A student writes the cell Pt(s) mid Fe²⁺(aq), Fe³⁺(aq) ∥ Cd²⁺(aq) mid Cd(s) and uses E°(Fe³⁺/Fe²⁺)…

Step 1. As written. $E^\circ_{cell} = E^\circ_{right} - E^\circ_{left} = -0.40\ \text{V} - 0.77\ \text{V} = -1.17\ \text{V}$.

Step 2. What the sign says. A negative emf means the reaction does not run from left to right. Iron(III) cannot oxidise iron(II) any further here, and cadmium ions cannot be reduced by iron(II).

Step 3. The cell that runs. Put cadmium, the lower couple, on the left: $\text{Cd}(s) \mid \text{Cd}^{2+}(aq) \parallel \text{Fe}^{3+}(aq), \text{Fe}^{2+}(aq) \mid \text{Pt}(s)$. Its emf is $0.77\ \text{V} - (-0.40\ \text{V}) = 1.17\ \text{V}$.