Calculate the e.m.f. of the following cell at 298K:
`2Cr(s)+3Fe^(2+)(0.1M)to2Cr^(3+)(0.01M)+3Fe(s)`
Given: `E_((Cr^(3+)//Cr))^(@)=-0.74V,E_((Fe^(2+)//Fe))^(@)=-0.44V`.

Explanation

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

### Step 1: Write the half-reactions
The overall cell reaction is:
\[ 2Cr(s) + 3Fe^{2+}(0.1M) \rightarrow 2Cr^{3+}(0.01M) + 3Fe(s) \]

The half-reactions are:
1. Oxidation (anode):
\[ 2Cr(s) \rightarrow 2Cr^{3+} + 6e^- \]
2. Reduction (cathode):
\[ 3Fe^{2+} + 6e^- \rightarrow 3Fe(s) \]

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

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

Here, the cathode is the reduction of \( Fe^{2+} \) and the anode is the oxidation of \( Cr \):
\[ E^\circ_{cell} = (-0.44 \, V) - (-0.74 \, V) \]
\[ E^\circ_{cell} = -0.44 + 0.74 \]
\[ E^\circ_{cell} = 0.30 \, V \]

### Step 4: Calculate the reaction quotient \( Q \)
The reaction quotient \( Q \) for the reaction is given by:
\[ Q = \frac{[Cr^{3+}]^2}{[Fe^{2+}]^3} \]

Substituting the concentrations:
\[ Q = \frac{(0.01)^2}{(0.1)^3} \]
\[ Q = \frac{0.0001}{0.001} \]
\[ Q = 0.1 \]

### Step 5: Use the Nernst equation to calculate the cell potential at non-standard conditions
The Nernst equation is:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q \]

Where \( n \) is the number of moles of electrons transferred, which is 6 in this case.

Substituting the values:
\[ E_{cell} = 0.30 - \frac{0.0591}{6} \log(0.1) \]

Calculating \( \log(0.1) \):
\[ \log(0.1) = -1 \]

Now substituting this back into the equation:
\[ E_{cell} = 0.30 - \frac{0.0591}{6} \times (-1) \]
\[ E_{cell} = 0.30 + \frac{0.0591}{6} \]
\[ E_{cell} = 0.30 + 0.00985 \]
\[ E_{cell} = 0.30985 \, V \]

### Step 6: Round the final answer
Rounding to two decimal places:
\[ E_{cell} \approx 0.31 \, V \]

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

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