Values of \(E^o\) reduction are given below:

\(Sn^{2+}|Sn = -0.14V\), \(Zn^{2+}|Zn = -0.76 V\), \(Cr^{3+}|Cr = -0.74V\), \(Mg^{2+}|Mg = -2.36 V\)

The ion which will get deposited at cathode in an electrochemical cell would be

\(Sn^{2+}\)

The correct answer is option 1. \(Sn^{2+}\).

In electrochemistry, the reduction potential (\(E^\circ\)) measures the tendency of a species to gain electrons and undergo reduction. The more positive the \(E^\circ\) value, the greater the tendency for reduction to occur.

In an electrochemical cell, reduction occurs at the cathode (the electrode where reduction takes place), while oxidation occurs at the anode (the electrode where oxidation takes place). The species that gets reduced at the cathode is the one with the most positive standard reduction potential (\(E^\circ\)). This is because it has the highest tendency to gain electrons and undergo reduction.

Let us analyze the given options:

1. \(Sn^{2+} \rightarrow Sn\) with \(E^\circ = -0.14 \, \text{V}\)

2. \(Zn^{2+} \rightarrow Zn\) with \(E^\circ = -0.76 \, \text{V}\)

3. \(Cr^{3+} \rightarrow Cr\) with \(E^\circ = -0.74 \, \text{V}\)

4. \(Mg^{2+} \rightarrow Mg\) with \(E^\circ = -2.36 \, \text{V}\)

Among these options, \(Sn^{2+}\) has the least negative \(E^\circ\) value (-0.14 V), indicating the highest tendency to gain electrons and undergo reduction. Therefore, \(Sn^{2+}\) will be reduced and deposited at the cathode in the electrochemical cell.

Hence, the correct answer is option 1 \(Sn^{2+}\).