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 ...

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Standard XI

Business Studies

Multinational Corporation and Its Features

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

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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 $0.32 = \frac{0.0591}{2} l o g K$
$K = 10^{\frac{0.32}{0.0295}}$

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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 $0.32 = \frac{0.0591}{2} l o g K$
$K = 10^{\frac{0.32}{0.0295}}$