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Significance of Standard Half-Cell Potential


The standard half-cell potential of the reaction
 
is a measure of the reduction tendency of Mn+ to M relative to that of H+ to H2. The standard potential of the latter reaction, i.e.
 
is taken to be zero (reference level). A positive potential implies that the ion Mn+ can be more easily reduced to M relative to that of H+ ion to H2. On the other hand, a negative potential implies that the ion Mn+ is more difficult to reduce as compared to the H+ ion.
If is positive, then the cell
Pt| H2(1.01325 bar)| H+(1 mol dm-3)||Mn+(1 mol dm-3)| M
will have a positive emf and, therefore, the cell reaction


will be spontaneous and, hence, Mn+ can be reduced to M by hydrogen gas. On the other hand, if is negative, the cell emf will also be negative and, hence, the cell reaction will not be spontaneous. Thus, Mn+ cannot be reduced to M by hydrogen gas. In the present case, if the cell as given by above equation is written in the reverse direction, i.e.

M| Mn+(1 mol dm-3)||H+(1 mol dm-3)| H2(1.013 25 bar)| Pt

the cell will have a positive emf and, hence, the reaction


will be a spontaneous reaction. Thus, H+ ions can be reduced to hydrogen gas by the metal M.
Taking the typical examples of Ag+|Ag and Zn2+|Zn half-cells, we find that


Hence we conclude that under standard conditions, Ag+ ions can be reduced to silver by hydrogen gas, whereas Zn2+ cannot be reduced. Alternatively, zinc metal can reduce H+ ions to hydrogen gas, whereas silver metal cannot reduce H+ ions.

Active metals such as Zn, Na and Mg have highly negative standard potentials indicating that their compounds are not reduced by hydrogen. However, the metal itself can be oxidized by H+ to yield H2. Noble metals such as Cu, Ag and Au have positive Eos and thus their compounds are readily reduced by H2 gas; the metals themselves are not oxidized in the presence of H+ ions.

If a pair of half-cells is coupled to make a cell, the half-cell with a more positive potential will constitute the positive terminal, whereas the half-cell with a less positive potential will constitute the negative terminal. In other words, the half-cell of the higher positive potential (which stands lower in the table) will constitute the RHC with the reduction half-cell reaction and the half-cell of less positive potential will constitute the LHC with the oxidation half-cell reaction. Consider, for example, the two half-cells Ni2+| Ni and Ag+| Ag. Their reduction potentials are


Since is more positive than , it is obvious, that the silver electrode will constitute the positive terminal (i.e. RHC) and nickel electrode the negative terminal (i.e. LHC). Thus, the required cell will have positive emf and hence the cell producing spontaneous cell reaction would be
Ni| Ni2+(aq)|| Ag+(aq)| Ag

Problem
Answer whether, under standard conditions, the following reactions are possible or not.
(a) Will zinc reduce Cu2+ to Cu?
(b) Will copper reduce Ag+ to Ag?
(c) Will Fe3+ be reduced to Fe2+ by Sn2+?
(d) Would you use silver spoon to stir a solution of Cu(NO3)2?
  1. The reaction will be
    Reduction: Cu2+ + 2e-  Cu
    Oxidation: Zn Zn2+ + 2e-
    The cell giving these half-cell reactions would be
    Zn| Zn2+|| Cu2+| Cu
    with

    = 0.337 V - (-0.763 V)
    = 1.100 V
    Since is positive, the reduction of Cu2+ to Cu by Zn is possible.
  2. The reactions would be
    Reduction: Ag+ + e- Ag
    Oxidation:
    The cell giving these half-cell reactions would be
    Cu| Cu2+| Ag+| Ag
    with

    = 0.799 V - 0.337 V
    = 0.462 V
    Since is positive, the reduction of Ag+ to Ag by Cu is possible.
  3. The reactions would be
    Reduction: Fe3+ + e- Fe2+
    Oxidation: Sn2+

    Sn4+ + 2e
    The cell giving these half-cell reactions would be
    Pt| Sn4+, Sn2+|| Fe3+, Fe2+| Pt
    with

    = 0.771 V - 0.150 V = 0.621 V
    since Eo is+ve, the reaction is feasible.
  4. The reactions would
    Reduction: Cu2+ + 2e-
    Cu
    Oxidation: Ag
    Ag+ + e
    The cell giving these half-cell reactions would be
    Ag| Ag+|| Cu2+| Cu
    with

    = 0.377 V - 0.799 V = - 0.462 V
    Since is negative, the cell reaction will not be spontaneous. Hence, we can use the silver spoon to stir a solution of Cu(NO3)2.




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