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

Explanation

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

**Step 1: Identify the half-reactions.**
The overall 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 (at the anode):
\[ 2Cr_{(s)} \rightarrow 2Cr^{3+}_{(aq)} + 6e^{-} \]
- Reduction (at the cathode):
\[ 3Cd^{2+}_{(aq)} + 6e^{-} \rightarrow 3Cd_{(s)} \]

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

**Step 3: Identify the anode and cathode.**
In a galvanic cell:
- The anode is where oxidation occurs (Cr in this case).
- The cathode is where reduction occurs (Cd in this case).

**Step 4: Use the formula for standard cell potential.**
The standard cell potential \( E^\circ_{cell} \) is calculated using the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]

Substituting the values:
\[ E^\circ_{cell} = E^\circ_{Cd^{2+}/Cd} - E^\circ_{Cr^{3+}/Cr} \]
\[ E^\circ_{cell} = (-0.04 \, V) - (-0.74 \, V) \]

**Step 5: Perform the calculation.**
\[ E^\circ_{cell} = -0.04 + 0.74 \]
\[ E^\circ_{cell} = 0.34 \, V \]

**Final Answer:**
The standard cell potential of the galvanic cell is \( 0.34 \, V \).

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