For the electrochemical cell, M| M^ + || X^ - |X, E^o ( M^ + /M) …

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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