A) \[S{{O}_{2}}\] is evolved
B) lead is formed
C) lead sulphate is consumed
D) sulphuric acid is consumed
Correct Answer: D
Solution :
[d] The reactions occurring operation of battery are anode: \[Pb(s)+HS{{O}^{-}}_{4}+{{H}_{2}}O\ to PbS{{O}_{4}}(s)+2{{e}^{-}}+{{H}_{3}}{{O}^{+}};\]\[{{E}^{o}}_{OP}=+0.356V\] Cathode:\[Pb{{O}_{2}}+HSO_{4}^{-}+3{{H}_{3}}{{O}^{+}}+2{{e}^{-}}\to PbS{{O}_{4(S)}}+5{{H}_{2}}O;\]\[{{E}^{o}}_{RP}=+1.685V\] Net change: \[Pb(s)+2HSO_{4}^{-}+2{{H}_{3}}{{O}^{+}}+Pb{{O}_{2}}(s)\to \] \[2PbS{{O}_{4}}(s)+4{{H}_{2}}O;{{E}^{o}}=2.041V\]
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