The electrode potential of M2+/M of 3d-series elements shows positive value for :
Solution
In the electrode potential series, only copper have positive value for electrode potential because copper has lower tendency than hydrogen to form ions. So, if standard hydrogen electrode (E<sub>Cell</sub> = 0) is connected to copper half-cell, the copper with be relatively less negative or less number of electrons.<br/><br/>$E_{(C{u^{2 + }}/Cu)}^o$ = + 0.34 V; $E_{(F{e^{2 + }}/Fe)}^o$ = $-$ 0.41 V<br/><br/>$E_{(C{o^{2 + }}/Co)}^o$ = $-$ 0.28 V; $E_{(Z{n^{2 + }}/Zn)}^o$ = $-$ 0.76 V<br/><br/>$\therefore$ Electrode potential of Cu $E_{(C{u^{2 + }}/Cu)}^o$ show positive value.
About this question
Subject: Chemistry · Chapter: Electrochemistry · Topic: Electrochemical Cells
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