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Standard XII

Chemistry

Electrolysis and Electrolytes

A silver elec...

Question

A silver electrode is immersed in saturated $A g_{2} S O_{4} (a q)$ . The potential difference between the silver and the standard hydrogen electrode is found to be 0.711 V. If the $K_{s p} (A g_{2} S O_{4})$ is $x \times 10^{- 3}$ .

[Given, $E_{A g^{+} / A g}^{o} = 0.799 V$ ]

The value of $100 x$ is:

164

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134

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106

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Solution

The correct option is D 106
The expression for the emf of the cell is
$E_{c e l l} = E_{c e l l}^{0} - \frac{0.0592}{n} l o g \frac{1}{K_{s p}}$
Substitute values in the above expression
$0.711 = 0.799 - \frac{0.0592}{2} l o g \frac{1}{K_{s p}}$
$l o g \frac{1}{K_{s p}} = 2.9729$
$\frac{1}{K_{s p}} = 939.66$
$K_{s p} = 1.06 \times 10^{- 3}$
$x = 1.06$
$100 x = 106$

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Similar questions

Q. A silver electrode is immersed in saturated

A g_{2} S O_{4 (a q)}

. The potential difference between silver and the standard hydrogen electrode is found to be

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Q. A silver electrode is immersed in saturated

A g_{2} S O_{4} (a q .)

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E_{A g^{+} / A g}^{\circ} = 0.799 V)

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A silver electrode is immersed in saturated Ag2SO4(aq). The potential difference between the silver and the standard hydrogen electrode is found to be 0.711 V. If the Ksp(Ag2SO4) is x×10−3.[Given, EoAg+/Ag=0.799V]The value of 100x is:

The correct option is D 106The expression for the emf of the cell is Ecell=E0cell−0.0592nlog1KspSubstitute values in the above expression0.711=0.799−0.05922log1Ksplog1Ksp=2.97291Ksp=939.66Ksp=1.06×10−3x=1.06100x=106