Calculate `E^(0-)` for each cell and write the equation for each cell precess. Explaing the significance of any negative. `E^(c-)` value.
`.` `Cd(s)|Cd^(2+)(1.0M)||AgNO_(3)(1.0M)|Ag(s)`
Given `E^(c-)._(Cd)=-0.40V,
``E^(c-)._(Ag)`=+0.80V

Explanation

To solve the problem, we need to calculate the standard cell potential \( E^{0}_{cell} \) for the given electrochemical cell and write the half-reaction equations for both the anode and cathode processes. We will also discuss the significance of any negative \( E^{0} \) value.

### Step 1: Identify the Anode and Cathode
In the given cell notation:
\[ \text{Cd(s)} | \text{Cd}^{2+}(1.0M) || \text{AgNO}_{3}(1.0M) | \text{Ag(s)} \]

- **Anode**: The left side (Cd) where oxidation occurs.
- **Cathode**: The right side (Ag) where reduction occurs.
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