Consider the cell Zn - Zn 2- aq- Cl - aq - Cl 2 g - 1 pt s Given - Zn -2-4 M - Cl -0-5 M - P Cl 2-0-1 barE Zn - Zn 2-0-76 V - E Cl 2 - Cl -1-32 VIf 2-303 RT - F -0-06A- E cell -2-17 VB- Ecell -0-56 VC- Cell is non-spontaneous at the given conditionD- None of these

The correct option is D None of theseCell reaction Zn(s)+Cl_2(g)→ Zn^2+(aq.)+2Cl^ (aq.) E^∘_cell=E^∘_oxidation potential+E^∘_reduction potential E^∘_cell=E^∘_Z ...

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Byju's Answer

Standard XII

Chemistry

Gibb's Free Energy

Consider the ...

Question

Consider the cell $Z n / Z n^{2 +} (a q .) | | C l^{-} (a q) / C l_{2} (g) | 1 p t (s)$
Given $[Z n^{+ 2}] = 4 M; [C l^{-}] = 0.5 M, (P_{C l_{2}}) = 0.1 b a r$
$E_{Z n / Z n^{2 +}}^{\circ} = 0.76 V; E_{C l_{2} / C l^{-}}^{\circ} = 1.32 V$
If $\frac{2.303 R T}{F} = 0.06$

E_{c e l l} = 2.17 V

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E_{c e l l}^{\circ} = 0.56 V

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Cell is non-spontaneous at the given condition

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None of these

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Solution

The correct option is D None of these
Cell reaction
$Z n (s) + C l_{2} (g) \to Z n^{2 +} (a q .) + 2 C l^{-} (a q .)$
$E_{c e l l}^{\circ} = E_{o x i d a t i o n p o t e n t i a l}^{\circ} + E_{r e d u c t i o n p o t e n t i a l}^{\circ}$
$E_{c e l l}^{\circ} = E_{Z n / Z n^{2 +}}^{\circ} + E_{C l_{2} / C l^{-}}^{\circ}$
$= 0.76 + 1.32 v$
$E_{c e l l}^{\circ} = 2.08 V$
$E_{c e l l} = E_{c e l l}^{\circ} - \frac{0.06}{2} l o g_{10} \frac{[Z n^{+ 2}] [C l^{-}]^{2}}{[Z n^{(s)}] P_{C l_{2}} (g)}$
$= 2.08 - \frac{0.06}{2} l o g_{10} \frac{4 \times (0.5)^{2}}{0.1}$
$= 2.05 V$
$E_{c e l l} = + v e$
$Δ G$ is negative, so the cell is spontaneous at the given conditions.

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Similar questions

Q. For the cell,

Z n (s) | Z n^{2 +} (a q) | | C u^{2 +} (a q) | C u (s)

The cell potential

E_{c e l l}

can be increased:

Q. For the cell

Z n (s) | Z n^{2 +} (a q) | | C u^{+ 2} (a q) | C u (s)

. The cell potential

E_{c e l l}

can be increased.

Z n (s) + C u^{2 +} (a q .) ⇌ C u (s) + Z n^{2 +} (a q .)

Reaction quotient

Q = \frac{[Z n^{2 +}]}{[C u^{2 +}]}

, variation

E_{c e l l}

with

Q

is of the type with

O A = 1.10 v o l t . E_{c e l l} = 1.1591 v o l t

when:

Z n + C u^{2 +} (a q) ⇌ C u + Z n^{2 +} (a q)

Reaction quotient is

Q = \frac{[Z n^{2 +}]}{[C u^{2 +}]}

. Variation of

E_{c e l l}

with log Q is of the type with OA = 1.10

V . E_{c e l l}

will be

1.1591 V

when:

Electra has trouble finding the correct Nernst equation for the cell $Z n (s) | Z n^{2 +} | | C u^{2 +} | C u (s) a t 25^{\circ} C$
Can you help her?

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Consider the cell Zn/Zn2+(aq.)||Cl−(aq)/Cl2(g)|1pt(s) Given [Zn+2]=4M;[Cl−]=0.5M,(PCl2)=0.1 bar E∘Zn/Zn2+=0.76V;E∘Cl2/Cl−=1.32V If 2.303 RTF=0.06

Consider the cell $Z n / Z n^{2 +} (a q .) | | C l^{-} (a q) / C l_{2} (g) | 1 p t (s)$
Given $[Z n^{+ 2}] = 4 M; [C l^{-}] = 0.5 M, (P_{C l_{2}}) = 0.1 b a r$
$E_{Z n / Z n^{2 +}}^{\circ} = 0.76 V; E_{C l_{2} / C l^{-}}^{\circ} = 1.32 V$
If $\frac{2.303 R T}{F} = 0.06$