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The E^(@)(M^(2+)//M) value for copper is...

The `E^(@)(M^(2+)//M)` value for copper is positive (+0.34 V). What is possibly the reason for this ?

Text Solution

Verified by Experts

`E^(@)(M^(2+)//M)` for any metal depends upon the sum of the enthalpy changes taking place in the following steps :
`M(s)+Delta_(a)H to M(g),(Delta_(a)H=" cnthalpy of atomisation")`
`M(g)+Delta_(i)H to M^(2+)(g),(Delta_(i)H=" ionisation enthalpy")`
`M^(2+)(g)aq to M^(2+)(aq)+Delta_("hyd")H(Delta_("hyd")H="hydration enthalpy")`
Copper possesses a high enthalpy of atomisation (i.e., energy absorbed) and low enthalpy of hydration, (i.e., energy released). Hence, `E^(@)(Cu^(2+)//Cu)` is positive.
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Knowledge Check

  • Ag^(+)(aq)+e^(-)rarrAg(s) E^(@)=0.80V Cu^(2+)(aq)+2e^(-)rarrCu(s) E^(@)=0.34V Which expression gives the voltage for this cell if [Cu^(2+)]=1.00M and [Ag^(+)]=0.010M ?

    A
    0.46V+0.0591V
    B
    0.46V+2xx0.0591V
    C
    0.46V-0.0591V
    D
    0.46V-2xx0.0591V

  • E^(@) of Fe^(2 +) //Fe = - 0.44 V, E^(@) of Cu //Cu^(2+) = -0.34 V . Then in the cell

    A
    `Cu^(2+)` Oxidizes Fe
    B
    `Fe^(2+)` oxidizes Cu
    C
    Cu Reduces `Fe^(2+)`
    D
    Fe reduces `Cu^(2+)`

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    Consider the standard electrode potential values (M2+/M) of the elements of the first transition series. {:(Ti,,V,,Cr,,Mn,,Fe,,CO,,Ni,,Cu,,Zn),(-1.63,,-1.18,,-0.90,,-1.18,,-0.44,,-0.28,,-0.25,,+0.34,,-0.76):} Explain : (i) E^(@) value for copper is positive. (ii) E^(@) value of Mn is more negative as expected from the trend. (iii) Cr^(2+) is a stronger reducing agent than Fe^(2+) .

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    If E^(@) for copper electrode is +0.34 V , will you calculate e.m.f, value when the solution in contact with it is 0.1 M in copper ions . How does e.m.f. for copper electrode change when concentration of Cu^(2+) ion in the solution is decreased ?

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