Calculate the emf of the following concentration cell at 25-C- Ags - AgNO3 0-01 M - AgNO3 0-05 M - Ags-

The correct option is D 0.0414 V E = E^∘ 0.0591nlog_10Q (∵ E^∘ = 0 for all concentration cells) = 0 0.05911log_10 (0.010.05 ) = 0. ...

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Standard XII

Chemistry

EMF

Calculate the...

Question

Calculate the emf of the following concentration cell at $25^{\circ} C$ ;
$A g (s) | A g N O_{3} (0.01 M) ∥ A g N O_{3} (0.05 M) | A g (s)$ .

- 0.414 V

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0.828 V

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0.414 V

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0.0414 V

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Solution

The correct option is D $0.0414 V$
$E = E^{\circ} - \frac{0.0591}{n} {log}_{10} Q$
$(∵ E^{\circ} = 0$ for all concentration cells)
$= 0 - \frac{0.0591}{1} {log}_{10} (\frac{0.01}{0.05})$
$= 0.0414 V$ .

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Similar questions

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Q. Calculate the cell potential (

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is 0.788 V. Calculate the solubility product of AgI.

Calculate the emf of the following cell at $25^{\circ} C$ .
$A g (s) | A g N O_{3} (0.01 {mol kg}^{- 1}) | | A g N O_{3} (0.05 {mol kg}^{- 1}) | A g (s)$

Q. Calculate the emf of the following cell at

25^{\circ} C

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[Given,

E_{c e l l}^{0} = + 0.46 V

and

log 10^{n} = n

]

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Calculate the emf of the following concentration cell at 25∘C;Ag(s)|AgNO3(0.01 M)∥AgNO3(0.05 M)|Ag(s).

The correct option is D 0.0414 VE=E∘−0.0591nlog10Q(∵E∘=0 for all concentration cells)=0−0.05911log10(0.010.05)=0.0414 V.

Calculate the emf of the following concentration cell at $25^{\circ} C$ ;
$A g (s) | A g N O_{3} (0.01 M) ∥ A g N O_{3} (0.05 M) | A g (s)$ .

The correct option is D $0.0414 V$
$E = E^{\circ} - \frac{0.0591}{n} {log}_{10} Q$
$(∵ E^{\circ} = 0$ for all concentration cells)
$= 0 - \frac{0.0591}{1} {log}_{10} (\frac{0.01}{0.05})$
$= 0.0414 V$ .