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A half-cell consisting of palladium rod dipping into a $1\ M$ $\ce{Pd(NO_3)_2}$ solution is connected with a standard hydrogen half-cell. The cell voltage is $0.99\ volt$ and the platinum electrode in the hydrogen half-cell is the anode. Determine $E^ο$ for the reaction $\ce{Pd\longrightarrow Pd^{2+} + 2e}$.

Now the question is: can I find a numerical value of the needed $E^ο$, if I can't use half-cell potential of $\ce{Pt^{2+}\!−\!Pt}$?

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    $\begingroup$ You have the overall cell voltage, so you should be able to leave the E0 as unknown and solve for it. $\endgroup$ – jonsca Aug 14 '13 at 10:42
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    $\begingroup$ Careful, in the standard hydrogen electrode the redox reaction is not $Pt^{2+} + 2e^- -> Pt$. $\endgroup$ – bobthechemist Aug 14 '13 at 14:17

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