Calculate e.m.f. of the following cell at 298 K,
`2Cr(s)+3Fe^(2+)(0.1 M) to 2Cr^(3+)(0.01 M)+3Fe(s)`
`("Given" : E_((Cr^(3+)//Cr))^(@)=-0.74" V ", E_((Fe^(2+)//Fe))^(@)=-0.44" V ")`

Explanation

To calculate the electromotive force (e.m.f.) of the given electrochemical cell reaction at 298 K, we will follow these steps:

### Step 1: Identify the half-reactions and determine the anode and cathode
The given cell reaction is:
\[ 2Cr(s) + 3Fe^{2+}(0.1 \, M) \rightarrow 2Cr^{3+}(0.01 \, M) + 3Fe(s) \]

- **Oxidation half-reaction** (anode):
\[ 2Cr(s) \rightarrow 2Cr^{3+} + 6e^- \]

- **Reduction half-reaction** (cathode):
\[ 3Fe^{2+} + 6e^- \rightarrow 3Fe(s) \]

### Step 2: Determine the standard reduction potentials
From the problem statement, we have:
- \( E^\circ_{(Cr^{3+}/Cr)} = -0.74 \, V \)
- \( E^\circ_{(Fe^{2+}/Fe)} = -0.44 \, V \)

### Step 3: Calculate the standard cell potential \( E^\circ_{cell} \)
Using the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]

Substituting the values:
\[ E^\circ_{cell} = (-0.44 \, V) - (-0.74 \, V) \]
\[ E^\circ_{cell} = -0.44 + 0.74 \]
\[ E^\circ_{cell} = 0.30 \, V \]

### Step 4: Calculate the e.m.f. using the Nernst equation
The Nernst equation is given by:
\[ E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log \left( \frac{[products]}{[reactants]} \right) \]

Where:
- \( n \) = number of moles of electrons transferred in the balanced equation
- \( [products] \) = concentrations of products
- \( [reactants] \) = concentrations of reactants

### Step 5: Determine \( n \)
From the half-reactions:
- Chromium loses 6 electrons (2 Cr → 2 Cr³⁺)
- Iron gains 6 electrons (3 Fe²⁺ → 3 Fe)

Thus, \( n = 6 \).

### Step 6: Calculate the concentrations for the Nernst equation
- Concentration of products:
\[ [Cr^{3+}]^2 = (0.01)^2 = 0.0001 \]

- Concentration of reactants:
\[ [Fe^{2+}]^3 = (0.1)^3 = 0.001 \]

### Step 7: Substitute values into the Nernst equation
Now substituting into the Nernst equation:
\[ E_{cell} = 0.30 \, V - \frac{0.059}{6} \log \left( \frac{(0.01)^2}{(0.1)^3} \right) \]

Calculating the logarithm:
\[ \frac{(0.01)^2}{(0.1)^3} = \frac{0.0001}{0.001} = 0.1 \]
\[ \log(0.1) = -1 \]

Now substituting this back:
\[ E_{cell} = 0.30 \, V - \frac{0.059}{6} \times (-1) \]
\[ E_{cell} = 0.30 \, V + \frac{0.059}{6} \]
\[ E_{cell} = 0.30 \, V + 0.009833 \]
\[ E_{cell} \approx 0.3098 \, V \]

### Final Step: Round the answer
Rounding to two decimal places, we get:
\[ E_{cell} \approx 0.31 \, V \]

### Summary
The e.m.f. of the cell at 298 K is approximately **0.31 V**.

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