A) 107.88 gm
B) 1.6 gm
C) 0.8 gm
D) 21.60 gm
Correct Answer: D
Solution :
At cathode: \[A{{g}^{+}}+{{e}^{-}}\to Ag\] At Anode: \[2O{{H}^{-}}\to {{H}_{2}}O+\frac{1}{2}{{O}_{2}}+2{{e}^{-}}\] \[{{E}_{Ag}}=\frac{108}{1}=108;\,\,{{E}_{{{O}_{2}}}}=\frac{\frac{1}{2}\times 32}{2}=8\] \[\frac{{{W}_{Ag}}}{{{E}_{Ag}}}=\frac{{{W}_{{{O}_{2}}}}}{{{E}_{{{O}_{2}}}}}\]; \[{{W}_{Ag}}=\frac{1.6\times 108}{8}=21.6\,\,gm\].You need to login to perform this action.
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