The value of equilibrium constant Kf for the reaction- Zn2-aq-4OH-aq- ZnOH42-aq is represented in scientific notation as p-10q- then q is-Given - Zn2-aq-2e- Zns- E0-0-76 V ZnOH42-aq-2e- Zns-4OH-aq- E0-1-36V 2-303RTF-0-06

The correct option is B 20 Δ G^o= RTlnK_eq logK_eq=nFE^oRT× 2.303⇒2× 0.60.06⇒ 20 K=10^20 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Effective Equilibrium Constant

The value of ...

Question

The value of equilibrium constant $(K_{f})$ for the reaction: $Z n^{2 +} (a q) + 4 O H^{-} (a q) ⇌ Z n (O H)_{4}^{2 -} (a q)$ is represented in scientific notation as $p \times 10^{q},$ then q is:
Given : $Z n^{2 +} (a q) + 2 e^{-} \to Z n (s); E^{0} = - 0.76 V$
$Z n (O H)_{4}^{2 -} (a q) + 2 e^{-} \to Z n (s) + 4 O H^{-} (a q); E^{0} = - 1.36 V$
$2.303 \frac{R T}{F} = 0.06$

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

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

Open in App

Solution

The correct option is B 20
$Δ G^{o} = - R T l n K_{e q}$
$l o g K_{e q} = \frac{n F E^{o}}{R T \times 2.303} \Rightarrow \frac{2 \times 0.6}{0.06} \Rightarrow 20$
$K = 10^{20}$

Suggest Corrections

Similar questions

Q. A button cell used in watches functions as following:

Z n (s) + A g_{2} O (s) + H_{2} O (l) ⇌ 2 A g (s) + Z n^{2 +} (a q) + 2 O H^{-} (a q)

If half cell potentials are:

Z n^{2 +} (a q) + 2 e \to Z n (s); E^{0} = - 0.76 V

A g_{2} O (s) + H_{2} O (l) 2 e \to 2 A g (s) + 2 O H^{-} (a q), E^{0} = 0.34 V

The cell potential will be:

Q. Which of the following statement(s) is/are true?

Z n^{2 +} (a q) + 2 e^{-} \to Z n (s); E^{0} = - 0.76 V

C u^{2 +} (a q) + 2 e^{-} \to C u (s); E^{0} = 0.34 V

$E^{0}$ for the cell, $Z n | Z n {^{2 +}}_{(a q)} | | C u {^{2 +}}_{(a q)} | C u$ is $1.10 V$ at $25^{0} C$ . The equilibrium constant for $Z n + C u {^{2 +}}_{(a q)} \to Z n {^{2 +}}_{(a q)} + C u$ the cell reaction is of the order of :

Q. Which of the following reaction will provide maximum voltage?
Given:

Z n^{2 +} (a q) + 2 e^{-} \to Z n (s); E^{0} = - 0.76 V

C u^{2 +} (a q) + 2 e^{-} \to C u (s); E^{0} = 0.34 V

A g^{+} (a q) + e^{-} \to A g (s); E^{0} = 0.80 V

2 H^{+} (a q) + 2 e^{-} \to H_{2} (g); E^{0} = 0 V

N i^{2 +} (a q) + 2 e^{-} \to N i (s); E^{0} = - 0.25 V

Q. A button cell used in watches functions as following:

Z n (s) + A g_{2} O (s) + H_{2} O (l) ⇌ 2 A g (s) + Z n^{2 +} (a q) + 2 O H^{-} (a q)

If half cell potentials are:

Z n^{2 +} (a q) + 2 e^{-} \to Z n (s); E^{\circ} = - 0.76 V

A g_{2} O (s) + H_{2} O (l) + 2 e^{-} \to 2 A g (s) + 2 O H^{-} (a q), E^{\circ} = 0.34 V

The cell potential will be:

Join BYJU'S Learning Program

The value of equilibrium constant (Kf) for the reaction: Zn2+(aq)+4OH−(aq)⇌Zn(OH)2−4(aq) is represented in scientific notation as p×10q, then q is:Given : Zn2+(aq)+2e−→Zn(s);E0=−0.76V Zn(OH)2−4(aq)+2e−→Zn(s)+4OH−(aq);E0=−1.36V 2.303RTF=0.06

The correct option is B 20ΔGo=−RTlnKeqlogKeq=nFEoRT×2.303⇒2×0.60.06⇒20K=1020

The correct option is B 20
$Δ G^{o} = - R T l n K_{e q}$
$l o g K_{e q} = \frac{n F E^{o}}{R T \times 2.303} \Rightarrow \frac{2 \times 0.6}{0.06} \Rightarrow 20$
$K = 10^{20}$