Question

For the electrochemical cell,
M | [M⁺ || X⁻ | X, E°(M⁺/M) = 0.44 V and
E° (X/X⁻) =0.33 V. From this data one can deduce that:
(a) M + X → M⁺ + X⁻ is spontaneous reaction
(b) M⁺ + X⁻ → M + X is spontaneous reaction
(c) E꜀ₑₗₗ = 0.77V
(d) E꜀ₑₗₗ = – 0.77V

Answer 1

Answer:

(d) E cell= -0.77V

Explanation:

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Answer 2

(d) option is correct

Explanation:

(a) $M+X\rightarrow M^{+}+X^{-}$

here $M\rightarrow M^{+}$  increase by +1 charge so, oxidation

$X\rightarrow X^{-}$ reduction of 1 charge so, reduction

$E^{0}_{cell}= E_{LRP}-E_{RRP}$

Where $E^{0}_{cell}$ is cell potential

$E_{LRP}$= reduction potential of anode

$E_{RRP}$= reduction potential of cathode

for given equation in (a)

$E^{0}_{cell}$= $E_{M/M^{+}}-E_{X/X^{-}}$

$E^{0}_{cell}=-0.44-0.33$

$E^{0}_{cell}=-0.77$

So,$\Delta G=-nFE^{0}_{cell}$

$\Delta G$ will be positive as $E^{0}_{cell}$ is negative ,so reaction will non-spontaneous

(b) $M^{+} +X^{-} \rightarrow M+X$

by same method as above calculate $E^{0}_{cell}$

$E^{0}_{cell}= E_{LRP}-E_{RRP}$

$E^{0}_{cell}$= $E_{X/X^{-}}-E_{M/M^{+}}$

$E^{0}_{cell}$ = -0.11

$\Delta G$ will be positive as $E^{0}_{cell}$ is negative ,so reaction will non-spontaneous again.

so by above we can conclude that (d) is correct option.