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Calculate the e.m.f. of the following cell at `298K :` `Pt(s)|Br_(2)(l)|Br^(-)(0.010M)||H^(+)(0.030M)|H_(2)(g)(1"bar")|Pt(s)` Given `: E_((1)/(2)Br_(2)//Br^(-))^(@)=+1.08V`.A. `-1.288V`B. `1.288V`C. `0.128V`D. `-128V` |
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Answer» Correct Answer - 2 Cell reaction `:` `2Br^(-)+2H^(+)rarrBr_(2)+H_(2)` Nernst eqn. `E_(cell)=E_(cell)^(@)-(0.0591)/(2)log.(1)/([Br^(-2)][H^(+)]^(2))` `:. " "E_(cell)=(0-1.08)-(0.0591)/(2)log.(1)/((0.01)^(2)(0.03)^(2))=-1.08-(0.0591)/(2)log(1.111xx10^(7))` `=-1.08-(0.0591)/(2)(7.0457)=-1.08-0.208=-1.288V` Thus, oxidation will occur at the hydrogen electrode and reduction on the `Br_(2)` electrode `E_(cell)=1.288V`. |
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