A cell is written as Pt(s) mid Sn²⁺(aq), Sn⁴⁺(aq) ∥ Br⁻(aq) mid Br₂(l) mid Pt(s)…

Step 1. The anode. The left electrode holds tin(II) and tin(IV) ions. Tin(II) is oxidised: $\text{Sn}^{2+} \rightarrow \text{Sn}^{4+} + 2e^-$.

Step 2. The cathode. Bromine is reduced: $\text{Br}_2 + 2e^- \rightarrow 2\text{Br}^-$.

Step 3. The cell reaction. The electrons already balance. $\text{Sn}^{2+} + \text{Br}_2 \rightarrow \text{Sn}^{4+} + 2\text{Br}^-$, with $n = 2$.

Step 4. The emf. $E^\circ_{cell} = 1.09\ \text{V} - 0.15\ \text{V} = 0.94\ \text{V}$.