What is the cell potential of a cell in which the following reaction occurs. `Ni(s)+Cu^(2+)(1M)rarrSn^(2+)(1M)|Fe` `E_(Ni)//Ni)^(2+)=-0.25V.E_(Cu)^(@+
What is the cell potential of a cell in which the following reaction occurs.
`Ni(s)+Cu^(2+)(1M)rarrSn^(2+)(1M)|Fe`
`E_(Ni)//Ni)^(2+)=-0.25V.E_(Cu)^(@+)=0.34V`
A. 0.295V
B. `+0.59V`
C. `0.885V`
D. `0.442V`.
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Correct Answer - B
Anode half reaction: `Ni(s)rarrNi^(2+)+2e^(-)`
Cathode half reaction : `Cu^(2+)aq+2e^(-)rarrCu(s)`
`therefore `The cell is represented as
`Ni|Ni^(2+)(1M)||Cu^(2+)(1M)|Cu(s)`
`E^(@)=E_("red"("cathode"))^(@)-E_("red"("anode"))^(@)`
`=0.34V-(-0.25)V=+0.59V`
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