For the cell- Crs - Cr3- -Fe2- - Fes G- - -nFE- the value of n will be-

The correct option is D 6Cr → Cr ^3+ +3e ^ ........(1)Fe ^2+ +2e ^ → Fe.........(2)(1)x2 and (2)x 3, #160;2Cr +3Fe ^2+ ...

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Gibb's Free Energy

For the cell:...

Question

For the cell:
$C r (s) | C r^{3 +} | | F e^{2 +} | F e (s)$
$△ G^{\circ} = - n F E^{\circ}$ ; the value of $n$ will be:

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Solution

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C r (s) | C r^{3 +} | | F e^{2 +} | F e (s)

n

E = E^{\circ} - \frac{0.059}{n} {log}_{10} Q

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

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

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

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)

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