A) 15.2
B) 7.6
C) 11.4
D) 22.8
Correct Answer: B
Solution :
Anodic half cell: \[P{{b}^{+2}}\,+\,2{{H}^{2}}O\,\to \,Pb{{O}_{2}}\,+\,4{{H}^{+}}+\,2{{e}^{-}}\] \[\frac{{{n}_{p{{b}^{+2}}}}}{1}\,\,=\,\,\frac{n{{e}^{-}}}{2}\] \[\frac{{{n}_{PbS{{O}_{4}}}}}{1}\,\,=\,\,\frac{0.05}{2}\] \[{{W}_{PbS{{O}_{4}}}}\,\,=\,\,\frac{0.05}{2}\,\,\times \,\,303\,\,=\,\,7.6\,\,g\]
You need to login to perform this action.
You will be redirected in
3 sec
