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Calculate Delta(r)G^(@) and emf (E) that...

Calculate `Delta_(r)G^(@)` and emf (E) that can be obtained from the following cell under the standard conditions at `25^(@)C? Zn(s)|Zn^(2+)(aq)||Sn^(2+)(aq)|Sn(s)`
`["Given : "E_(Zn^(2+)|Zn)^(@)=-0.76V , E_(Sn^(2+)|Sn)^(@)=-0.14V and F="96500 C.mol."^(-1)]`

Text Solution

Verified by Experts

`E_("cell")^(@)=E_("cathode")^(@)-E_("anode")^(@)=E_(Sn^(2)|Sn)^(@)-E_(Zn^(2+)|Zn)^(@)`
`=[-0.14-(-0.76)]V=0.62V`
`{:("Anode reaction : "Zn(s)rarrZn^(2+)(aq)+2e),("Cathode reaction :"Sn^(2+)(aq)+2erarrSn(s)),(bar("Cell reaction :"Zn(s)+Sn^(2+)(aq)rarrZn^(2+)(aq)+Sn(s)" ")):}`
So, for the cell reaction, n = 2
We have the relation `Delta_(r)G^(@)=-nFE_("cell")^(@)`. Substituting the values of n, F and `E_("cell")^(@)`, we get
`Delta_(r)G^(@)=-2xx96500xx0.62J=-119.66kJ`
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Knowledge Check

  • The standard free energy of the reaction Fe(s)+Sn^(2+)(aq, 1M)rarr Fe^(2+)(aq, 1M)+Sn(s) is ("given : at "25^(@)C, E_(Fe^(2+)|Fe)^(@)=-0.44V and E_(Sn^(2+)|Sn)^(@)=-0.14V)-

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