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Use standard reduction potentials to calculate the standard free energy change in $kJ$ for the reaction:

$C{o}^{2+}(aq)+Cd(s)Co(s)+C{d}^{2+}(aq)$

$C{o}^{2+}(aq)+2e^{-}Co(s)E_{\text{r}\text{e}\text{d}}^o=-0\mathrm{.}280V$

$C{d}^{2+}(aq)+2e^{-}Cd(s)E_{\text{r}\text{e}\text{d}}^0=-0\mathrm{.}403V$

$G=,kJ$

$K$ for this reaction would be

than one.