For the electrochemical cell,M|M^{+}||X^{-}|X | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The cell reaction is $M + X 	o M^{+} + X^{-}$.For this cell,the anode reaction is $M 	o M^{+} + e^{-}$ $(E^{\circ}_{ox} = -E^{\circ}(M^{+}|M) =

Vedclass · Accepted Answer

$M^{+} + X^{-} 	o M + X$ is the spontaneous reaction

Vedclass · Answer

(B) The cell reaction is $M + X 	o M^{+} + X^{-}$.For this cell,the anode reaction is $M 	o M^{+} + e^{-}$ $(E^{\circ}_{ox} = -E^{\circ}(M^{+}|M) = -0.44 \; V)$ and the cathode reaction is $X + e^{-} 	o X^{-}$ $(E^{\circ}_{red} = E^{\circ}(X|X^{-}) = 0.33 \; V)$.$E^{\circ}_{cell} = E^{\circ}_{red} + E^{\circ}_{ox} = 0.33 \; V + (-0.44 \; V) = -0.11 \; V$.Since $E^{\circ}_{cell}$ is negative,the forward reaction $M + X 	o M^{+} + X^{-}$ is non-spontaneous.Conversely,the reverse reaction $M^{+} + X^{-} 	o M + X$ has $E^{\circ}_{cell} = +0.11 \; V$,which is positive,making it spontaneous.

For the electrochemical cell,$M|M^{+}||X^{-}|X,$ given $E^{\circ}(M^{+}|M) = 0.44 \; V$ and $E^{\circ}(X|X^{-}) = 0.33 \; V$. From this data,one can deduce that:

Similar Questions

Consider the following diagram in which an electrochemical cell is coupled to an electrolytic cell. What will be the polarity of electrodes $A$ and $B$ in the electrolytic cell?

Which of the following hot aqueous solutions is used to dip carbon rods of $H_2 - O_2$ fuel cell?

The cell reaction $Mn + 2Ag_{(C_1)}^{+(aq)} \to Mn_{(C_2)}^{+2_{(aq)}} + 2Ag$ is represented by

Which of the following curves represents the variation of $\lambda_{M}$ with $\sqrt{C}$ for $AgNO_{3}$?

The acid used in lead storage battery is

Vedclass Test Series

Exam Paper Generator

Online Exam Module