The standard potentials of two couples are E°(Sm³⁺/Sm²⁺) = -1.55 V and E°(Yb³⁺/Yb²⁺) = -1.05 V…

Step 1. The emf for samarium. The hydrogen ion is reduced, so it is the cathode. $E^\circ = 0.00\ \text{V} - (-1.55\ \text{V}) = 1.55\ \text{V}$.

Step 2. The emf for ytterbium. $E^\circ = 0.00\ \text{V} - (-1.05\ \text{V}) = 1.05\ \text{V}$.

Step 3. Gibbs energy. For samarium, $\Delta G^\circ = -1 \times 96.5\ \text{kJ/(V mol)} \times 1.55\ \text{V} = -149.6\ \text{kJ/mol}$. For ytterbium, $\Delta G^\circ = -96.5 \times 1.05 = -101.3\ \text{kJ/mol}$.