While Fe3- is stable- Mn3- is not stabe in acid solution because-Mn3-e- - Mn2- E- -1-5 V- O2-4H- - 4e- - 2H2O- E- - 1-23 V- Fe3- - e- - Fe2- E- 0-77V

Mn^3+ oxides H_2 O to O_2 because the standard reduction potential of (Mn^3+ + e^ → Mn^2+; E^∘ = +1.5 V) is greater than that (O_2+4H^+ +4e^ → 2H_2O; E^∘ = + ...

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Standard XII

Chemistry

Electrode Potential

While Fe3+ is...

Question

While $F e^{3 +}$ is stable, $M n^{3 +}$ is not stabe in acid solution because:
$(M n^{3 +} + e^{-} \to M n^{2 +}, E^{\circ} = 1.5 V; O_{2} + 4 H^{+} + 4 e^{-} \to 2 H_{2} O, E^{\circ} = + 1.23 V; F e^{3 +} + e^{-} \to F e^{2 +}, E^{\circ} = 0.77 V)$

$O_{2} o x i d i s e s M n^{2 +} t o M n^{3 +}$

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$O_{2} o x i d i s e s M n^{2 +} a n d F e^{2 +} t o F e^{3 +}$

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$F e_{3 +} o x i d i s e s H_{2} O t o O_{2}$

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$M n^{3 +} o x i d i s e s H_{2} O t o O$

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Solution

The correct option is D
$M n^{3 +} o x i d i s e s H_{2} O t o O$

$M n^{3 +}$ oxides $H_{2} O$ to $O_{2}$ because the standard reduction potential of $(M n^{3 +} + e^{-} \to M n^{2 +}; E^{\circ} = + 1.5 V)$ is greater than that $(O_{2} + 4 H^{+} + 4 e^{-} \to 2 H_{2} O; E^{\circ} = + 1.23 V)$

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Similar questions

While $F e^{3 +}$ is stable, $M n^{3 +}$ is not stabe in acid solution because:
$(M n^{3 +} + e^{-} \to M n^{2 +}, E^{\circ} = 1.5 V; O_{2} + 4 H^{+} + 4 e^{-} \to 2 H_{2} O, E^{\circ} = + 1.23 V; F e^{3 +} + e^{-} \to F e^{2 +}, E^{\circ} = 0.77 V)$

Q. Redox reactions play a pivotal role in chemistry and biology. The values of standard redox potential

(E_{⊖})

of two half cell reactions decided which way the reaction is expected to proceed. A simple example is a Daniell cell in which zinc goes into solution and copper gets deposited. Given below are a set of half cell reactions (acidic medium) along with their

E_{⊖}

(V with respect to normal hydrogen electrode) values. Using this data, obtain correct explanations for the following question

I_{2} + 2 e^{-} \to 2 I^{⊖}; E^{⊖} = 0.54

C l_{2} + 2 e^{-} \to 2 C l^{⊖}; E^{⊖} = 1.36

M n^{3 +} + e^{-} \to M n^{2 +}; E^{⊖} = 1.50

F e^{3 +} + e^{-} \to F e^{2 +}; E^{⊖} = 0.77

O_{2} + 4 H^{⨁} + 4 e^{-} \to 2 H_{2} O; E^{⊖} = 1.23

While

F e^{3 +}

is stable,

M n^{3 +}

is not stable in acid solution because

E^{\circ}

for the half cells :

M n O_{2} + 4 H^{+} + 2 e^{-} \to M n^{2 +} + 2 H_{2} O i s 1.23 V F e^{3 +} + e^{-} \to F e^{2 +} i s 0.77 V

Which of the following statements is true?

The standard reduction potential data at $25^{\circ} C$ is given below.

$E^{\circ} (F e^{3 +} / F e^{2 +}) = 0.77 V; E^{\circ} (F e^{2 +} / F e) = - 0.44 V; E^{\circ} (C u^{2 +} / C u) = 0.34 V; E^{\circ} (C u^{+} / C u) = + 0.52 V; E^{\circ} (O_{2} (g) + 4 H^{+} + 4 e^{-}) \to 2 H_{2} O = + 1.23 V; E^{\circ} (O_{2} (g) + 2 H_{2} O + 4 e^{-}) \to 4 O H^{-} = + 0.40 V E^{\circ} (C r^{3 +} / C r) = - 0.74 V; E^{\circ} (C r^{2 +} / C r) = + 0.91 V$

Match $E^{\circ}$ of the rebox pair in Column I with the values given in Column II and select the correct answer using the code given below the lists.

$\begin{matrix} C o l u m n I & C o l u m n I I P . E^{\circ} (F e^{3 +} / F e) & 1. - 0.18 V Q . E^{\circ} (4 H_{2} O ⇌ 4 H^{+} + 4 O H^{-}) & 2. - 0.4 V R . E^{\circ} (C u^{2 +} + C u \to 2 C u^{+}) & 3. - 0.04 V S . E^{\circ} (C r^{3 +} / C r^{2 +}) & 4. - 0.83 V \end{matrix}$

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While Fe3+ is stable, Mn3+ is not stabe in acid solution because: (Mn3++e−→Mn2+, E∘=1.5 V;O2+4H++4e−→2H2O, E∘=+1.23V;Fe3++e−→Fe2+,E∘=0.77V)

The correct option is D Mn3+ oxidises H2O to O Mn3+ oxides H2O to O2 because the standard reduction potential of (Mn3++e−→Mn2+; E∘=+1.5 V) is greater than that (O2+4H++4e−→2H2O;E∘=+1.23V)

While $F e^{3 +}$ is stable, $M n^{3 +}$ is not stabe in acid solution because:
$(M n^{3 +} + e^{-} \to M n^{2 +}, E^{\circ} = 1.5 V; O_{2} + 4 H^{+} + 4 e^{-} \to 2 H_{2} O, E^{\circ} = + 1.23 V; F e^{3 +} + e^{-} \to F e^{2 +}, E^{\circ} = 0.77 V)$

The correct option is D
$M n^{3 +} o x i d i s e s H_{2} O t o O$

$M n^{3 +}$ oxides $H_{2} O$ to $O_{2}$ because the standard reduction potential of $(M n^{3 +} + e^{-} \to M n^{2 +}; E^{\circ} = + 1.5 V)$ is greater than that $(O_{2} + 4 H^{+} + 4 e^{-} \to 2 H_{2} O; E^{\circ} = + 1.23 V)$