A galvanic cell consists of one half cell that is made up of an inert graphite electrode in a solution containing 1.0 M Fe$^{2+}$ (aq) and 1.0 M Fe$^{3+}$ (aq) at 25°C - VCE - SSCE Chemistry - Question 8 - 2005 - Paper 1

To determine which option provides a positive polarity for the electrode in the second half cell, we need to identify the half-reaction that would occur.

1. Understanding the Half Cell Reactions: The existing half cell contains a reduction reaction involving Fe$^{3+}$ being reduced to Fe$^{2+}$:

   $$$$

ightarrow Fe^{2+}$$
This reaction has a positive electrode potential because it involves reduction.

2. Analyzing the Options:
   • A. Lead in Pb$^{2+}$: The reduction of Pb$^{2+}$ to Pb is less favorable compared to Ag.
   • B. Silver in Ag$^{+}$: The half-reaction is: $$$$

ightarrow Ag$$ This reaction has a higher standard reduction potential and can serve as a positive electrode when coupled with the Fe$^{3+}$/Fe$^{2+}$ half cell.

• C. Iron in Fe$^{2+}$: This does not provide positive polarity as it can only oxidize, not reduce with the existing ions.
• D. Inert Graphite in Br$^{-}$: This does not involve metals in favorable redox potential for galvanic cells.

3. Conclusion: The silver electrode in a 1.0 M Ag$^{+}$ solution can effectively be used as the second half cell, ensuring that the polarity remains positive.