Given- E-Cr3- Cr - -0-72 V- E-Fe2- Fe - -0-42 V-The potential for the cell Cr-Cr3-0-1 M-Fe2-0-01 M-Fe is-

The correct option is D 0.26 V Cr|Cr^3+(0.1 M)||Fe^2+(0.01 M)|Fe Oxidation half cell Cr→ Cr^3+ + 3e^ ]× 2 Reduction half cell Fe^2+ + 2e^ → F ...

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Given, E∘Cr...

Question

Given, $E_{C r^{3 +} / C r}^{\circ} = - 0.72 V, E_{F e^{2 +} / F e}^{\circ} = - 0.42 V$ .
The potential for the cell $C r | C r^{3 +} (0.1 M) | | F e^{2 +} (0.01 M) | F e$ is:

0.26 V

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0.399 V

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- 0.399 V

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- 0.26 V

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Solution

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Similar questions

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

E_{C r^{3 +} |}^{o}

_{Cr}

= - 0.72 V, E_{F e^{2 +} | F e}^{o} = - 0.42

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

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

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