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STD 12 - 2. Electrochemistry — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 2. Electrochemistry4 Marks
MCQ
For the electrochemical cell, $M|{M^ + }||{X^ - }|X,$ ${E^o}({M^ + }/M)$ $= 0.44\, V$  and ${E^o}(X/{X^ - })$ $= 0.33\,V$. From this data one can deduce that
  • A
    $M\, + \,X\, \to {M^ + } + {X^ - }$ is the spontaneous reaction
  • ✓
    ${M^ + } + {X^ - } \to M + X$ is the spontaneous reaction
  • C
    ${E_{cell}}$$= 0.77 V$
  • D
    ${E_{cell}}$ $= -0.77 V$
✓

Answer

Correct option: B.
${M^ + } + {X^ - } \to M + X$ is the spontaneous reaction
b
(b) For the given cell $M|{M^ + }||{X^ - }|X$, the cell reaction is derived as follows:

$RHS$: reduction $X + {e^ - } \to {X^ - }$       …..$(i)$

$LHS$: Oxidation $M \to {M^ + } + {e^ - }$     …..$(ii)$

Add $(i)$ and $(ii)$ $M + X \to {M^ + } + {X^ - }$

The cell potential = $ - 0.11\,V$

Since ${E_{cell}} = -ve$, the cell reaction derived above is not spontaneous. In fact, the reverse reaction will occur spontaneously.

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