At 298 K,some standard electrode potentials | vedclass

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(D) The cell reaction is: $X_{(s)} + Y^{2+}(0.1 \ M) \longrightarrow X^{2+}(0.001 \ M) + Y_{(s)}$According to the Nernst equation: $E_{	ext{cell}} =

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$A, B, C$

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(D) The cell reaction is: $X_{(s)} + Y^{2+}(0.1 \ M) \longrightarrow X^{2+}(0.001 \ M) + Y_{(s)}$According to the Nernst equation: $E_{	ext{cell}} = E^{\circ}_{	ext{cell}} - \frac{0.0591}{2} \log \frac{[X^{2+}]}{[Y^{2+}]}$Here,$\frac{[X^{2+}]}{[Y^{2+}]} = \frac{0.001}{0.1} = 0.01 = 10^{-2}$Therefore,$E_{	ext{cell}} = E^{\circ}_{	ext{cell}} - \frac{0.0591}{2} \log(10^{-2}) = E^{\circ}_{	ext{cell}} + 0.0591 \approx E^{\circ}_{	ext{cell}} + 0.06 \ V$Since $X$ undergoes dissolution,$X$ acts as the anode. For the reaction to be spontaneous,$E_{	ext{cell}} > 0$.$E^{\circ}_{	ext{cell}} = E^{\circ}_{Y^{2+}/Y} - E^{\circ}_{X^{2+}/X}$$(A) \ Cd$ and $Ni: E^{\circ}_{	ext{cell}} = -0.24 - (-0.40) = +0.16 \ V; E_{	ext{cell}} = +0.16 + 0.06 = +0.22 \ V > 0$$(B) \ Cd$ and $Fe: E^{\circ}_{	ext{cell}} = -0.44 - (-0.40) = -0.04 \ V; E_{	ext{cell}} = -0.04 + 0.06 = +0.02 \ V > 0$$(C) \ Ni$ and $Pb: E^{\circ}_{	ext{cell}} = -0.13 - (-0.24) = +0.11 \ V; E_{	ext{cell}} = +0.11 + 0.06 = +0.17 \ V > 0$$(D) \ Ni$ and $Fe: E^{\circ}_{	ext{cell}} = -0.44 - (-0.24) = -0.20 \ V; E_{	ext{cell}} = -0.20 + 0.06 = -0.14 \ V The reaction is spontaneous when $E_{	ext{cell}} > 0$. Thus,combinations $(A), (B),$ and $(C)$ are correct.

$Pb^{2+} / Pb$	$-0.13 \ V$
$Ni^{2+} / Ni$	$-0.24 \ V$
$Cd^{2+} / Cd$	$-0.40 \ V$
$Fe^{2+} / Fe$	$-0.44 \ V$

Similar Questions

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