominator of $$Q$$, so $$Q$$ decreases and $$\ln Q$$ decreases. Hence $$E_{cell}$$ increases.

Case 5: Increase in $$[Cl^{-}]$$. This raises the denominator of $$Q$$, so $$Q$$ decreases and $$\ln Q$$ decreases. Hence $$E_{cell}$$ increases.

Therefore, the cell potential increases in Cases 3, 4 and 5, corresponding to decreasing $$[Fe^{2+}]$$, increasing $$[Fe^{3+}]$$ and increasing $$[Cl^{-}]$$. These are options C, D and E. Hence the correct choice is Option C.