Write the cell reaction and calculate the standard E° of the cell:

Zn | Zn²⁺ (1M) || Cd²⁺ (1M) | Cd

Given:
E°(Zn/Zn²⁺) = 0.763 V
E°(Cd/Cd²⁺) = 0.403 V

Reverse the signs of oxidation potential to get the values of the reduction or electrode potentials.
E(Zn,Zn²⁺) = 0.763 V ∴ E°(Zn²⁺/Zn) = 0.763 V
E(Cd,Cd²⁺) = 0.403 V ∴ E°(Cd²⁺/Cd) = −0.403 V

Since Zn²⁺/Zn electrode is at lower potential, it acts as the anode, while Cd²⁺/Cd electrode with higher potential acts as the cathode.
Zn loses electrons and Cd²⁺ ions accept them.

Cell reaction:
Zn + Cd²⁺ ⟶ Zn²⁺ + Cd

Standard cell potential:
E°_cell = E°(Cd) − E°(Zn²⁺/Zn)
= −0.403 − (−0.763)
= +0.360 V