The standard reduction potential values of three metallic cations, $X, Y, Z$ are $0.52, - 3.03$ and $- 1.18 V$ respectively. The order of reducing power of the corresponding metals is:

Y > Z > X

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X > Y > Z

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Z > Y > X

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Z > X > Y

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Solution

The correct option is A $Y > Z > X$
Lower the value of $E^{\circ}$ , stronger the reducing agent.
Hence,
$Reducing power : Y (E^{\circ} = - 3.03 V) > Z (E^{\circ} = - 1.18 V) > X (E^{\circ} = 0.52 V)$

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The standard reduction potential values of three metallic cations, X,Y,Z are 0.52,−3.03 and −1.18 V respectively. The order of reducing power of the corresponding metals is:

The correct option is A Y>Z>XLower the value of E∘, stronger the reducing agent. Hence, Reducing power:Y(E∘=−3.03 V)>Z(E∘=−1.18 V)>X(E∘=0.52 V)

The standard reduction potential values of three metallic cations, $X, Y, Z$ are $0.52, - 3.03$ and $- 1.18 V$ respectively. The order of reducing power of the corresponding metals is:

The correct option is A $Y > Z > X$
Lower the value of $E^{\circ}$ , stronger the reducing agent.
Hence,
$Reducing power : Y (E^{\circ} = - 3.03 V) > Z (E^{\circ} = - 1.18 V) > X (E^{\circ} = 0.52 V)$