E deg(Ag+/Ag) = 0.80 V and E deg for [Ag(NH3)2]+ + e- -> Ag + 2NH3 is 0.37 V at 298 K. The formation constant K_f of [Ag(NH3)2]+ is closest to (take 2.303RT/F = 0.0591 V)

JEE Main levelSingle correct optionDifficulty 2 of 3ec103

$E^{\circ} (Ag X^{+} / Ag) = 0.80 V$ and $E^{\circ}$ for $[Ag (NH X_{3}) X_{2}] X^{+} + e X^{-} Ag + 2 NH X_{3}$ is $0.37 V$ at 298 K. The formation constant $K_{f}$ of $[Ag (NH X_{3}) X_{2}] X^{+}$ is closest to (take $2.303 RT / F = 0.0591 V$ )

Subtracting the half reactions gives

Ag X^{+} + 2 NH X_{3} [Ag (NH X_{3}) X_{2}] X^{+}

with

E^{\circ} = 0.80 - 0.37 = 0.43 V

and

n = 1

lo g K_{f} = \frac{0.43}{0.0591} = 7.28

, so

K_{f} = 1 0^{7.28} \approx 1.9 \times 1 0^{7}

. Complexation lowers the free

[Ag X^{+}]

, which is why the complex couple has the lower potential.