For the cell reaction-Zn s - Cu 2- aq - Cu s - Zn 2- aq E cell -1-10 V - G - is-A- 212-3 kJB- 106-05 kJC- -212-3 kJD- -424-6 kJ

The correct option is C 212.3 kJG^∘ = nFE_cell^∘ = 2 × 96500 × 1.10 = 212300 J = 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:
$Z n (s) + C u^{2 +} (a q) \to C u (s) + Z n^{2 +} (a q)$
$E_{c e l l}^{\circ} = 1.10 V, △ G^{\circ}$ is:

212.3 kJ

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106.05 kJ

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-212.3 kJ

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-424.6 kJ

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Solution

The correct option is C -212.3 kJ
$△ G^{\circ} = - n F E_{c e l l}^{\circ} = - 2 \times 96500 \times 1.10 = - 212300 J = - 212.3 kJ$

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Similar questions

Q. 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}

Q. 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}^{\circ} = 1.10 V, △ G^{\circ}

is:

Q. Select the correct statement for a galvanic cell involving the following reaction:

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

E^{\circ}

for the electrochemical cell,

Z n (s) | Z n^{2 +} 1 M (a q .) ∥ C u^{2 +} 1 M (a q .) | C u (s)

1.10 V

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. The equilibrium constant for the cell reaction,

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

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) E∘cell=1.10 V,△G∘ is:

The correct option is C -212.3 kJ△G∘=−nFE∘cell=−2×96500×1.10=−212300 J=−212.3 kJ

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

The correct option is C -212.3 kJ
$△ G^{\circ} = - n F E_{c e l l}^{\circ} = - 2 \times 96500 \times 1.10 = - 212300 J = - 212.3 kJ$