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What ratio of Pb^(2+) to Sn^(2+) concent...

What ratio of `Pb^(2+)` to `Sn^(2+)` concentration is needed to reverse the following cell reaction ?
`Sn(s)+Pb^(2+) (aq.) rarr Sn^(2+) (aq.)+Pb(s)`
`E_(Sn^(2+)//Sn)^(@)=-0.136` volt and `E_(Pb^(2+)//Pb)^(@)=-0.126` volt

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Verified by Experts

The correct Answer is:
`([Pb^(2+)])/([Sn^(2+)])=0.458`
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