Given $E_{C r^{3 +} |}^{o}$ $_{Cr}$ $= - 0.72 V, E_{F e^{2 +} | F e}^{o} = - 0.42$ V.

The potential for the cell

Cr $| C r^{3 +} (0.1 M) ∥ F e^{2 +} (0.01 M) |$ Fe is:

- 0.339 V

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- 0.26 V

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0.26V

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0.339V

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Solution

The correct option is C $0.26V$
Cell reaction;
$2 C r (s) + 3 F e^{2 +} (a q) \to 2 C r^{3 +} (a q) + 3 F e (s)$

Now,
$E_{c e l l} = {E^{o}}_{F e^{2 +} | F e} - {E^{o}}_{C r^{3 +} | C r} - \frac{0.0591}{n} l o g \frac{[C r^{3 +}]^{2}}{[F e^{2 +}]^{3}}$

$E_{c e l l} = - 0.42 - (- 0.72) - \frac{0.0591}{6} l o g \frac{(0.1)^{2}}{(0.01)^{3}} = 0.26 V$

Option C is correct.

Suggest Corrections

Similar questions

E_{C r^{3 +} / C r}^{\circ} = - 0.72 V, E_{F e^{2 +} / F e}^{\circ} = - 0.42 V

C r | C r^{3 +} (0.1 M) | | F e^{2 +} (0.01 M) | F e

C r | C r^{3 +} (0.1 M) | | F e^{2 +} (0.01 M) | F e

E_{C r^{3 +} / C r}^{\circ} = - 0.75 v o l t; E_{F e^{2 +} / F e}^{\circ} = - 0.45 v o l t)

E_{\frac{C r^{3 +}}{C r}}^{\circ} = - 0.72 V E_{\frac{F e^{2 +}}{F e}}^{\circ} = - 0.42 V

\frac{C r}{C r^{3 +}} (0.1 M) | | F e^{2 +} (0.01 M) | F e i s

Given, $E_{C r^{3 +} / C r}^{0} = - 0.72 V, E_{F e^{2 +} / F e}^{0} = - 0.42 V$

The potential for the cell

C r | C r^{3 +} (0.1 M) | | F e^{2 +} (10.0 M) | F e

E_{C r^{3 +} / C r}^{\circ} = - 0.72, E_{F e^{2 +} / F e}^{\circ} = - 0.42 V

C r | C r^{3 +} (0.1 M) | | F e^{2 +} (0.01 M) | F e

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