InterviewSolution
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InterviewSolution
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The standard reduction potential for the half-cell, `NO_(3(aq.))^(-)+2H_((aq.))^(+)+e^(-)rarrNO_(2(g))+2H_(2)O` is `0.78V`. (i) Calculate the reduction potential in 8M `H^(+)`. (ii) What will be the reduction potential of the half-cell in a neutral solution. Assume all the other species to be at unit concentration |
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Answer» (i) The redox reaction is : `NO_(3)^(-)(aq)+2H^(+)(aq)+e^(-) to NO_(2)(g)+H_(2)O(l)` According to Nernst equation : `E=E^(@)-(0.0591)/(n)"log"([NO_(2)(g)][H_(2)O(l)])/([NO_(3)^(-)(aq)][H^(+)(aq)]^(2))` According to available data, `[NO_(2)(g)]=1,[H_(2)O(l)]=1, [NO_(3)^(-)(aq)]=1,n=1,E^(@)=0.78" V", H^(+)=8M` `:. " " E=0.78-(0.0591)/(1)"log"(1)/((8)^(2))=0.78+0.0591 log 8^(2)=0.78+0.0591(2xx"log "8)` `=0.78+0.0591xx2xx0.9031=0.78+0.1067=0.8867" V"` (ii) For neutral solution `(H^(+)]=10^(-7)M` `E=0.78-(0.0591)/(1)"log"(1)/((10^(-7))^(2))=0.78-0.0591(log 10^(14))` `=0.78-0.0591(14" log "10)=0.78-0.0591xx14" "(because "log" 10=1)` `=0.78-0.8274=-0.0474" V"`. |
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