When aqueous $A g N O_{3}$ is electrolyzed using Pt electrodes, if 3.2 g of $O_{2}$ is liberated at the anode, the Amount of Ag that will be deposited at the cathode in grams is (at wt. of Ag = 108)

32 g

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1.6 g

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4.32 g

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43.2 g

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Solution

The correct option is D
43.2 g

$A g^{+} + e \to A g;$

$Eq. wt of 'Ag' E_{A g} = \frac{108}{1} = 108 g (a t o m i c m a s s o f A g = 108)$

$20 H^{-} \to O_{2} + 2 H^{+} + 4 e;$

Eq. wt of $^{'} O_{2}^{'}$ $E_{o 2}$ = $\frac{32}{4} = 8 g$ $(M o l a r m a s s o f O_{2} = 32)$

$When same quantity of electricity is passes$

$no. of eqts of 'Ag' = no. of eqts of^{'} O_{2}^{'}$

$\frac{W_{A g}}{E_{A g}} = \frac{W_{O_{2}}}{E_{O_{2}}}; \frac{W_{A g}}{108} = \frac{3.2}{8}; W_{A g} = 43.2 g$

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When aqueous AgNO3 is electrolyzed using Pt electrodes, if 3.2 g of O2 is liberated at the anode, the Amount of Ag that will be deposited at the cathode in grams is (at wt. of Ag = 108)

The correct option is D 43.2 g Ag++e→Ag; Eq. wt of 'Ag'EAg=1081=108g (atomic mass of Ag=108) 20H−→O2+2H++4e; Eq. wt of ′O′2 Eo2 = 324=8g (Molar mass of O2=32) When same quantity of electricity is passes no. of eqts of 'Ag' = no. of eqts of ′O′2 WAgEAg=WO2EO2;WAg108=3.28;WAg=43.2g

When aqueous $A g N O_{3}$ is electrolyzed using Pt electrodes, if 3.2 g of $O_{2}$ is liberated at the anode, the Amount of Ag that will be deposited at the cathode in grams is (at wt. of Ag = 108)