What will be the E0 value for the given half cell reaction- Sn4-aq - 4e- - SnsGiven-E0 value for the half cell reaction- Sn4-aq - 2e- - Sn2-aq- E0 - 0-15 VSn2-aq - 2e- - Sns- E0 - -0-14 V

Δ G^0 = nFE^0 Sn^4+(aq) + 2e^ → Sn^2+(aq)....(1); Δ G^0_1 = 2F × 0.15 Sn^2+(aq) + 2e^ → Sn(s) ......(2); Δ G^0_2 = +2F × 0.14 Adding equation (1) and (2), ...

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Standard XII

Chemistry

Equilibrium Constant and Standard Free Energy Change

What will be ...

Question

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,
$S n^{4 +} (a q) + 2 e^{-} \to S n^{2 +} (a q); E^{0} = 0.15 V$
$S n^{2 +} (a q) + 2 e^{-} \to S n (s); E^{0} = - 0.14 V$

+ 0.001 V

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+ 0.005 V

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- 0.001 V

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- 0.005 V

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Solution

The correct option is B $+ 0.005 V$
$Δ G^{0} = - n F E^{0}$

$S n^{4 +} (a q) + 2 e^{-} \to S n^{2 +} (a q) . . . . (1); Δ G_{1}^{0} = - 2 F \times 0.15$
$S n^{2 +} (a q) + 2 e^{-} \to S n (s) . . . . . . (2); Δ G_{2}^{0} = + 2 F \times 0.14$

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

Thus,
$Δ G_{3}^{0} = Δ G_{1}^{0} + Δ G_{2}^{0}$
$- 4 F E^{0} = - 2 F \times 0.15 + 2 F \times 0.14$
$- 4 E^{0} = - 2 \times 0.15 + 2 \times 0.14$
$E^{0} = + 0.005 V$

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Similar questions

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. 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

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. 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? Sn4+(aq)+4e−→Sn(s) Given: E0 value for the half cell reaction, Sn4+(aq)+2e−→Sn2+(aq); E0=0.15 V Sn2+(aq)+2e−→Sn(s); E0=−0.14 V

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,
$S n^{4 +} (a q) + 2 e^{-} \to S n^{2 +} (a q); E^{0} = 0.15 V$
$S n^{2 +} (a q) + 2 e^{-} \to S n (s); E^{0} = - 0.14 V$