The following two redox reactions are given- i MnO4-8H-5e- Mn2-4H2O - Eo-X1 Vii MnO2-4H-2e- Mn2-2H2O - Eo-X2 VFind Eo for the following reaction- MnO4-4H-3e- MnO2-2H2O

The correct option is B X_1 X_2 i) MnO_4^ +8H^++5e^ → Mn^2++4H_2O ; E^o=X_1V ......(1)ii) MnO_2+4H^++2e^ → Mn^2++2H_2O ;E^o=X_2V ......(2)R ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

EMF

The following...

Question

The following two redox reactions are given:
i) $M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O; E^{o} = X_{1} V$
ii) $M n O_{2} + 4 H^{+} + 2 e^{-} \to M n^{2 +} + 2 H_{2} O; E^{o} = X_{2} V$
Find $E^{o}$ for the following reaction.
$M n O_{4}^{-} + 4 H^{+} + 3 e^{-} \to M n O_{2} + 2 H_{2} O$

X_{2} - X_{1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

X_{1} - X_{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{5 X_{1} - 2 X_{2}}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2 X_{1} - 5 X_{2}}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $X_{1} - X_{2}$
i) $M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O; E^{o} = X_{1} V$ ......(1)
ii) $M n O_{2} + 4 H^{+} + 2 e^{-} \to M n^{2 +} + 2 H_{2} O; E^{o} = X_{2} V$ ......(2)
Reverse equation (2)
iii) $M n^{2 +} + 2 H_{2} O \to M n O_{2} + 4 H^{+} + 2 e^{-}, E^{o} = - X_{2} V$ ......(3)
Add equation (3) to equation (1),
$M n O_{4}^{-} + 4 H^{+} + 3 e^{-} \to M n O_{2} + 2 H_{2} O$
$E^{o}$ for the above reaction is $X_{1} - X_{2}$

Suggest Corrections

Similar questions

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}

Q. For the following half cell reactions,

E^{o}

values are also given:

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

Select the correct statements.

Q. Calculate the value of

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

M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O

;

E^{\circ} = 1.51 V

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

;

E^{\circ} = 1.23 V

E_{R P}^{o}

and their respective half reactions for some changes are given below:

(a)

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

(b)

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

(c)

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

(acid med.)
(d)

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

(acid med.)
(e)

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

(acid med.)

The most stable oxidation state of Mn is :

$M n {O_{4}}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O E^{o} = 1.51 V$
The quantity of electricity required in Faraday to reduce five moles of $M n O_{4}^{-}$ is ____.

Join BYJU'S Learning Program

The following two redox reactions are given: i) MnO−4+8H++5e−→Mn2++4H2O ; Eo=X1 Vii) MnO2+4H++2e−→Mn2++2H2O ;Eo=X2 VFind Eo for the following reaction.MnO−4+4H++3e−→MnO2+2H2O

The correct option is B X1−X2i) MnO−4+8H++5e−→Mn2++4H2O;Eo=X1V......(1)ii) MnO2+4H++2e−→Mn2++2H2O;Eo=X2V......(2)Reverse equation (2)iii) Mn2++2H2O →MnO2+4H++2e−,Eo=−X2V......(3)Add equation (3) to equation (1),MnO−4+4H++3e−→MnO2+2H2OEo for the above reaction is X1−X2