Given standard reduction potentials:
$E^\circ(\text{Sn}^{4+}/\text{Sn}^{2+}) = +0.15\text{ V}$
$E^\circ(\text{Cr}^{3+}/\text{Cr}) = -0.74\text{ V}$

• The electrode with the lower standard reduction potential acts as the anode (undergoes oxidation): $\text{Cr}(s) \rightarrow \text{Cr}^{3+}(aq) + 3e^-$.
• The electrode with the higher reduction potential acts as the cathode (undergoes reduction): $\text{Sn}^{4+}(aq) + 2e^- \rightarrow \text{Sn}^{2+}(aq)$.
• Cell convention: $\text{Anode} | \text{Anodic electrolyte} || \text{Cathodic electrolyte} | \text{Cathode}$.