A $1.0 M$ solution of $C d^{2 +}$ is added to excess iron and the system is allowed to reach the concentration of $C d^{2 +}$ ?
$C d^{2 +} (a q) + F e (s) \to C d (s) + F e^{2 +} (a q); E^{o} = 0.037$

0.195

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.097

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.053

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.145

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $0.053$
$C d^{2 +} (a q) + F e (s) ⇌ C d (s) + F e^{2 +} (a q)$
At eqm. $1 - x$ $-$ $-$ $x$
At equilibrium $E^{\circ} = \frac{0.0591}{2} l o g (\frac{F e^{2 +}}{C d^{2 +}})$
$0.037 = \frac{0.0591}{2} l o g (\frac{x}{1 - x})$
$x = [F e^{2 +}] \Rightarrow 0.947 M$

$∴ [C d^{2 +}] = 0.053 M$

Hence option (C) is correct.

Suggest Corrections

Similar questions

C d (N O_{3})_{2}

A g N O_{3}

C d (N O_{3})_{2}

A g N O_{3}

C d (N O_{3})_{2}

A g N O_{3}

E_{A g^{+} / A g}^{0} = 0.77 V a n d E_{C d^{2 +} / C d}^{0} = - 0.40

C d^{2 +} (a q) + Z n (s) \to Z n^{2 +} (a q) + C d (s)

C d^{2 +} (a q) + 4 N H_{3} (a q) ⇌ C d (N H_{3})_{4}^{2 +} (a q)

2 \times 10^{x}

\frac{(x + 1)}{5}

E_{C d^{2 +} | C d}^{\circ} = - 0.04 V; E_{C d (N H_{3})_{4}^{2 +} | C d}^{\circ} = - 0.61 V

Join BYJU'S Learning Program