## Chemistry: The Molecular Science (5th Edition)

The more positive the value of E°, more easily the substance on the left-hand side of the half-reaction can be reduced. The more negative the value of E°, the more easily the substance on the right-hand side can be oxidised. But, in our case, $Co^{2+}$ is on the left-hand side of the half-reaction with E° value more negative than $Pt$ which is on the right-hand side of the half-reaction with more positive E° value. Therefore, $Pt$ cannot reduce $Co^{2+}$.