Explanation:
Let us assume that the solution contains both [tex]Fe^{2+}[/tex] and [tex]Fe^{3+}[/tex] which has a cell voltage of 0.719 V.
Therefore, voltage of cell contains both Ag/AgCl reference electrode where [tex]Fe^{2+}/Fe^{3+}[/tex] electrode is 0.719 V.
As, [tex]E_{cathode} - E_{anode}[/tex] = 0.719 V
It is known that potential of the silver-silver chloride reference electrode is 0.197 V.
Hence, [tex]E_{anode} [/tex] = 0.197 V. Now, calculate [tex]E_{cathode} [/tex] as follows.
[tex]E_{cathode} - E_{anode}[/tex] = 0.719 V
[tex]E_{cathode} - 0.197 V[/tex] = 0.719 V
[tex]E_{cathode}[/tex] = 0.916 V
Now, voltage of the cell that contains both calomel reference electrode and [tex]Fe^{2+}/Fe^{3+}[/tex] electrode as follows.
[tex]E_{anode}[/tex] = calomel electrode = 0.241 V
Voltage of cell = [tex]E_{cathode} - E_{anode}[/tex]
= 0.916 V - 0.241 V
= 0.675 V
Thus, we can conclude that 0.675 V is the new voltage.