The emf of the cell Zn - Zn 2-0-1 M F e 2- Fe 0-01 M is 0-2905 V- Equilibrium constant for the cell reaction isA- 100-32 - 0-591B- 100-32 - 0-0295C- 100-26 - 0-0295D- 100-32 - 0-295

The correct option is B E=E^∘ 0.0591/n log Q 0.2905 = E^∘ 0.0591/n log [0.1/0.01] E^∘ = 0.2905 + 0.0295 = +0.32 V Therefore 0.32 = 0.0591/2 log K K = 10 ^0 ...

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

Byju's Answer

Standard IX

Chemistry

Equilibrium Constant from Nernst Equation

The emf of th...

Question

The emf of the cell $Z n | Z n^{2 +} (0.1 M) | | F e^{2 +} | F e (0.01 M)$ is $0.2905 V$ . Equilibrium constant for the cell reaction is

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

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
$E = E^{\circ} - \frac{0.0591}{n} l o g Q$
$0.2905 = E^{\circ} - \frac{0.0591}{n} l o g [\frac{0.1}{0.01}]$
$E^{\circ} = 0.2905 + 0.0295 = + 0.32 V$
Therefore $0.32 = \frac{0.0591}{2} l o g K$
$K = 10^{\frac{0.32}{0.0295}}$

Suggest Corrections

Similar questions

Q. Zn

| Z n^{2 +} (a = 0.1 M) ∥ F e^{2 +} (a = 0.01 M) |

F e

. The emf of the above cell is

0.2905 V

Equilibrium constant for the cell reaction is:

Q. The emf of the cell

Z n | Z n^{2 +} (0.1 M) | | F e^{2 +} | F e (0.01 M)

0.2905 V

. Equilibrium constant for the cell reaction is

Q. The emf of the cell,

Z n | Z n^{2 +} (0.01 M) | | F e^{2 +} (0.001 M) | F e

298 K

0.2905 V

then the value of equilibrium constant for the cell reaction is:

Z n | {Z n}^{2 +} (a = 0.1 M) ∥ {F e}^{2 +} (a = 0.01 M) | F e

. The emf of the above cell is

0.2905 V

. Equilibrium constant for the cell reaction is

The EMF of the cell

$Z n | Z n^{2 +} (0.01 M) | | F e^{2 +} (0.001 M) | F e$ at $298 K$ is $0.2905$ then the value of equilibrium constant for the cell reaction is :

Join BYJU'S Learning Program

The emf of the cell Zn|Zn2+(0.1M) || Fe2+|Fe(0.01M) is 0.2905 V. Equilibrium constant for the cell reaction is

The correct option is B E=E∘−0.0591n log Q 0.2905=E∘−0.0591nlog[0.10.01] E∘=0.2905+0.0295=+0.32 V Therefore 0.32=0.05912log K K=100.320.0295

The emf of the cell $Z n | Z n^{2 +} (0.1 M) | | F e^{2 +} | F e (0.01 M)$ is $0.2905 V$ . Equilibrium constant for the cell reaction is

The correct option is B
$E = E^{\circ} - \frac{0.0591}{n} l o g Q$
$0.2905 = E^{\circ} - \frac{0.0591}{n} l o g [\frac{0.1}{0.01}]$
$E^{\circ} = 0.2905 + 0.0295 = + 0.32 V$
Therefore $0.32 = \frac{0.0591}{2} l o g K$
$K = 10^{\frac{0.32}{0.0295}}$