Test Index

Electrochemistry

Section: Chemistry

Question : 8 of 30

Marks: +1, -0

298\ \mathrm{K}

\mathrm{Pb}^{2+}/\mathrm{Pb}\;-0.13\ \mathrm{V}

\mathrm{Ni}^{2+}/\mathrm{Ni}\;-0.24\ \mathrm{V}

\mathrm{Cd}^{2+}/\mathrm{Cd}\;-0.40\ \mathrm{V}

\mathrm{Fe}^{2+}/\mathrm{Fe}\;-0.44\ \mathrm{V}

0.001\ \mathrm{M}

\mathrm{X}^{2+}

0.1\ \mathrm{M}

\mathrm{Y}^{2+}

\mathrm{X}

\mathrm{Y}

298\ \mathrm{K}

\mathrm{X}

\mathrm{X}

\mathrm{Y}

\mathrm{R}=8.314\ \mathrm{J}\ \mathrm{K}^{-1}\ \mathrm{mol}^{-1}

\mathrm{F}=96500\ \mathrm{C}\ \mathrm{mol}^{-1}

Solution:

\mathrm{X}(\mathrm{s}) \rightarrow \mathrm{X}^{+2}(0.001\ \mathrm{M})+2\mathrm{e}^{-}

(anode)

\mathrm{Y}^{+2}(0.1\ \mathrm{M})+2\mathrm{e}^{-} \rightarrow \mathrm{Y}(\mathrm{s})

(cathode)

──────────────────────

E_{\;\text{cell}\;}=E_{\;\text{cell}\;}^{o}-\;\frac{0.06}{2}\log\;\frac{\mathrm{X}^{+2}}{\mathrm{Y}^{+2}}

E_{\;\text{cell}\;}=E_{\;\text{cell}\;}^{\circ}+0.06

(a)

\mathrm{Cd}(\mathrm{X})

and

\mathrm{Ni}(\mathrm{Y})\quad E_{\;\text{cell}\;}^{\circ}=+0.4-0.24 ;\quad E_{\;\text{cell}\;}=0.22

(b)

\mathrm{Cd}(\mathrm{X})

and

\mathrm{Fe}(\mathrm{Y})\quad E_{\;\text{cell}\;}^{c}=-0.04 ;\quad E_{\;\text{cell}\;}=0.02

(c)

\mathrm{Ni}(\mathrm{X})

and

\mathrm{Pb}

(Y)

E_{\;\text{cell}\;}^{\circ}=0.11 ;\quad E_{\;\text{cell}\;}=0.17

(d)

\mathrm{Ni}(\mathrm{X})

and

\mathrm{Fe}

(Y)

E_{\;\text{cell}\;}^{\circ}=-0.2 ;\quad E_{\;\text{cell}\;}=-0.14

Since in (a) (b) (c), Ecell is positive, hence answer is (a) (b) (c).

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