The standard reduction potential for \(Sn^{4+}/Sn^{2+}\) is \(+ 0.15 V\) and for \(Cr^{3+}/Cr\) is \(- 0.74 V\). These two half-cells, coupled in their standard states, are connected to make a cell. The galvanic cell can be correctly represented by:

\(Cr(s)|Cr^{3+}(aq)||Sn^{4+} (aq) | Sn^{2+}(aq)\)

The correct answer is option 3. \(Cr(s)|Cr^{3+}(aq)||Sn^{4+} (aq) | Sn^{2+}(aq)\).

Let us break down the explanation:

In a galvanic cell, oxidation occurs at the anode and reduction occurs at the cathode. The half-cell with the higher standard reduction potential (more positive) is considered the cathode, where reduction occurs, while the half-cell with the lower standard reduction potential (less positive) is considered the anode, where oxidation occurs.

Given:

Standard reduction potential for \(Sn^{4+}/Sn^{2+}\) is \(+0.15 \, \text{V}\).

Standard reduction potential for \(Cr^{3+}/Cr\) is \(-0.74 \, \text{V}\).

1. We identify that \(Sn^{4+}/Sn^{2+}\) has the higher standard reduction potential (\(+0.15 \, \text{V}\)), making it the cathode half-reaction.

2. \(Cr^{3+}/Cr\) has the lower standard reduction potential (\(-0.74 \, \text{V}\)), indicating it undergoes oxidation and therefore should be the anode half-reaction.

To represent the galvanic cell correctly:

The cathode half-reaction (\(Sn^{4+}(aq) + 2e^- \rightarrow Sn^{2+}(aq)\)) is written on the left side.

The anode half-reaction (\(Cr(s) \rightarrow Cr^{3+}(aq) + 3e^-\)) is written on the right side.

So, the correct representation of the galvanic cell is:

\[ Cr(s)|Cr^{3+}(aq)||Sn^{4+}(aq) | Sn^{2+}(aq) \]

This arrangement ensures that electrons flow from the anode (Cr) to the cathode (Sn), as expected in a galvanic cell.