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If electrode potential of following cell: Pt_(s)|Fe_(aq)^(2+),Fe_(aq)^(3+)||MnO_(4(aq))^(-),Mn_(aq)^(2+),H_(aq)^(+)|Pt_(s) is X then calculate value of 20X. [Given: E_(MnO_(4)^(-)|Mn^(2+) = 1.51 V, E_(Fe^(3+)|Fe^(2+) = 0.78 V,(2.303 RT)/F = 0.06]

Answer»


SOLUTION :`-0.12 - (0.0591)/2 log (1/X) = -0.24`
`log 1/X = (0.12 XX 2)/0.06 =4`
`X = 10^(-4)`


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