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Given the cell: Cds|CdOH2s|NaOHaq,0.01  M|H2g,1 bar|Pts with Ecell=0.0  V If ECd2+|Cd∘=−0.39  V, then Ksp of CdOH2 is 10−5x :( take RT/F=0.06) Then find the value of x

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Detailed Solution

Ecell =0.06(5−4)=+0.06VEcell =EH+∣H2−EOH−Cd(OH)2Cd (or) ECell ∘=0.03log×Cd2+OH−2Kw2∵Ecell =0∴Ecell =0.03log×Cd2+OH−2Kw2⇒log⁡KspKw2=0.390.03=13∴Ksp=1013×10−142=10−15
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