Why cation present below in electrochemical series can discharge before the elements present above them in an electrochemical cell?

Why cation present below in electrochemical series can discharge before the elements present above them in an electrochemical cell?
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Ans: This is because their ability to be reduced (reduction potential) is more than the cation present above them in electrochemical series. So they take electrons before the other cations and get discharged.
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Related questions

During electrolysis of aqeous solution of NaCl, there are two type of anions are present Cl- and –OH, Cl- is discharged first than –OH,although it is present below Cl- in electrochemical series. Give reason.

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Why gold ,silver and pt are used in making ornaments?

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Why carbon electrode is used during electrolysis of Nacl, instead of Pt electrode?

Description : Why carbon electrode is used during electrolysis of Nacl, instead of Pt electrode?

Last Answer : Ans: During electrolysis of Nacl, chlorine gas is produced at anode. Thus to avoid it’s reaction with platinum we use carbon electrode.

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Description : Two half-reactions of an electrochemical cell are given below : MnO-4 (aq) + 8H+ (aq) + 5e- → Mn2+ (aq) + 4H2O (I), E° = 1.51 V Sn2+ (aq) → Sn4+ (aq) + 2e- ... the redox equation from the standard potential of the cell and predict if the reaction is reactant favoured or product favoured. -Chemistry

Last Answer : The reactions can be represented at anode and at cathode in the following ways : At anode (oxidation) : Sn2+ → = Sn4+ (aq) + 2e- ] 5 E° = + 0.15 V At cathode (reduction) : MnO-4(aq) + 8H+ ... = E°cathode - E°anode = 1.51 - 0.15 = + 1.36 V ∴ Positive value of E°cell favours formation of product.

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Description : Two half-reactions of an electrochemical cell are given below : MnO-4 (aq) + 8H+ (aq) + 5e- → Mn2+ (aq) + 4H2O (I), E° = 1.51 V Sn2+ (aq) → Sn4+ (aq) + ... -reactions and calculate the cell potential from the standard potentials and predict if the reaction is reactant or product favoured. -Chemistry

Last Answer : The reactions can be represented at anode and at cathode in the following ways : At anode (oxidation) : Sn2+ → Sn4+ (aq) + 2e- ] 5 E° = + 0.15 V Af cathode (reduction) : MnO-4(aq) + 8H+ ( ... = E°cathode - E°anode = 1.51 - 0.15 = + 1.36 V ∴ Positive value of E°cell favours formation of product.

Using the electrochemical series, how can I explain why manufacturers coat steel nails with zinc rather than copper?

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