An electric current is passed through silver voltameter connected to a water voltameter. The cathode of the silver voltameter weighed $0.108 g$ more at the end of the electrolysis. The volume of oxygen evolved at STP is :

56 c m^{3}

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550 c m^{3}

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5.6 c m^{3}

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11.2 c m^{3}

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22.4 c m^{3}

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Solution

The correct option is D $5.6 c m^{3}$
$A g^{+} + e^{-} \to A g$
$2 O^{2 -} \to O_{2} + 4 e^{-}$
$4 A g^{+} + 2 O^{2 -} \to O_{2} + 4 A g$ ...(i)
Amount of silver generated = $0.108 g$
No. of mole of $A g = \frac{0.108}{108} = 0.001 m o l e$
No. of mole of $O_{2}$ generated from equation (i) $\frac{0.001}{4}$
volume of $O_{2} = \frac{0.001}{4} \times 22.4 \times 10^{3} c m^{3}$
$= 5.6 c m^{3}$

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An electric current is passed through silver voltameter connected to a water voltameter. The cathode of the silver voltameter weighed 0.108 g more at the end of the electrolysis. The volume of oxygen evolved at STP is :

The correct option is D 5.6 cm3Ag++e−→Ag2O2−→O2+4e−4Ag++2O2−→O2+4Ag ...(i)Amount of silver generated = 0.108gNo. of mole of Ag=0.108108=0.001moleNo. of mole of O2 generated from equation (i) 0.0014volume of O2=0.0014×22.4×103cm3 =5.6cm3

An electric current is passed through silver voltameter connected to a water voltameter. The cathode of the silver voltameter weighed $0.108 g$ more at the end of the electrolysis. The volume of oxygen evolved at STP is :