The standard emf of the following electrodes are- E-Fe3-Fe2- - - 0-77 V- E-Sn2- Sn - -0-14 Vunder standard conditions- the potential for the reaction- Sns - 2Fe3-aq- 2Fe2-aq - Sn2-aq is-

The correct option is A 1.68 V E_0 for reaction #160; #160; #160; #160; #160; #160; =2 [ E^0Fe^3+/Fe^2+ ]+1 [ E^0sn/S^2+ ] ...

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Standard XII

Chemistry

Solubility Product

The standard ...

Question

The standard emf of the following electrodes are;

$E_{F e^{3 +} / F e^{2 +}}^{\circ} = + 0.77 V; E_{S n^{2 +} / S n}^{\circ} = - 0.14 V$

under standard conditions, the potential for the reaction,

$S n_{(s)} + 2 F e_{(a q)}^{3 +} \to 2 F e_{(a q)}^{2 +} + S n_{(a q)}^{2 +}$ is:

1.68 V

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1.40 V

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0.91 V

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0.63 V

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Solution

The correct option is A $1.68 V$
$E_{0}$ for reaction

$= 2 [E^{0} F e^{3 +} / F e^{2 +}] + 1 [E^{0} s n / S^{2 +}]$

$= 2 (0.77) + 2 (- (- 0.14))$

$= 1.68 V$

Hence, the correct option is $A$

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Similar questions

Consider the following $E^{0}$ values:

$E_{F e^{3 +} / F e^{2 +}}^{0} = + 0.77 V$ ; $E_{S n^{2 +} / S n}^{0} = - 0.14 V$

Under standard conditions the EMF for the reaction

$S n_{(s)} + 2 F e {^{3 +}}_{(a q)} \to 2 F e {^{2 +}}_{(a q)} + S n {^{2 +}}_{(a q)}$ is :

Q. Consider the following

E^{0}

values:

E^{0_{F e 3 + / F e 2 +}} = + 0.77 V

E_{S n^{2 +} / S n}^{0} = - 0.14 V

Under standard conditions the potential for the reaction

S n (s)

+ 2 F e^{3 +} (a q) \to 2 F e^{2 +} (a q) + S n^{2 +} (a q)

is:

Q. Consider the following

E^{0}

values:

E_{F e^{3 +} / F e^{2 +}}^{0}

= 0.77V,

E_{S n^{2 +} / S n}^{0}

= -0.14V

Under standard conditions the potential for the reaction, is:

S n (s) + 2 F e^{3 +} (a q) \to 2 F e^{2 +} (a q) + S n^{2 +} (a q)

Q. The standard electrode potentials for the reactions

A g^{+} (a q .) + e^{-} \to A g (s)

S n^{2 +} (a q .) + 2 e^{-} \to S n (s)

25^{o} C

are

0.80 v o l t

and

- 0.14 v o l t

respectively. The

e m f

of the cell,

S n | S n^{2 +} (1 M) | | A g^{+} (1 M) | A g

is :

Q. The standard reduction potential for

F e^{2 +} | F e

and

S n^{2 +} | S n

electrodes are

- 0.44 V

and

- 0.14 V

respectively. For the cell reaction,

F e^{2 +} + S n \to F e + S n^{2 +}

, the standard

E . M . F .

is:

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The standard emf of the following electrodes are;E∘Fe3+/Fe2+=+0.77 V;E∘Sn2+/Sn=−0.14 Vunder standard conditions, the potential for the reaction,Sn(s)+2Fe3+(aq)→2Fe2+(aq)+Sn2+(aq) is:

The correct option is A 1.68 VE0 for reaction =2[E0Fe3+/Fe2+]+1[E0sn/S2+] =2(0.77)+2(−(−0.14)) =1.68 VHence, the correct option is A

The standard emf of the following electrodes are;

$E_{F e^{3 +} / F e^{2 +}}^{\circ} = + 0.77 V; E_{S n^{2 +} / S n}^{\circ} = - 0.14 V$

under standard conditions, the potential for the reaction,

$S n_{(s)} + 2 F e_{(a q)}^{3 +} \to 2 F e_{(a q)}^{2 +} + S n_{(a q)}^{2 +}$ is:

The correct option is A $1.68 V$
$E_{0}$ for reaction

$= 2 [E^{0} F e^{3 +} / F e^{2 +}] + 1 [E^{0} s n / S^{2 +}]$

$= 2 (0.77) + 2 (- (- 0.14))$

$= 1.68 V$

Hence, the correct option is $A$