For the given reactions:Sn^{2+} + 2e^{-} righ | vedclass

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(C) Given:$(1) \ Sn^{2+} + 2e^{-}$ $ightarrow Sn, \ E^{\circ}_{1} = -0.140 \ V, \ \Delta G^{\circ}_{1} = -nFE^{\circ}_{1} = -2 	imes F 	imes (-0.1

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$16$

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(C) Given:$(1) \ Sn^{2+} + 2e^{-}$ $ightarrow Sn, \ E^{\circ}_{1} = -0.140 \ V, \ \Delta G^{\circ}_{1} = -nFE^{\circ}_{1} = -2 	imes F 	imes (-0.140) = +0.280F$$(2) \ Sn^{4+} + 4e^{-}$ $ightarrow Sn, \ E^{\circ}_{2} = 0.010 \ V, \ \Delta G^{\circ}_{2} = -nFE^{\circ}_{2} = -4 	imes F 	imes (0.010) = -0.040F$We need the potential for $Sn^{4+} + 2e^{-} ightarrow Sn^{2+}$.This can be obtained by $(2) - (1)$:$(Sn^{4+} + 4e^{-}) - (Sn^{2+} + 2e^{-}) ightarrow Sn - Sn$$Sn^{4+} + 2e^{-} ightarrow Sn^{2+}$$\Delta G^{\circ}_{3} = \Delta G^{\circ}_{2} - \Delta G^{\circ}_{1} = -0.040F - 0.280F = -0.320F$Since $\Delta G^{\circ}_{3} = -nFE^{\circ}_{Sn^{4+}/Sn^{2+}}$,where $n=2$:$-2FE^{\circ}_{Sn^{4+}/Sn^{2+}} = -0.320F$$E^{\circ}_{Sn^{4+}/Sn^{2+}} = 0.160 \ V = 16 	imes 10^{-2} \ V$The value is $16$.

Similar Questions

The standard electrode potentials are: $K^{+}/K = -2.93 \ V$,$Ag^{+}/Ag = 0.80 \ V$,$Hg^{2+}/Hg = 0.79 \ V$,$Mg^{2+}/Mg = -2.37 \ V$,$Cr^{3+}/Cr = -0.74 \ V$. Arrange these metals in the increasing order of their reducing power.

For the following cell,the standard electrode potential $E_{cell}^0$ is . . . . . . .
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$Zn | Zn^{2+} || Cu^{2+} | Cu$ (in $V$)

$I_2 + 2e^{-} \to 2I^{-}$; $E^{o} = 0.54 \ V$
$Cl_2 + 2e^{-} \to 2Cl^{-}$; $E^{o} = 1.36 \ V$
$Mn^{3+} + e^{-} \to Mn^{2+}$; $E^{o} = 1.50 \ V$
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Which of the following is a correct statement?

Which of the following metals will not release $H_{2(g)}$ upon reaction with an acid?

The $EMF$ of the cell $Ni | Ni^{2+} (1.0 \ M) || Au^{3+} (1.0 \ M) | Au$ is ............ $V$. (Given: $E^o_{Ni^{2+}/Ni} = -0.25 \ V$; $E^o_{Au^{3+}/Au} = 1.50 \ V$)

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