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Electrochemical Series

Which one is ...

Question

Which one is the most powerful oxidizing agent?

C l_{2} + 2 e^{-} \to 2 C l^{-}, E^{o} = 1.36 V

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N a^{+} + e^{-} \to N a, E^{o} = - 2.71 V

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M n O {_{4}}^{-} + 2 H_{2} O + 3 e^{-} \to M n O_{2} + 4 O H^{-}, E^{o} = 0.6 V

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H_{2} O_{2} + 2 H^{+} + 2 e^{-} \to 2 H_{2} O, E^{o} = 1.78 V

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Solution

The correct option is C $H_{2} O_{2} + 2 H^{+} + 2 e^{-} \to 2 H_{2} O, E^{o} = 1.78 V$
The one which gets reduced most easily is the best oxidising agent:
$∴$ according to the given SRP values,
$H_{2} O_{2}$ has highest SRP value.
$∴$ it is the best reducing agent.

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Similar questions

M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O; E^{o} = X_{1} V

M n O_{2} + 4 H^{+} + 2 e^{-} \to M n^{2 +} + 2 H_{2} O; E^{o} = X_{2} V

E^{o}

M n O_{4}^{-} + 4 H^{+} + 3 e^{-} \to M n O_{2} + 2 H_{2} O

E^{o}

M n^{2 +} + 2 H_{2} O \to M n O_{2} + 4 H^{+} + 2 e^{-}

E^{o} = - 1.23 V

M n O_{4}^{-} + 4 H^{+} + 3 e^{-} \to M n O_{2} + 2 H_{2} O

E^{o} = + 1.70 V

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

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

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

I_{2 (g)} + e^{-} \to 2 I_{(a q)}^{-}; E^{o} = + 0.53 V

M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O; E^{O} = 1.51 V; Δ G_{1}^{0} = - 5 \times 1.51 \times F

M n O_{2} + 4 H^{+} + 2 e^{-} \to M n^{2 +} + 2 H_{2} O; E^{O} = 1.23 V; Δ G_{2}^{0} = - 5 \times 1.23 \times F

E_{M n O_{4}^{-} | M n O_{2}}^{\circ}

C o^{3 +} + e^{-} \to C o^{2 +}; E^{o} = + 1.81 V

P b^{4 +} + 2 e^{-} \to P b^{2 +}; E^{o} = + 1.67 V

C e^{4 +} + e^{-} \to C e^{3 +}; E^{o} = + 1.61 V

B i^{3 +} + 3 e^{-} \to B i; E^{o} = + 0.20 V

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