InterviewSolution
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InterviewSolution
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(a) Define the term molar conductivity . How is it related to conductivity of the related solution ? (b) One half-cell in a voltaic cell is constructed from a silver wire dipped in silver nitrate solution of unknown concentration . Its other half-cell consists of a zinc electrode dipping in `1.0`M solution of `Zn(NO_(3))_(2)` . A voltage of `1.48` V is measured for this cell . Use this information to calculate the concentration of silver nitrate solution used. `(E_(Zn^(2+)|Zn)^(@) = -0.76V , E_(Ag^(+)|Ag)^(@) = + 0.80V`). |
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Answer» (b) Half cell reactions : At Cathod : `" " 2Ag^(+) (aq) + 2e^(-) to 2Ag(s)` At Anode : `" " Zn(s) to Zn^(2+) (aq) + 2e^(-)` The cell can be represented as Zn(s) `|Zn^(2+) (aq) || Ag^(+) (aq) |Ag(s)` The cell reaction is Zn(s) + `2Ag^(+)(aq) to Zn^(2+) (aq) + 2 Ag(s)` `therefore " " E_("cell")^(@) = E_("cathod2")^(@) - E_("anode")^(@)` ` = 0.80 - (-0.76) = 0.80 + 0.76 = 1.56V` `therefore " " E_("cell") = E_("cell")^(@) - (0.059)/(n) "log" ([Zn^(2+)])/([Ag^(+)]^(2))` `1.48 = 1.56 - (0.059)/(2) "log" (1)/([Ag^(+)]^(2))` `(0.059)/(2) "log" (1)/([Ag^(+)]^(2)) = 1.56 - 1.48` log `[Ag^(+)]^(-2) = (0.08 xx 2)/(0.059)` `-2 "log"[Ag^(+)] = (0.08 xx 2)/(0.059)` `therefore " " log [Ag^(+)] = -1.356 = -2 + 2 - 1.356` `[Ag^(+)] = "anti-log" (2644) = 4.406 xx 10^(-2) M` |
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