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Given $E^{\circ}{_{Cr3+/Cr}}= -0.72 V,E^{\circ}{_{Fe2+/Fe}}= -0.42V$

The potential for the cell

$Cr\mid Cr^{3+}\left ( 0.1M \right )\parallel Fe^{2+}\left ( 0.01M \right )\mid Fe$ is

Option 1)
$-0.26V$

Option 2)
$0.26V$

Option 3)
$0.339 \: V$

Option 4)
$-0.339\: V$

Answers (1)

best_answer

As we learnt in

Electrode Potential(Nerst Equation) -

$M^{n+}(aq)+ne^{-}\rightarrow M(s)$

- wherein

$E_{(M^{n+}/M)}=E_{(M^{n+}/M)}^{0}-\frac{RT}{nf}ln\frac{[M]}{[M^{n+}]}$

$[M^{n+}]$ is concentration of species

F= 96487 C $moi^{-1}$

R= 8.314 $JK^{-1}moi^{-1}$

T= Temperature in kelvin

$Cr\rightarrow Cr^{3+}+3e^{-}\: \: \: \: \: \:\: \: \: \: \: \: \: E^{\circ}{_{red}}= -0.72V$

$\frac{Fe^{2+}+2e^{-}\rightarrow Fe}{2Cr+3Fe^{2+}\rightarrow 2Cr^{3+}+3Fe}\; \; \; \; \; \; \; E^{\circ}{_{red}}= -0.42\: V$

$E^{\circ}{_{cell}}= E^{\circ}{_{cathode}}-E^{\circ}{_{anode}}= -0.42-\left ( -0.72 \right )$

$E^{\circ}{_{cell}}= 0.3$

According to Nernst equation

$E{_{cell}}= E^{\circ}{_{cell}}-\frac{0.059}{n_{cell}}\log_{10}\frac{\left [ Cr^{3+} \right ]^{2}}{\left [ Fe^{2+} \right ]^{3}}$

$E_{cell}=0.3-\frac{0.059}{6}log_{10}\frac{\left ( 0.1 \right )^{2}}{\left ( 0.01 \right )^{3}}$

$E_{cell}=0.3-\frac{0.059}{6}log_{10}10^{4}$

$E_{cell}=0.3-0.039$

$\therefore \; \; E_{cell}=0.261\, V$

Option 1)

$-0.26V$

This option is incorrect.

Option 2)

$0.26V$

This option is correct.

Option 3)

$0.339 \: V$

This option is incorrect.

Option 4)

$-0.339\: V$

This option is incorrect.

Posted by

divya.saini

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