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Explain the following (i) It isnot possible to determine the absoulte value of singel electrode potential (ii) Iron undergoes oxidation more readily than copper (iii)In an electrochemical cell anelectrode with lower electrode potential acts as the reductin agent (iv)When a coper rod is placed in silver nitrate solutoion the solution becomes hot but the revrse not true (v) Iron reacts with dilute H_(2)SO_(4) to evolve H_(2) gas but Ag does not (vi) The colour of KI solution containg starch turns blue when it is shaken with cold CI_(2) water Explain why? |
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Answer» Solution :(i) The potential difference between two electrodes can be determined by connecting them to a voltmeter therefore it is not possible to determine thepotential of a single electrode because a single electro constitiues a HALF cell and a half cell reaction cannot TAKE place independely An electrode in a half cell cannot lose or gain electrons by itself for transfer of electrons one half cell to beconnected to some other half cel thus we cannot determine the absolute value of electrode potential ofa single eletrode the half cell with some STANDARD electrode as the reference electrode (ii) The electrode potential of iron `(E_(Fe^(2+)//Fe)^(@)=-0.44)` is lower than that of copper and henceFe has greater tendency to get CONVERTED in to `Fe^(2+)` ions than Cu in other words UNDERGOES oxidation more readily than copper (iv) copper has lower electrode potential than silver therefore Cu releases electrons and tets oxidised to `Cu^(2+)` ions while `Ag^(+)` ions accept these electrons and get reduced to Ag metal `Cu(s)+2Ag^(+)(aq)rarrCu^(2+)(aq)+2Ag(s)` (v) Fe has lower electrode potential `(E_(Fe^(2+)//Fe)^(@)=-0.44)` than that of hydrogen therefore Fe is a better reducing agent than `H_(2)` and hence reduce `H^(+)` ions to produce `H_(2)` gas `Fe(s)+2H^(+)(a)rarrFe^(2+)(Aq)+H_(2)(g)` In contrast s has higher electrode potential than hydrogen therefore `H_(2)` is a better reducing agent than ag in other words Agcannot reduce to colourless `CI^(-)` ions while `I^(-)` ions get oxidised to violet coloured iondine `CI_(2)(aq)+2I^(-)(aq)rarr2CI^(-)(aq)+I_(2)` (s) |
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