A) 0.4 gms
B) 4 gms
C) 10 gms
D) 2 gms
Correct Answer: A
Solution :
Anode \[Zn\to Z{{n}^{2+}}+2{{e}^{-}}\] |
Cathode \[2{{H}^{+}}+2{{e}^{-}}\to {{H}_{2}}\] |
Cell : \[Zn+2{{H}^{+}}\,\,\,\,{{H}_{2}}+Z{{n}^{2+}}\] |
\[E_{cell}^{0}=0-(-0.76)=0.76\,V\] \[\therefore \,\,\,0.701=0.76-\frac{0.059}{2}\,\,\log \,\,\frac{0.01\times 1}{{{[{{H}^{+}}]}^{2}}}\] |
\[\therefore \,\,\,[{{H}^{+}}]={{10}^{-2}}M\] |
\[\therefore \,\,\,NaOH\]Required is 0.01 mole = 0.4 gms. |
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