# Voltage applied when charging a galvanic cell

A working galvanic cell is constructed using an iron electrode in an iron(II) nitrate solution and a silver electrode in a silver(I) nitrate solution. When the cell runs down, it can be charged up again. Which of the following scenarios will occur during the charging process?

$\ce{[Eº(Ag+(aq)/Ag(s)) = +0.80 V and Eº(Fe^2+(aq)/Fe(s) = –0.41 V]}$

A. A minimum of 1.21 volts is applied to the cell and electrons move from the silver electrode to the iron electrode.

B. A minimum of 1.21 volts is applied to the cell and electrons move from the iron electrode to the silver electrode.

C. A minimum of 0.39 volts is applied to the cell and electrons move from the silver electrode to the iron electrode.

D. A minimum of 0.39 volts is applied to the cell and electrons move from the iron electrode to the silver electrode.

E. A minimum of 2.01 volts is applied to the cell and electrons move from the silver electrode to the iron electrode.

I know reduction of silver is spontaneous and oxidation of iron is spontaneous from the values given, so reduction occurs at the silver electrode and oxidation occurs at the iron electrode, so electrons flow from iron to silver.

In the reverse reaction, electrons will now flow from silver to iron. However, I'm unsure about the stoichiometry of the electron transfer occuring during charging:

$\ce{2Ag -> 2Ag+ + 2e- -1.60V}$

$\ce{Fe^2+ + 2e- -> Fe -0.41V}$

I multiplied the oxidation of silver by 2, as two electrons must be supplied for iron to reduce, hence getting -2.01V, and answer E. However, the solutions state A, which suggests the oxidation of silver was no multiplied by 2. Why is this?