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The standard reduction potential of Pb a...

The standard reduction potential of Pb and Zn electrodes are -0.12 6 and -0.763 volts respectively . The e.m.f of the cell
`Zn|Zn^(2+)(0.1M)||Pb^(2+)(1M)Pb` is

A

0.637V

B

`lt0.637V`

C

`gt0.637V`

D

`0.889`

Text Solution

Verified by Experts

The correct Answer is:
C
`E_("cell")^(@)=E_("cathode")^(@)-E_("anode")^(@)`
The two half cell reactions and overall reaction are
`ZnrarrZn^(2+)+2e^(-)`
`Pb^(2+)+2e^(-)rarrPb`
`Zn+Pb^(2+)rarrZn^(2+)+Pb`
`E_("cell")^(@)=E_("cell")^(@)-(0.0591V)/(n)"log"([Zn^(2+)])/([Pb^(2+)])`
Here `E_("cell")^(@)=E_((Pb//Pb))^(@)-E_((Zn^(2+)//Zn))^(@)`
`=-0.126V-(-0.763V)`
`=0.763V-0.126V=0.637V`
Here n=2
`[Zn^(2+)]=0.1M`
`[Pb^(2+)]=1.0M`
`therefore=0.637V-(0.0591V)/(2)"log"(0.1)/(1.0)`
`=0.637V+(0.0591V)/(2)gt0.637V`
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