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Application of Electrolysis

The standard ...

Question

The standard reduction potentials at 298K for the following half cell reaction are given below:

$Z n^{2 +} (a q) + 2 e^{-} ⇌ Z n (s); E^{o} = - 0.762 V$
$C r^{3 +} (a q) + 3 e^{-} ⇌ C r (s); E^{o} = + 0.762 V$
$2 H^{+} (a q) + 2 e^{-} ⇌ H_{2} (g); E^{o} = + 0.00 V$
$F e^{3 +} (a q) + e^{-} ⇌ F e^{2 +} (a q); E^{o} = + 0.770 V$

Based on the above reactions, select the strongest reducing agent?

Z n (s)

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C r (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 $Z n (s)$
An electrode with low SRP acts as a strong reducing agent. So, the reducing power is maximum for one with maximum oxidation potential.

Oxidation potential of Zn is $0.762$ V, oxidation potential of Cr is $- 0.762$ V, oxidation potential of $H_{2}$ is $0$ V, and oxidation potential of $F e^{2 +}$ is $- 0.770$ V.

$∴$ Zn is the strongest reducing agent.

Hence, the correct option is $A$

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Similar questions

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^{2 +} (a q) + 2 e^{-} ⇌ Z n (s);

- 0.762 V

C r^{3 +} (a q) + 3 e^{-} ⇌ C r (s);

- 0.74 V

2 H^{+} (a q) + 2 e^{-} ⇌ H_{2} (g);

+ 0.00 V

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

+ 0.77 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

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

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); 0.00

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

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