Question:

X, Y and Z represent three electrodes $Al^{3+}/Al$, $Cu^{2+}/Cu$ and $Ag^{+}/Ag$ with $E^{\circ}$ values -1.66, 0.34 and 0.80 V respectively. The correct order of oxidising power of these three electrodes is

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More Positive $E^{\circ}$ = Stronger Oxidising Agent (likes to gain electrons). More Negative $E^{\circ}$ = Stronger Reducing Agent.
Updated On: Jun 3, 2026
  • $X>Y>Z$
  • $Z>Y>X$
  • $X=Y=Z$
  • $Y>Z>X$
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The Correct Option is B

Solution and Explanation

Step 1: Concept
The standard reduction potential ($E^{\circ}$) measures the tendency of a chemical species to gain electrons and be reduced.

Step 2: Meaning
A higher (more positive) $E^{\circ}$ value signifies a greater tendency to undergo reduction, which directly translates to a stronger oxidising power.

Step 3: Analysis
The given electrodes and their corresponding $E^{\circ}$ values are: * X ($Al^{3+}/Al$): $-1.66\text{ V}$ * Y ($Cu^{2+}/Cu$): $+0.34\text{ V}$ * Z ($Ag^{+}/Ag$): $+0.80\text{ V}$ Comparing these numerical values: $$+0.80\text{ V} > +0.34\text{ V} > -1.66\text{ V}$$ $$Z > Y > X$$

Step 4: Conclusion
Therefore, the correct order of decreasing oxidising strength is $Z > Y > X$.

Final Answer: (B)
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