Given below are the half cell reactions-Mn2-aq-2e- Mns- E0-1-18 V2Mn-3aq -e- Mn2-aq- E0-1-51 VThe E0 for 3Mn2-aq- Mns -2Mn3-aq cell reaction will be-

The standard reduction potential of the half cells are given. E^0_Mn^2+(aq)/Mn(s)= 1.18 V E^0_Mn^3+(aq)/Mn^2+(aq)= +1.51 V Δ G^0 = nFE^0 ∴ Mn^2+(aq) + 2e^ → ...

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

Byju's Answer

Standard XII

Chemistry

Faraday's Second Law

Given below a...

Question

Given below are the half cell reactions:
$M n^{2 +} (a q) + 2 e^{-} \to M n (s); E^{0} = - 1.18 V$
$2 (M n^{+ 3} (a q) + e^{-} \to M n^{2 +} (a q)); E^{0} = + 1.51 V$

The $E^{0}$ for $3 M n^{2 +} (a q) \to M n (s) + 2 M n^{3 +} (a q)$ cell reaction will be:

+ 0.33 V

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

- 2.69 V

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

+ 2.69 V

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

- 0.33 V

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

Open in App

Solution

The correct option is B $- 2.69 V$
The standard reduction potential of the half cells are given.
$E_{M n^{2 +} (a q) / M n (s)}^{0} = - 1.18 V$
$E_{M n^{3 +} (a q) / M n^{2 +} (a q)}^{0} = + 1.51 V$

$Δ G^{0} = - n F E^{0}$
$∴$
$M n^{2 +} (a q) + 2 e^{-} \to M n (s) . . . . . . (1); Δ G_{1}^{0} = 2 F \times 1.18 2 M n^{3 +} (a q) + 2 e^{-} \to 2 M n^{2 +} (a q) . . . (2); Δ G_{2}^{0} = - 2 F \times 1.51$

Reverse equation (2)
$2 M n^{2 +} (a q) \to 2 M n^{3 +} (a q) + 2 e^{-} . . . . (3); Δ G_{3}^{0} = + 2 F \times 1.51$

Adding equation (1) and (3), we get,
$3 M n^{2 +} (a q) \to M n (s) + 2 M n^{3 +} (a q) . . . . . . (4); Δ G_{4}^{0} = - 2 F \times E^{0}$

$Δ G_{4}^{0} = Δ G_{1}^{0} + Δ G_{3}^{0}$
$- 2 F \times E^{0} = 2 F \times 1.18 + + 2 F \times 1.51$
$- E^{0} = 2.69 V$
$E^{0} = - 2.69 V$

Suggest Corrections

Similar questions

Q. Given below are the half -cell

M n^{2 +} + 2^{e -} \to M n; E^{0} = - 1.18 V

2 (M n^{3 +} + e^{-} \to M n^{2 +}); E^{0} = + 1.51 V

The

E^{0}

for

3 M n^{2 +} \to M n + 2 M n^{3 +}

will be:

Given below are the half-cell reactions:

M n^{2 +} + 2 e^{-} \to M n; E^{0} = - 1.18 V

2 (M n^{3 +} + e^{-} \to M n^{2 +}); E^{0} = + 1.51 V

The

E^{0}

for

3 M n^{2 +} \to M n + 2 M n^{3 +}

will be:

Q. For

M^{2 +} / M and M^{3 +} / M^{2 +}

systems,

E^{0}

values for some metals are as follows:

C r^{2 +} (a q) + 2 e^{-} \to C r (s); E^{0} = - 0.91 V C r^{3 +} (a q) + e^{-} \to C r^{2 +} (a q); E^{0} = - 0.41 V

M n^{2 +} (a q) + 2 e^{-} \to M n (s); E^{0} = - 1.2 V M n^{3 +} (a q) + e^{-} \to M n^{2 +} (a q); E^{0} = + 1.51 V

F e^{2 +} (a q) + 2 e^{-} \to F e (s); E^{0} = - 0.44 V F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q); E^{0} = + 0.77 V

Which of the following statement(s) is/are correct regarding the given standard reduction potentials?

Q. Given below are the half - cell reactions

M n^{2 +} + 2 e^{-} \to M n; E^{\circ} = - 1.18 e V

2 (M n^{3 +} + e^{-} \to M n^{2 +}); E^{0} = + 1.51 e V

The

E^{\circ}

for

3 M n^{2 +} \to M n + 2 M n^{3 +}

will be

Q. What will be the

E^{0}

value for the given half cell reaction?

C r^{2 +} (a q) + 2 e^{-} \to C r (s)

Given:

E^{0}

value for the half cell reaction,

C r^{3 +} (a q) + e^{-} \to C r^{2 +} (a q); E^{0} = - 0.41 V

C r^{3 +} (a q) + 3 e^{-} \to C r (s); E^{0} = - 0.74 V

Join BYJU'S Learning Program

Explore more

Faraday's Second Law

Standard XII Chemistry

Join BYJU'S Learning Program

Given below are the half cell reactions: Mn2+(aq)+2e−→Mn(s); E0=−1.18 V 2(Mn+3(aq)+e−→Mn2+(aq)); E0=+1.51 V The E0 for 3Mn2+(aq)→Mn(s)+2Mn3+(aq) cell reaction will be:

Given below are the half cell reactions:
$M n^{2 +} (a q) + 2 e^{-} \to M n (s); E^{0} = - 1.18 V$
$2 (M n^{+ 3} (a q) + e^{-} \to M n^{2 +} (a q)); E^{0} = + 1.51 V$

The $E^{0}$ for $3 M n^{2 +} (a q) \to M n (s) + 2 M n^{3 +} (a q)$ cell reaction will be: