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Chemistry Question on Galvanic Cells

In a cell, the following reactions take place
Fe2+→Fe3++e−Fe^{2+} → Fe^{3+} + e^-
E°Fe3+/Fe2+=0.77 VE°_{Fe^{3+}/Fe^{2+}} = 0.77\ V
2I−→I2+2e−2I^- → I_2 + 2e^-
EI2/I−=0.54 VE_{I_2/I^-} = 0.54\ V
The standard electrode potential for the spontaneous reaction in the cell is x×10–2Vx×10^{–2} V at 208 K208\ K. The value of xx is _______. (Nearest Integer)

Answer

E°cell=E°cathode−E°anodeE°_{cell} = E°_{cathode} - E°_{anode}
E°cell=0.77–0.54E°_{cell}= 0.77 – 0.54
E°cell=0.23 VE°_{cell}= 0.23\ V
E°cell=23×10–2VE°_{cell}= 23×10^{–2} V

So, the answer is 2323.