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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:
$Ag^{+} + e^{-} \rightarrow Ag \quad E^0 = x \ V$
$Fe^{2+} + 2e^{-} \rightarrow Fe \quad E^0 = y \ V$
$Fe^{3+} + 3e^{-} \rightarrow Fe \quad E^0 = z \ V$

Similar Questions

Given the standard electrode potentials $E^{\circ}_{Fe^{3+}/Fe^{2+}} = +0.77 \ V$ and $E^{\circ}_{Sn^{2+}/Sn} = -0.14 \ V$,calculate the standard cell potential $E^{\circ}_{cell}$ for the reaction: $Sn_{(s)} + 2Fe^{3+}_{(aq)} \rightarrow 2Fe^{2+}_{(aq)} + Sn^{2+}_{(aq)}$ (in $V$)

The standard electrode potential for the two electrodes $A^{+}/A$ and $B^{+}/B$ are respectively $0.5 \ V$ and $0.75 \ V$. The $EMF$ of the given cell $A | A^{+}(a = 1) || B^{+}(a = 1) | B$ will be $.......... \ V$. (in $.25$)

The $emf$ of a galvanic cell constituted with the electrodes $Zn^{2+} | Zn$ $(-0.76 \ V)$ and $Fe^{2+} | Fe$ $(-0.41 \ V)$ is

Give the symbolic representation of the following half-cells (electrodes):
$(i)$ $2H^{+}_{(aq)} + 2e^- \to H_{2_{(g)}}$
$(ii)$ $Br_{2_{(aq)}} + 2e^- \to 2Br^{-}_{(aq)}$
$(iii)$ $2Br^{-}_{(aq)} \to Br_{2_{(aq)}} + 2e^-$

$E^{\circ}$ of $Cu$ is $+0.34 \ V$ while that of $Zn$ is $-0.76 \ V$. Explain.

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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$