$2C{H_3}C{H_2}C{H_2}Br + 2Na\xrightarrow{{ether}}$ $C{H_3}{(C{H_2})_4}C{H_3} + 2NaBr$
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$Sn ^{2+}+2 e ^{-} \rightarrow Sn$
$Sn ^{4+}+4 e ^{-} \rightarrow Sn$
The electrode potentials are; $E _{ Sn ^{2+} / Sn }^{\circ}=-0.140\, V$ and $E _{ Sn ^{4+} / Sn }^{\circ}=0.010\, V$. The magnitude of standard electrode potential for $Sn ^{4+} / Sn ^{2+}$ i.e. $E _{ Sn ^{4+} / Sn ^{2+}}^{\circ}$ is $.....\times 10^{-2}\, V$. (Nearest integer)
Above conversion can be achieved by