The Eo for the reaction- 10Cl- aq - 2MnO4- aq - 16H- aq - 5Cl2g - 2Mn2- aq - 8H2Ol is 0-15 V- The K c for the reaction is-

The correct option is A 2.65 × 10^25 According to equation E = E^o RTnF ln #160;Q. at eqm, E=0 E^o = RTnF ln K_c E^o #160;=0.0 ...

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Nernst Equation

The Eo for ...

Question

The $E^{o}$ for the reaction,
$10 C l^{-} (a q) + 2 M n O_{4}^{-} (a q) + 16 H^{+} (a q) \to 5 C l_{2} (g) + 2 M n^{2 +} (a q) + 8 H_{2} O (l)$ is $0.15 V$ . The K $_{c}$ for the reaction is:

2.65 \times 10^{25}

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4.9 \times 10^{12}

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1.2 \times 10^{5}

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3.4 \times 10^{2}

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Solution

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10 C l^{-} (a q) + 2 M n O_{4}^{-} (a q) + 16 H^{+} (a q) \to 5 C l_{2} (g) + 2 M n^{2 +} (a q) + 8 H_{2} O (l)

E^{o}

25^{o}

C u_{(s)} \to C u_{(a q)}^{2 +} + 2 e^{-}; E^{o} = - 0.325 V

C u_{(a q)}^{+} + e^{-} \to C u_{(s)}; E^{o} = 0.50 V

E^{o}

2 C u_{(a q)}^{+} \to C u_{(a q)}^{2 +} + C u_{(s)}

{C u}^{+} (a q)

2 {C u}^{+} (a q) ⇌ {C u}^{2 +} (a q) + C u (s)

E^{o}

E_{{C u}^{2 +} / C u}^{o} = 0.34 V

E_{{C u}^{2 +} / {C u}^{+}}^{o} = 0.15 V

M n (O H)_{2} (s) + M n O_{4}^{-} (a q) \to M n O_{2} (s)

E_{R P}^{o}

M n O_{4}^{-} (a q) + 8 H^{+} + 5 e \to M n^{2 +} (a q) + 4 H_{2} O (l); E^{o} = + 1.52 V

M n O_{4}^{2 -} (a q) + 4 H^{+} + 2 e \to M n O_{2} (s) + 2 H_{2} O (l); E^{o} = + 2.26 V

2 M n O_{4}^{-} (a q) + 2 e \to 2 M n O_{4}^{2 -} (a q); E^{o} = + 0.56 V

M n^{3 +} + e \to M n^{2 +}; E^{o} = + 1.51 V

M n^{4 +} + e \to M n^{3 +}; E^{o} = 0.95 V

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