Calculate the standard cell potential of galvanic cell in which the following reaction takes place
`2Cr_(s)+3Cd_(aq)^(+2)rarr2cr_(aq)^(+3)+3Cd_(s)`
Given `E_(Cr^(+3)//Cr)=-0.74(V)E^(@)_(Cd^(+2)//Cd)=-0.04(V)`

Options

A. 0.74V
B. 1.14V
C. 0.34V
D. `0.34V`

Answer

To calculate the standard cell potential of the galvanic cell for the given reaction:

**Step 1: Identify the half-reactions.**
The overall cell reaction is:
\[ 2Cr_{(s)} + 3Cd^{2+}_{(aq)} \rightarrow 2Cr^{3+}_{(aq)} + 3Cd_{(s)} \]

From this reaction, we can identify the oxidation and reduction half-reactions:
- **Oxidation (Anode):**
\[ 2Cr_{(s)} \rightarrow 2Cr^{3+}_{(aq)} + 6e^{-} \]
- **Reduction (Cathode):**
\[ 3Cd^{2+}_{(aq)} + 6e^{-} \rightarrow 3Cd_{(s)} \]

**Step 2: Write the standard reduction potentials.**
The standard reduction potentials given are:
- For the chromium half-reaction:
\[ E^\circ_{Cr^{3+}/Cr} = -0.74 \, V \]
- For the cadmium half-reaction:
\[ E^\circ_{Cd^{2+}/Cd} = -0.04 \, V \]

**Step 3: Determine the standard cell potential.**
The standard cell potential \( E^\circ_{cell} \) is calculated using the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]

Here, the cathode is where reduction occurs (Cd) and the anode is where oxidation occurs (Cr):
\[ E^\circ_{cell} = E^\circ_{Cd^{2+}/Cd} - E^\circ_{Cr^{3+}/Cr} \]

Substituting the values:
\[ E^\circ_{cell} = (-0.04 \, V) - (-0.74 \, V) \]

**Step 4: Simplify the equation.**
This simplifies to:
\[ E^\circ_{cell} = -0.04 \, V + 0.74 \, V \]
\[ E^\circ_{cell} = 0.74 \, V - 0.04 \, V \]
\[ E^\circ_{cell} = 0.70 \, V \]

**Step 5: Final answer.**
Thus, the standard cell potential of the galvanic cell is:
\[ E^\circ_{cell} = 0.70 \, V \]

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