ElectrochemistrymediumMCQ SINGLE

For the electrochemical cell, , V and V. From this data, one can deduce thatElectrochemistry Chemistry Question

Question

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

Answer: B

💡 Solution & Explanation

The standard reduction potential of $M^+$ to $M$ is +0.44 V, and the standard reduction potential of $X$ to $X^-$ is +0.33 V. For the reaction $M^+ + X^- \rightarrow M + X$, the cell potential is $E^\circ_{cathode} - E^\circ_{anode} = E^\circ_{M^+\|M} - E^\circ_{X\|X^-} = 0.44 - 0.33 = +0.11$ V. Since $E^\circ$ is positive, this reaction is spontaneous. Therefore, correct answer is B.

💬
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes
Ask on WhatsApp →

Practice 22,000+ questions like this

AI-adaptive practice, video lectures, and full JEE Advanced Chemistry content — all in one place.

JEE Advanced · JEE Mains · NEET · IChO · AP Chemistry