If the equilibrium constant for the reaction Cd2+(aq)+4NH3(aq)⇌CdNH34 2+(aq) is 10x the value of x is - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

Solution:

${Cd}^{2 +} + 2 e^{-} ⇌ Cd, {ΔG}_{1}^{\circ}$

$\frac{Cd + 4 {NH}_{3} ⇌ {[Cd {({NH}_{3})}_{4}]}^{2 +} + 2 e^{-}, Δ G_{2}^{\circ}}{{Cd}^{2 +} + 4 {NH}_{3} ⇌ {[Cd {({NH}_{3})}_{4}]}^{2 +}, Δ G_{3}^{\circ}}$

$\begin{array}{l} {ΔG}_{1}^{\circ} + {ΔG}_{2}^{\circ} = {ΔG}_{3}^{\circ} \\ - 2 \times F \times (- 0 .40) - 2 \times F (+ 0 .61) = RTln K_{f} \\ ∴ \ln K_{f} = \frac{2 F}{RT} (0 .61 - 0 .4) \\ = - 2 \times F (- 0 .4) - 2 F (0 .61) \\ \log K_{f} = \frac{2 \times 0 .21}{0 .0591} = 7 \Rightarrow K_{f} = 10^{7} \end{array}$

$If the equilibrium constant for the reaction {Cd}^{2 +} (aq) + 4 {NH}_{3} (aq) ⇌ {[Cd {({NH}_{3})}_{4}]}_{}^{2 +} (aq) is 10^{x}$
the value of x is

Solution:

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