Zns- Cl21 atm- Zn2- 2 Cl-The E- of the cell is 2-12 V- To increase E-

Zn(s)+ Cl_2(1 atm)→ Zn^2++ 2 Cl^ The EMF of the cell is nbsp; E_cell= E^∘_cell RTnFln[Zn^2+][Cl^ ]^2P_Cl_2 To increase the E_cell, the ln[Zn^2+][Cl^ ]^2P_Cl ...

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Standard XII

Chemistry

Spontaneity

Zns+ Cl21 atm...

Question

$Z n (s) + C l_{2} (1 atm) \to Z n^{2 +} + 2 C l^{-}$
The $E^{\circ}$ of the cell is $2.12 V$ . To increase $E$ ,

Z n^{2 +}

concentration should be increased

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Z n^{2 +}

concentration should be decreased

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C l^{-}

concentration should be increased

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Partial pressure of

C l_{2}

should be increased

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Solution

The correct option is B $Z n^{2 +}$ concentration should be decreased
$Z n (s) + C l_{2} (1 atm) \to Z n^{2 +} + 2 C l^{-}$
The $E M F$ of the cell is
$E_{cell} = E_{c e l l}^{\circ} - \frac{R T}{n F} ln \frac{[Z n^{2 +}] [C l^{-}]^{2}}{P_{C l_{2}}}$
To increase the $E_{cell}$ , the $ln \frac{[Z n^{2 +}] [C l^{-}]^{2}}{P_{C l_{2}}}$ term should decrease. So $Z n^{2 +}$ concentration should be decreased and partial pressure of $C l_{2}$ should increased,

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Similar questions

Z n (s) + C l_{2} (1

atm

) \to Z n^{2 +} + 2 C l^{-} : E^{o} = 2.12 V

. To increase the e.m.f. of the above cell:

Q. For the cell :

Z n | Z n^{2 +} | | C u^{2 +} | C u

E_{z n}^{⊖} = 0.736 V; E_{(c u)}^{⊖} = - 0.35 V

Calculate the cell EMF.

Q. The standard reduction potential

E^{\circ}

, for the half reaction are:-

Z n \to Z n^{2 +} + 2 e^{-}; E^{\circ} = 0.76 V

C u \to C u^{2 +} + 2 e^{-}; E^{\circ} = 0.34 V

The emf for the cell reaction,

Z n (s) + C u^{2 +} \to Z n^{2 +} + C u (s)

is_____________.

Q. Calculate

△ G^{\circ}

for the following cell reaction:

Z n (s) + A g_{2} O (s) + H_{2} O (l) ⇌ Z n^{2 +} (a q) + 2 A g (s) + 2 O H^{-} (a q)

E_{A g^{+} / A g}^{\circ} = + 0.80 V

and

E_{Z n^{2 +} / Z N}^{\circ} = - 0.76 V

Q. The standard electrode potential of the half cells is given below:

Z n^{2 +} + 2 e^{-} \to Z n; E = - 0.76 V

F e^{2 +} + 2 e^{-} \to F e; E = - 0.44 V

The emf of the cell

F e^{2 +} = Z n^{-} \to Z n^{2 +} + F e

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Zn(s)+Cl2(1 atm)→Zn2++2Cl− The E∘ of the cell is 2.12 V. To increase E,

$Z n (s) + C l_{2} (1 atm) \to Z n^{2 +} + 2 C l^{-}$
The $E^{\circ}$ of the cell is $2.12 V$ . To increase $E$ ,