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A cell is constructed Pb^(2+) Pb and Ni^(2+), Ni electrodes. If E^(@) values of Pb and Ni electrodes are respectively -0.13 and -0.25V, write (a) the cell reaction and (b) cell notation. |
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Answer» Solution :Reduction potential of NI electrode is less and hence it REDUCES `Pb^(2+)`ions. The CELL reaction is written as, `Ni+Pb^(2+) RARR Pb+Ni^(2+)` The oxidation half reaction is,`Ni rarr Ni^(2+) +2e^(-)` The reduction half reaction is , `Pb^(2+)+2e^(-) rarr Pb` The cell is written as, `Ni, Ni_((aq))^(2+) //Pb_((aq))^(2+), Pb`. |
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