For the electrochemical cell- M - M - X - X - Pt- E O M - - M -0-44 V and E - X - X -0-33 V- From this data one can deduce thatA- M - X r M - X -is the spontaneous reactionB- Ecell -0-77 VC- E cell -0-77 VD- M - X - r M - X is the spontaneous reaction

The correct option is D M+ + X (r) M + X is the spontaneous reactionSRP of M+2/M is greater than of X/X ; hence X ion can reduce M+2 ion. ...

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For the electrochemical cell, M|M⁺| | X |X^-, Pt. E^o(M⁺/M) = 0.44 V and E^o(X/X^-) = 0.33 V. From this data one can deduce that

M + X (r) M⁺ + X^- is the spontaneous reaction

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M⁺ + X^-(r) M + X is the spontaneous reaction

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E_cell = 0.77 V

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E_cell = -0.77 V

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M | M^{+} | | X | X^{-}

E_{M^{+} / M}^{0} = 0.44 V

E_{X / X^{-}}^{0} = 0.33 V

M | M^{+} | | X^{-} | X, E_{M^{+} / M}^{o} = 0.44 V

E_{X / X^{-}}^{o} = 0.33 V

M | M^{+} ∥ X^{-} | X

E_{(M^{+} | M)}^{o} = 0.44 V

E_{(X | X^{-})}^{o} = 0.33 V

0.5 a m p

M | M^{+} ∥ X^{-} | X, E_{M^{+} / M}^{o} = 0.44 V a n d E_{X / X^{-}}^{o} = 0.33 V

(M | M^{\oplus}) | | (X^{⊖} | X), E_{(M^{\oplus} | M)}^{⊖} = 0.44 V,

E_{(x | X^{⊖})}^{⊖} = - 0.33 V

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