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: \[Zn+2{{H}^{+}}\rightleftharpoons {{H}_{2}}+Z{{n}^{2+}}\] |
\[;\,\,E_{cell}^{0}=0-(-\,0.76)=0.76\,V\] |
\[\therefore \] \[0.701=0.76-\frac{0.059}{2}\log \frac{0.01\times 1}{{{[{{H}^{+}}]}^{2}}}\] |
\[\therefore \] \[[{{H}^{+}}]={{10}^{-\,2}}M\] |
\[\therefore \] NaOH required is 0.01 mole = 0.4 gms. |
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