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Given the following standard electrode potentials , PbBr_2(s) + 2e^(-) to Pb(s)+ 2Br^(-) (aq) , E^@ - 0.248 VPb^(2+)(aq) + 2e^(-) to Pb(s) , E^(@) - 0.126 V . If the K_(sp) forPbBr_2 is 7.4 xx 10^(-x) , x is |
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Answer» <BR> Solution :`E_(Br)^(-)//PbBr_2 // Pb = E_(Pb^(+2)//Pb)^(0) + (0.059)/(2) log_(10)^(K_(SP)) - 0.248 = -0.126 + (0.059)/(2) log K_(SP)``K_(SP) = 7.4 xx 10^(-5) implies x=5` |
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