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Gibb's Free Energy

For the cell,...

Question

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:

By increasing

[C u^{2 +}]

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By increasing

[Z n^{2 +}]

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By decreasing

[C u^{2 +}]

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By decreasing

[Z n^{2 +}]

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Solution

The correct option is D By decreasing $[Z n^{2 +}]$
Cell reaction:
$Z n (s) + C u^{2 +} (a q) \to Z n^{2 +} (a q) + C u (s)$
We know,
$E_{c e l l} = E^{o} - \frac{R T}{n F} l n \frac{[Z n^{2 +}]}{[C u^{2 +}]}$
Hence on increasing $[C u^{2 +}]$ negative part of the equation will be small, which will increase $E_{c e l l}$ .
And on decreasing $[Z n^{2 +}]$ negative part of the equation will be small again which will increase $E_{c e l l}$ .

Suggest Corrections

Similar questions

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

E_{c e l l}

Z n (s) + C u^{2 +} (0.1 M) \to Z n^{2 +} (1 M) + C u (s)

E_{c e l l}^{0}

E_{c e l l}

(2.303 \frac{R T}{F} = 0.0591)

Z n | Z n^{2 +} (a q) (1.0 M) | | C u^{2 +} (a q) (1.0 M) | | C u

0.350

C u^{2 +} (a q) + 2 e^{-} \to C u

- 0.763

Z n^{2 +} (a q) + 2 e^{-} \to Z n

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

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

E_{c e l l}

Q

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

E^{\circ}

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

1.10 V

25^{\circ} C

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

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