Calculate the EMF of the cell containing chromium and cadmium electrodes (Given `E_(Cr^(3+)//Cr)^(@)=-0.74V,E_(Cd^(2+)//Cd)^(@)=-0.40V`)

Answer

To calculate the EMF (Electromotive Force) of the cell containing chromium and cadmium electrodes, we will follow these steps:

### Step 1: Identify the half-reactions
We need to determine which half-reaction is the reduction and which is the oxidation. The given standard electrode potentials are:
- \( E^\circ_{Cr^{3+}/Cr} = -0.74 \, V \)
- \( E^\circ_{Cd^{2+}/Cd} = -0.40 \, V \)

Since the more positive value corresponds to the reduction potential, cadmium (Cd) will undergo reduction, and chromium (Cr) will undergo oxidation.

### Step 2: Write the half-reactions
The half-reactions can be written as:
- Reduction half-reaction (cathode):
\[ Cd^{2+} + 2e^- \rightarrow Cd \]
- Oxidation half-reaction (anode):
\[ Cr \rightarrow Cr^{3+} + 3e^- \]

### Step 3: Determine the EMF of the cell
The EMF of the cell can be calculated using the formula:
\[ E_{cell} = E_{cathode} - E_{anode} \]

Here, the cathode is where reduction occurs (Cd), and the anode is where oxidation occurs (Cr).

### Step 4: Substitute the values
Substituting the values from the standard electrode potentials:
- \( E_{cathode} = E^\circ_{Cd^{2+}/Cd} = -0.40 \, V \)
- \( E_{anode} = E^\circ_{Cr^{3+}/Cr} = -0.74 \, V \)

Now, substituting these into the EMF equation:
\[ E_{cell} = (-0.40 \, V) - (-0.74 \, V) \]
\[ E_{cell} = -0.40 \, V + 0.74 \, V \]
\[ E_{cell} = 0.34 \, V \]

### Final Answer
The EMF of the cell is \( 0.34 \, V \).

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