For the cell reaction-Zn s - Cu 2- aq - Cu s - Zn 2- aq E cell 0-1-10 V - - G 0 isA- 212-3 KJB- 106-05 KJC- -212-3 KJD- 454-6 KJ

The correct option is C –212.3KJΔ G^0= nFE^0_cell Δ G^0= 2mol × 96500 C/mol × 1.10V = –212300 CV = 212.3 KJ. ...

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Byju's Answer

Standard XII

Chemistry

Equilibrium Constant from Nernst Equation

For the cell ...

Question

For the cell reaction:
$Z n (s) + C u^{2 +} (a q) \to C u (s) + Z n^{2 +} (a q)$
$E_{c e l l}^{0} = 1.10 V, Δ G^{0}$ is

212.3 KJ

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106.05KJ

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–212.3KJ

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454.6 KJ

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Solution

The correct option is C –212.3KJ
$Δ G^{0} = - n F E_{c e l l}^{0}$
$Δ G^{0} = - 2 m o l \times 96500 C / m o l \times 1.10 V$
= –212300 CV = -212.3 KJ.

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Similar questions

Q. For the cell reaction:

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is:

Q. For the cell reaction:

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

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will be:

$E^{0}$ for the cell, $Z n | Z n {^{2 +}}_{(a q)} | | C u {^{2 +}}_{(a q)} | C u$ is $1.10 V$ at $25^{0} C$ . The equilibrium constant for $Z n + C u {^{2 +}}_{(a q)} \to Z n {^{2 +}}_{(a q)} + C u$ the cell reaction is of the order of :

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For the cell reaction: Zn(s)+Cu2+(aq)→Cu(s)+Zn2+(aq) E0cell=1.10 V,ΔG0 is

The correct option is C –212.3KJΔG0=−nFE0cell ΔG0=−2mol×96500 C/mol×1.10V = –212300 CV = -212.3 KJ.

For the cell reaction:
$Z n (s) + C u^{2 +} (a q) \to C u (s) + Z n^{2 +} (a q)$
$E_{c e l l}^{0} = 1.10 V, Δ G^{0}$ is

The correct option is C –212.3KJ
$Δ G^{0} = - n F E_{c e l l}^{0}$
$Δ G^{0} = - 2 m o l \times 96500 C / m o l \times 1.10 V$
= –212300 CV = -212.3 KJ.