Two new metals, X and Y, are tested against cadmium, for which E°(Cd²⁺/Cd) = -0.40 V…

Step 1. Cadmium and X. Cadmium is negative, so X is the positive electrode: $E^\circ(\text{X}) = -0.40\ \text{V} + 0.65\ \text{V} = 0.25\ \text{V}$.

Step 2. X and Y. Y is negative, so X is positive. $E^\circ(\text{Y}) = 0.25\ \text{V} - 0.55\ \text{V} = -0.30\ \text{V}$.

Step 3. Cadmium and Y. Y has the higher potential, $-0.30\ \text{V}$ against $-0.40\ \text{V}$, so Y is the positive electrode. The emf is $-0.30\ \text{V} - (-0.40\ \text{V}) = 0.10\ \text{V}$.