E- for the cell is Zn -Zn2- aq- -Cu2- aq- Cu at 25-C- the equilibrium constant for the reaction Zn-Cu2- aq - Cu-Zn 2- aq is of the order of

E^∘_cell = 0.059/nlog K log K=1.10× 2/0.059 =37.2881 ⇒ K = 10^ 37 ...

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Byju's Answer

Standard XI

Chemistry

Equilibrium Constant from Nernst Equation

E∘ for the ce...

Question

$E^{\circ}$ for the cell is $Z n | Z n^{2 +} (a q) | | C u^{2 +} (a q) | C u$ at $25^{\circ} C$ , the equilibrium constant for the reaction $Z n + C u^{2 +} (a q) ⇌ C u + Z n^{2 +} (a q)$ is of the order of

10^{- 37}

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10^{- 28}

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10^{+ 18}

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10^{+ 17}

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Solution

The correct option is A $10^{- 37}$
$E_{c e l l}^{\circ} = \frac{0.059}{n} l o g K$
$l o g K = \frac{1.10 \times 2}{0.059} = 37.2881 \Rightarrow K = 10^{- 37}$

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E∘ for the cell is Zn |Zn2+ (aq)||Cu2+ (aq)| Cu at 25∘C, the equilibrium constant for the reaction Zn+Cu2+(aq)⇌Cu+Zn2+ (aq) is of the order of

The correct option is A 10−37E∘cell=0.059nlog K log K=1.10×20.059=37.2881⇒K=10−37

$E^{\circ}$ for the cell is $Z n | Z n^{2 +} (a q) | | C u^{2 +} (a q) | C u$ at $25^{\circ} C$ , the equilibrium constant for the reaction $Z n + C u^{2 +} (a q) ⇌ C u + Z n^{2 +} (a q)$ is of the order of

The correct option is A $10^{- 37}$
$E_{c e l l}^{\circ} = \frac{0.059}{n} l o g K$
$l o g K = \frac{1.10 \times 2}{0.059} = 37.2881 \Rightarrow K = 10^{- 37}$