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2. Electrochemistry — Chemistry STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceChemistry2. Electrochemistry1 Mark
MCQ
Given :

$E^o_{Fe^{3+} /Fe} = -0.036\,V,$  $E^o _{Fe^{2+} /Fe} = -0.439\,V$

The value of standard electrode potential for the change,

$Fe^{3+} (aq) + e^- \rightarrow Fe^{2+} (aq)$ will be ........ $V$.

  • A
    $0.385$
  • ✓
    $0.770$
  • C
    $-0.270$
  • D
    $-0.072$
✓

Answer

Correct option: B.
$0.770$
b
$F e^{3+}+3 e^{-} \rightarrow F e, E^{o}_{F e^{3} 4 / F e}$

$=-0.036 V \ldots(i)$

$F e^{2+}+2 e^{-} \rightarrow F e, E^{\circ}_{F e^{2} / / F e}$

$=-0.439 V \ldots(i i)$

we have to calculate

$F e^{3+}+e^{-} \rightarrow F e^{2+}, \Delta G=?$

To obtain this equation subtract equ $( ii )$ from $( i )$ we get

$F e^{3+}+e^{-} \rightarrow F e^{2+} \ldots(i i i)$

As we know that $\Delta G=-n F E$

Thus for reaction ( iii)

$\Delta G=\Delta G_{1}-\Delta G$

$-n F E^{\circ}=-n F E_{1}-\left(-n F E_{2}\right)$

$-n F E^{\circ}=n F E_{2}-n F E_{1}$

$-1 F E^{\circ}=2 \times 0.439 F-3 \times 0.036 F$

$- F E^{\circ}=0.770 F$

$\therefore \quad E^{\circ}=-0.770 V$

$O^{-}>F^{-}>N a^{+}>M g^{++}>A l^{3+}$

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