What will be the E0 value for the given half cell reaction- Cr2-aq - 2e- - CrsGiven-E0 value for the half cell reaction- Cr3-aq - e- - Cr2-aq- E0 - -0-41 VCr3-aq - 3e- - Crs- E0 - -0-74 V

Δ G^0 = nFE^0 Cr^3+(aq) + 3e^ → Cr(s)....(1); Δ G^0_1 = +3F × 0.74 Cr^3+(aq) + e^ → Cr^2+(aq)......(2); Δ G^0_2 = +F × 0.41 Reversing equation (2), we get, ...

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Byju's Answer

Standard VII

Chemistry

Equilibrium Constant from Nernst Equation

What will be ...

Question

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,
$C r^{3 +} (a q) + e^{-} \to C r^{2 +} (a q); E^{0} = - 0.41 V$
$C r^{3 +} (a q) + 3 e^{-} \to C r (s); E^{0} = - 0.74 V$

+ 0.33 V

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- 1.15 V

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- 0.33 V

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- 0.91 V

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Solution

The correct option is D $- 0.91 V$
$Δ G^{0} = - n F E^{0}$

$C r^{3 +} (a q) + 3 e^{-} \to C r (s) . . . . (1); Δ G_{1}^{0} = + 3 F \times 0.74$
$C r^{3 +} (a q) + e^{-} \to C r^{2 +} (a q) . . . . . . (2); Δ G_{2}^{0} = + F \times 0.41$
Reversing equation (2), we get,
$C r^{2 +} (a q) \to C r^{3 +} (a q) + e^{-} . . . . . (3); Δ G_{3}^{0} = - F \times 0.41$

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

$Δ G_{4}^{0} = Δ G_{1}^{0} + Δ G_{3}^{0}$
$- 2 F \times E^{0} = + 3 F \times 0.74 - F \times 0.41$
$2 \times E^{0} = - 3 \times 0.74 + 0.41$
$E^{0} = \frac{- 3 \times 0.74 + 0.41}{2}$
$E^{0} = \frac{- 1.81}{2}$
$E^{0} = - 0.91 V$

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Similar questions

Q. What will be the

E^{0}

value for the given half cell reaction?

F e^{3 +} (a q) + 3 e^{-} \to F e (s)

Given:

E^{0}

value for the half cell reaction,

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

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

Q. What will be the

E^{0}

value for the given half cell reaction?

S n^{4 +} (a q) + 4 e^{-} \to S n (s)

Given:

E^{0}

value for the half cell reaction,

S n^{4 +} (a q) + 2 e^{-} \to S n^{2 +} (a q); E^{0} = 0.15 V

S n^{2 +} (a q) + 2 e^{-} \to S n (s); E^{0} = - 0.14 V

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 +} (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:

Q. Choose the correct option(s) for the given half cell reactions:

F_{2} (g) + 2 e^{-} \to 2 F^{-} (a q); E^{0} = 2.87 V C l_{2} (g) + 2 e^{-} \to 2 C l^{-} (a q); E^{0} = 1.36 V F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q); E^{0} = 0.77 V

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What will be the E0 value for the given half cell reaction? Cr2+(aq)+2e−→Cr(s) Given: E0 value for the half cell reaction, Cr3+(aq)+e−→Cr2+(aq); E0=−0.41 V Cr3+(aq)+3e−→Cr(s); E0=−0.74 V

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,
$C r^{3 +} (a q) + e^{-} \to C r^{2 +} (a q); E^{0} = - 0.41 V$
$C r^{3 +} (a q) + 3 e^{-} \to C r (s); E^{0} = - 0.74 V$