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(a) Calculate $E^{\circ}_{cell}$ for the following reaction at 298 K:

$2Cr(s)+3Fe^{2+}(0.001M)\rightarrow 2Cr^{3+}(0.001)+3Fe(s)$

Given: $E_{cell}=0.261V$

(b) Using the $E^{\circ}$ values of A and B, predict which one is better for coating the surface of iron $\left [ E^{\circ}\left ( Fe^{2+}/Fe \right ) \right =0.44V]$ to prevent corrosion and why?

Given: $E^{\circ}\left ( A^{2+}/A \right )=-2.37V, E^{\circ}\left ( B^{2+}/B \right )=-0.14V$

Answers (1)

(a)

$E^{\circ}_{cell}$ = ?

T = 298 K

$E_{cell}=0.261V$

$2Cr(s)+3Fe^{2+}(0.001M)\rightarrow 2Cr^{3+}(0.001)+3Fe(s)$

Cell $=Cr\left | Cr^{3+}(0.001M) \right |\left | Fe^{2+}(0.001M) \right |Fe$

Anode $=\left ( Cr(s)\rightarrow Cr^{3+}(aq)+3e^{-} \right )\times 2$

Cathode $=\left ( Fe^{2+}+2e^{-} \rightarrow Fe(s)\right )\times 3$

n = 6

$E_{cell}=E^{\circ}_{cell}-\frac{0.0591}{n}\log \frac{\left [ Cr^{3+} \right ]^{2}}{\left [ Fe^{2+} \right ]^{3}}$

$E_{cell}=0.261-\frac{0.0591}{6}\log \frac{(0.01)^{2}}{(0.01)^{3}}$

$E_{cell}=0.261-\frac{0.0591}{6}\log 10^{2}$

$E_{cell}=0.261-\frac{0.0591\times 2}{6}$

$E_{cell}=0.2413V$

(b)

$E^{\circ}\left ( A^{2+}/A \right )=-2.37V$

$E^{\circ}\left ( B^{2+}/B \right )=-0.14V$

$E^{\circ}\left ( Fe^{2+}/Fe \right ) =0.44V$

A is better for coating the surface of iron to prevent corrosion takes place when iron is oxidised therefore to prevent it element A with higer oxidation potential will oxidise faster.

Posted by

Sumit Saini

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