Titration of 25.00 mL of 0.050 M Sn2+ with 0.100 M Fe3+ in 1 M HCl to give Fe2+ and Sn4+ using Pt and Ag/AgCl electrodes.
The balanced titration reaction is follows:

Sn2+ + 2Fe3+ ⇌ Sn4+ + 2Fe2+

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! = 0.732 V; �!"!!
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! = 0.139 V

3.1.Write two half-reactions for the indicator electrode.
3.2.Write two Nernst equations for the cell voltage
3.3. Calculate electrical potential (E) at the following volumes of Fe3+: 1.0, 25.0, and 26.0 mL.