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Valency and Oxidation States

The standard ...

Question

The standard reduction potentials at 298 K for the following half reactions are given against each
$Z n^{2 +} (a q) + 2 e ⇌ Z n (s) - 0.762$
$C r^{3 +} (a q) + 2 e ⇌ C r (s) - 0.740$
$2 H^{+} (a q) + 2 e ⇌ H_{2} (g) 0.000$
$F e^{3 +} (a q) + 2 e ⇌ F e^{2 +} (a q) 0.770$
Which is the strongest reducing agent?

Zn(s)

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Cr(s)

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H_{2} (g)

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F e^{2 +} (a q)

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Solution

The correct option is A Zn(s)
More negative is the standard reduction potential least is the reduction tendency of that ion. Thus, the atom (correspondingly) has the largest oxidation tendency. Hence, it is the strongest reducing agent. Here, $Z n$ is the strongest reducing agent.

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Similar questions

Z n^{2 +} (a q) + 2 e^{-} ⇌ Z n (s); - 0.762

C r^{3 +} (a q) + 3 e^{-} ⇌ C r (s); - 0.740

2 H^{⨁} (a q) + 2 e^{-} ⇌ H_{2} (g); 0.000

F e^{3 +} (a q) + 3 e^{-} ⇌ F e^{2 +} (a q); 0.770

25^{\circ} C

Z n^{2 +} (a q .) + 2 e^{-} ⇌ Z n (s); - 0.762

C r^{3 +} (a q .) + 3 e^{-} ⇌ C r (s); - 0.740

2 H^{+} + 2 e^{-} ⇌ H_{2} (g); 0.00

F e^{3 +} + e^{-} ⇌ F e^{2 +}; 0.77

298 K

Z n^{2 +} (a q) + 2 e ⇌ Z n (s); - 0.762 V

C r^{3 +} (a q) + 3 e ⇌ C r (s); - 0.740 V

2 H^{+} (a q) + 2 e ⇌ H_{2} (g); 0.00 V

F e^{3 +} (a q) + e ⇌ F e^{2 +} (a q); 0.770 V

Z n_{(a q)}^{2 +} + 2 e^{-} ⇌ Z n_{(s)}

E^{\circ}

- 0.762 V

C r_{(a q)}^{3 +} + 3 e^{-} ⇌ C r_{(s)}

E^{\circ}

- 0.740 V

2 H_{(a q)}^{+} + 2 e^{-} ⇌ H_{2 (g)}

E^{\circ}

0.000 V

F e_{(a q)}^{3 +} + e^{-} ⇌ F e_{(s)}^{2 +}

E^{\circ}

0.770 V

298 K

{Z n}^{2 +} (a q) + 2 e^{-} ⇌ Z n (s) - 0.762

{C r}^{3 +} (a q) + 3 e^{-} ⇌ C r (s) - 0.74

2 H^{+} (a q) + 2 e^{-} ⇌ H_{2} (g) + 0.00

{F e}^{3 +} (a q) + e^{-} ⇌ {F e}^{2 +} (a q) + 0.77

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