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(B) The cell reaction is $Fe^{2+} + Ag^{+} \rightarrow Fe^{3+} + Ag$.This can be split into two half-reactions:Oxidation: $Fe^{2+} \rightarrow Fe^{3+}

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$x + 2y - 3z$

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(B) The cell reaction is $Fe^{2+} + Ag^{+} \rightarrow Fe^{3+} + Ag$.This can be split into two half-reactions:Oxidation: $Fe^{2+} \rightarrow Fe^{3+} + e^-$Reduction: $Ag^+ + e^- \rightarrow Ag$We are given:$(1)$ $Ag^+ + e^- \rightarrow Ag, \ E^0 = x \ V, \ \Delta G_1^0 = -F(x)$$(2)$ $Fe^{2+} + 2e^- \rightarrow Fe, \ E^0 = y \ V, \ \Delta G_2^0 = -2F(y)$$(3)$ $Fe^{3+} + 3e^- \rightarrow Fe, \ E^0 = z \ V, \ \Delta G_3^0 = -3F(z)$We need the potential for $Fe^{2+} \rightarrow Fe^{3+} + e^-$. Let this be $E^0_{ox}$.Using Hess's Law for $\Delta G^0$:$\Delta G^0(Fe^{2+}$ $\rightarrow Fe^{3+} + e^-) = \Delta G^0(Fe^{2+}$ $\rightarrow Fe) - \Delta G^0(Fe^{3+}$ $\rightarrow Fe)$$-1F(E^0_{ox}) = -2F(y) - (-3F(z))$$-F(E^0_{ox}) = -2Fy + 3Fz$$E^0_{ox} = 2y - 3z$Now,$E^0_{cell} = E^0_{red} + E^0_{ox} = x + (2y - 3z) = x + 2y - 3z$.

For the given cell reaction $Fe^{2+}_{(aq)} + Ag^{+}_{(aq)} \rightarrow Fe^{3+}_{(aq)} + Ag_{(s)}$,the standard cell potential is:
Given:
$Ag^{+} + e^{-} \rightarrow Ag \quad E^0 = x \ V$
$Fe^{2+} + 2e^{-} \rightarrow Fe \quad E^0 = y \ V$
$Fe^{3+} + 3e^{-} \rightarrow Fe \quad E^0 = z \ V$

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For the given cell reaction $Fe^{2+}_{(aq)} + Ag^{+}_{(aq)} \rightarrow Fe^{3+}_{(aq)} + Ag_{(s)}$,the standard cell potential is:Given:$Ag^{+} + e^{-} \rightarrow Ag \quad E^0 = x \ V$$Fe^{2+} + 2e^{-} \rightarrow Fe \quad E^0 = y \ V$$Fe^{3+} + 3e^{-} \rightarrow Fe \quad E^0 = z \ V$