Saved Bookmarks
| 1. |
Construct at cell consisting of `Ni_((ag))^(2+)| Ni_((s))` half cell and `H^(+) ||H_(2(g)) | Pt` half cell (a) Write the cell reaction ( b) Calculate emf of the cell if cell `[Ni^(2+)] = 0.1 ` `[ H^(+)] =0.05 M` and `E_(Ni)^(0) =- 0.257 V` |
|
Answer» Correct Answer - (a) cell reaction `: Ni_((s))^(2++) + 2H_((ag))^(+) to NI_((aq))^(+) to Ni_((aq))^(+) H_(2(g))` ( b) `E_("cell ") = 0.2096 V` Given : `E_(Ni^(+ + )//Ni) =- 0.257 V , [ H^(+) ] = 0.1 M` `[Ni ^(++) ] =0.1 M` (a) Half cells : (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` The formulation of the cell will be `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` |
|