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Arrange K^(+) , Zn^(2+) , H^(+) and Cu^(2+) ions in order of their tendency to be liberated at the cathode. [Given:mE_(Cu^(2+)|Cu)^(0) = +0.34 V , E_(2H^(+)|H_2)^(0) = 0.00 V, E_(Zn^(2+)|Zn = -.0.76 V, E_(K^(+)|K) = -2.93 V] |
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Answer» Solution :The increasing order of standard electrode potential of the given ions: `K_(K^(+)|K)^(0) lt E_(Zn^(2+)|Zn)^(0) lt E_(2H^(+)|H_(2))^(0) lt E_(Cu^(2+)|Cu)^(0)` . AMONG the given electrodes . `K^(+)|K` and `Cu^(2+)|Cu` have the lowest and highest reduction potential RESPECTIVELY. Thus , `Cu^(2+)` has the highest tendency to get REDUCED whereas K has the highest tendency to get oxidised. Thus the order of increasing tendencies to get liberated at cathode is : `K^(+) lt Zn^(2+) lt H^(+) lt Cu^(2+)` |
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