Given at $298 \ K$: $E^\ominus_{Fe^{2+}/Fe} = X \ V$; $E^\ominus_{Fe^{3+}/Fe} = Y \ V$. The $E^\ominus_{Fe^{3+}/Fe^{2+}}$ in Volt at $298 \ K$ is given by:

  • A
    $2X - 3Y$
  • B
    $3Y - 2X$
  • C
    $3Y + 2X$
  • D
    $Y + X$

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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 relation between $\Delta G$ and $E$ for a cell is $\Delta G = -nFE$. The cell reaction will be spontaneous if:

What is the $e.m.f.$ of the cell in $V$?
$Cr | Cr^{3+} (1.0 \ M) || Co^{2+} (1.0 \ M) | Co$
Given: $E^{o}$ for $Cr^{3+} | Cr = -0.74 \ V$ and $Co^{2+} | Co = -0.28 \ V$

What is $emf$?

Standard reduction potentials of the half-reactions are given below:
$F_{2(g)} + 2e^- \rightarrow 2F^-_{(aq)}$; $E^o = +2.85 \ V$
$Cl_{2(g)} + 2e^- \rightarrow 2Cl^-_{(aq)}$; $E^o = +1.36 \ V$
$Br_{2(l)} + 2e^- \rightarrow 2Br^-_{(aq)}$; $E^o = +1.06 \ V$
$I_{2(s)} + 2e^- \rightarrow 2I^-_{(aq)}$; $E^o = +0.53 \ V$
The strongest oxidising and reducing agents respectively are:

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