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Construct atcellconsisting of Ni_((ag))^(2+)| Ni_((s)) halfcell andH^(+) ||H_(2(g)) | Pthalf cell (a)Writethe cellreaction ( b)Calculateemfof the cell if cell[Ni^(2+)] = 0.1 [ H^(+)]=0.05 MandE_(Ni)^(0)=- 0.257 V |
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Answer» `[Ni ^(++) ]=0.1 M` (a) Halfcells : (i) `Ni_((ag))^(2+) | Ni_((s))" and" (II)H_((aq))^(+) | H_(2(s)) | Pt ` `E_(Ni^(2+)//Ni) = - 0.257 V , E_(H^(+) //H_(2) )^(0) E_(SHE)^(0)= 0.0V` Theformulation of thecell willbe `Ni_((s))| NI_((ag))^(2+) " || " H_((ag))^(+)|H_(2) (g , 1 atm ) |Pt` `(RHE (H_((ag)^(+))) +e^(-) to .(1)/(2) H_(2(g))xx2)/(Ni_((s) + 2H_((ag))^(+) to Ni_((aq))^(+) + H_(2(g))))"" underset"(Overall cell reaction )"underset"(Reduction at CATHODE )"("(Oxidation at anode)")` `( b) [Ni^(2+) ]=0.1 M :[ H^(+)]= 0.05 M , N=2` `E_(" cell" )^(0) = E_(H^(+)// H_(2))^(0) = E_(Ni^(2+)// Ni)^(0)=0.0 -(0.257)` `=0.256 V` `E_("cell ") = E_(" cell")^(0)- (0.0592)/(n)lgo _(10).([Ni^(2+)] xx [ H_(2)])/([H^(+) ]^(2))` `=0.257 -(0.0592)/(2) log_(10) .(0.1 xx1)/((0.05)^(2))` ` =0.257 - 0.0296 log _(10) 40` `=0.257 -0.0296 xx 1.6020` `=0.257 -0.04742` `=0.2096 V` `~= 0.2096V` |
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