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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^\circ ({M^ + }|M)$$ = 0.44\;V$ $E^\circ (X|{X^ - }) = 0.33\;V$ From this data, one can deduce that
  • A
    $E{^\circ _{cell}} = - 0.77\,V$
  • ✓
    ${M^ + } + {X^ - } \to M + X$ is the spontaneous reaction
  • C
    $M + X \to {M^ + } + {X^ - }$is the spontaneous reaction
  • D
    $E{^\circ _{cell}} = .77\;V$
✓

Answer

Correct option: B.
${M^ + } + {X^ - } \to M + X$ is the spontaneous reaction
b
(b)For ${M^ + } + {X^ - } \to M + X$

$E_{Cell}^0 = E_{Cathode}^0 + E_{Anode}^0$ $ = 0.44 - 0.33 = + 0.11\;V$

Since $E_{Cell}^0 = ( + )\;0.11\;V$ is positive hence this reaction should be spontaneous.

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