What is the EMF of the following cell?

\(Fe|Fe^{2+} (0.6M) || Sn^{2+}(0.2M)|Sn\); \(E^o_{Fe^{2+}/Fe} = 0.44 V\), \(E^o_{Sn^{2+}/Sn} = 0.14 V\)

0.566 V

The correct answer is option 1. 0.566 V.

To determine the EMF (electromotive force) of the given cell, we can use the Nernst equation. The cell reaction can be represented as:

\(\text{Fe} \, | \, \text{Fe}^{2+} (0.6\, \text{M}) \, || \, \text{Sn}^{2+} (0.2\, \text{M}) \, | \, \text{Sn}\)

The standard reduction potentials are given:

\( E^\circ_{\text{Fe}^{2+}/\text{Fe}} = -0.44 \, \text{V} \)

\( E^\circ_{\text{Sn}^{2+}/\text{Sn}} = 0.14 \, \text{V} \)

Reduction half-reaction (cathode):

\(\text{Sn}^{2+} + 2e^- \rightarrow \text{Sn} \quad E^\circ_{\text{Sn}^{2+}/\text{Sn}} = 0.14 \, \text{V}\)

Oxidation half-reaction (anode):

\(\text{Fe} \rightarrow \text{Fe}^{2+} + 2e^- \quad E^\circ_{\text{Fe}^{2+}/\text{Fe}} = -0.44 \, \text{V}\)

\(E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \)

\( E^\circ_{\text{cell}} = 0.14 \, \text{V} - (-0.44 \, \text{V}) = 0.14 \, \text{V} + 0.44 \, \text{V} = 0.58 \, \text{V} \)

Therefore, 

\(E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q  ----(1)\)

where \( n \) is the number of electrons transferred (2 in this case) and \( Q \) is the reaction quotient.

\(Q = \frac{[\text{Fe}^{2+}]}{[\text{Sn}^{2+}]} = \frac{0.6}{0.2} = 3 \)

Putting the values in equation (1), we get

\(E_{\text{cell}} = 0.58 \, \text{V} - \frac{0.0591}{2} \log 3 \)

\(E_{\text{cell}} = 0.58 \, \text{V} - 0.02955 \log 3 \)

Calculate the logarithm value: \(\log 3 \approx 0.477 \)

Calculate the correction term: \(0.02955 \times 0.477 \approx 0.0141\)

The EMF will be:

\(E_{\text{cell}} = 0.58 \, \text{V} - 0.0141 \, \text{V} \approx 0.566 \, \text{V}\)