Calculate the e.m.f. of the cell,
`Cr//Cr^(3+)(0.1 M) || Fe^(2+)(0.01 M)//Fe`
`"Given" : E_(Cr^(3+)//Cr)^(@)=-0.75" V ", E_(Fe^(2+)//Fe)^(@)=-0.45" V "`
`"Cell reaction" : 2Cr(s)+3Fe^(2+)(aq) to 2Cr^(3+)(aq)+3Fe(s)`
`{"Hint". E_(cell)=E_(cell)^(@)-(0.0591V)/(6)"log"([Cr^(3+)]^(2))/([Fe^(2+)]^(3)}`

Explanation

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

### Step 1: Identify the half-reactions and standard cell potential (E°cell)

The cell reaction is given as:
\[ 2 \text{Cr}(s) + 3 \text{Fe}^{2+}(aq) \rightarrow 2 \text{Cr}^{3+}(aq) + 3 \text{Fe}(s) \]

The standard reduction potentials are:
- For the chromium half-reaction:
\[ \text{Cr}^{3+} + 3e^- \rightarrow \text{Cr} \quad E^\circ = -0.75 \, \text{V} \]
- For the iron half-reaction:
\[ \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \quad E^\circ = -0.45 \, \text{V} \]

To find \( E^\circ_{cell} \), we use the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]

In this case, chromium is being oxidized (anode) and iron is being reduced (cathode):
\[ E^\circ_{cell} = E^\circ_{Fe} - E^\circ_{Cr} \]
\[ E^\circ_{cell} = (-0.45) - (-0.75) \]
\[ E^\circ_{cell} = -0.45 + 0.75 = 0.30 \, \text{V} \]

### Step 2: Apply the Nernst Equation

The Nernst equation is given by:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log \left( \frac{[\text{Cr}^{3+}]^2}{[\text{Fe}^{2+}]^3} \right) \]

Where:
- \( n \) is the number of moles of electrons transferred in the balanced equation. Here, \( n = 6 \) (2 Cr gives 6 electrons).
- The concentrations are given as:
- \([\text{Cr}^{3+}] = 0.1 \, \text{M}\)
- \([\text{Fe}^{2+}] = 0.01 \, \text{M}\)

### Step 3: Substitute values into the Nernst Equation

Substituting the values into the Nernst equation:
\[ E_{cell} = 0.30 - \frac{0.0591}{6} \log \left( \frac{(0.1)^2}{(0.01)^3} \right) \]

Calculating the logarithmic term:
\[ \frac{(0.1)^2}{(0.01)^3} = \frac{0.01}{0.000001} = 10000 \]
\[ \log(10000) = 4 \]

Now substituting this back into the equation:
\[ E_{cell} = 0.30 - \frac{0.0591}{6} \times 4 \]
\[ E_{cell} = 0.30 - \frac{0.0591 \times 4}{6} \]
\[ E_{cell} = 0.30 - \frac{0.2364}{6} \]
\[ E_{cell} = 0.30 - 0.0394 \]
\[ E_{cell} = 0.2606 \, \text{V} \]

### Step 4: Round the final answer

Rounding to two decimal places, we get:
\[ E_{cell} \approx 0.26 \, \text{V} \]

### Final Answer:
The e.m.f. of the cell is approximately **0.26 V**.

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