Question

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:

• A. Sn$^{2+}$(aq) | Sn$^{4+}$(aq) || Cr$^{3+}$(aq) | Cr(s)
• B. Sn$^{4+}$(aq) | Sn$^{2+}$(aq) || Cr$^{3+}$(aq) | Cr(s)
• C. Cr(s) | Cr$^{3+}$(aq) || Sn$^{4+}$(aq) | Sn$^{2+}$(aq)
• D. Cr(s) | Cr$^{3+}$(aq) | Sn$^{2+}$(aq) || Sn$^{4+}$(aq)

Hint:

In galvanic cells, the substance with the higher (more positive) reduction potential acts as the cathode, where reduction takes place. The substance with the lower (more negative) reduction potential is the anode, where oxidation occurs.

Answer 1

The Correct Option is C

The standard reduction potential for Sn$^{4+}$/Sn$^{2+}$ is +0.15 V and for Cr$^{3+}$/Cr is -0.74 V. The cell potential is calculated as: \[ \text{Cell Potential} = \text{E}^\circ_{\text{cathode}} - \text{E}^\circ_{\text{anode}} \] In this case, the cathode (where reduction occurs) will be the Sn$^{4+}$/Sn$^{2+}$ half-reaction, and the anode (where oxidation occurs) will be the Cr$^{3+}$/Cr half-reaction. Based on the reduction potential, Sn$^{4+}$ will be reduced and Cr will be oxidized.
Thus, the cell notation will be: \[ \text{Cr(s)} | \text{Cr}^{3+}(aq) || \text{Sn}^{4+}(aq) | \text{Sn}^{2+}(aq) \]