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Redox Reactions

Which is the ...

Question

Which is the strongest reducing agent among the following, if the standard electrode potentials are given respectively at $298 K$ ?
(correct answer +2, wrong answer -0.5)

$X^{+} + e^{-} \to X; E^{o} = - 3.05 V$

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Y_{2} + 2 e^{-} \to 2 Y^{-}; E^{o} = 2.87 V

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A^{+} + e^{-} \to A; E^{o} = - 0.52 V

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A^{2 +} + 2 e^{-} \to A; E^{o} = 0.34 V

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Solution

The correct option is A
$X^{+} + e^{-} \to X; E^{o} = - 3.05 V$

Lesser is the standard reduction electrode potential value, it is easy to donate electrons and hence, stronger is the reducing agent.

Suggest Corrections

Similar questions

F_{2} (g) + 2 e^{-} \to 2 F^{-} (a q); E^{0} = + 2.85 V

C l_{2} (g) + 2 e^{-} \to 2 C l^{-} (a q); E^{0} = + 1.36 V

B r_{2} (l) + 2 e^{-} \to 2 B r^{-} (a q); E^{0} = + 1.06 V

I_{2} (s) + 2 e^{-} \to 2 I^{-} (a q); E^{0} = + 0.53 V

Given standard electrode potentials:

$F e^{3 +} + 3 e^{-} \to F e; E^{0} = - 0.036 V$

$F e^{2 +} + 2 e^{-} \to F e; E^{0} = - 0.440 V$

The standard electrode potential $E^{o}$ for $F e^{3 +} + e^{-} \to F e^{2 +}$ is:

M g^{2 +} + 2 e^{-} \to M g; E^{0} = - 2.37 V

Z n^{2 +} + 2 e^{-} \to Z n; E^{0} = - 0.76 V

F e^{2 +} + 2 e^{-} \to F e; E^{0} = - 0.44 V

M^{2 +} / M and M^{3 +} / M^{2 +}

E^{0}

C r^{2 +} (a q) + 2 e^{-} \to C r (s); E^{0} = - 0.91 V C r^{3 +} (a q) + e^{-} \to C r^{2 +} (a q); E^{0} = - 0.41 V

M n^{2 +} (a q) + 2 e^{-} \to M n (s); E^{0} = - 1.2 V M n^{3 +} (a q) + e^{-} \to M n^{2 +} (a q); E^{0} = + 1.51 V

F e^{2 +} (a q) + 2 e^{-} \to F e (s); E^{0} = - 0.44 V F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q); E^{0} = + 0.77 V

C u^{2 +} + 2 e^{-} ⟶ C u

E^{0} = 0.337 V

C u^{2 +} + e^{-} ⟶ C u^{+}

E^{0} = 0.153 V

E^{0}

C u^{+} + e^{-} ⟶ C u

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