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  • Help me please, - Redox Reaction and Electrochemistry - JEE Main-2

Given :E^{0}_{Fe^{3+}/Fe}= -0.036 V,E^{0}_{Fe^{2+}/Fe}= -0.439 \: V.The value of standard electrode potential for the change, Fe^{3+}_{\left ( aq \right )}+e^{-}\rightarrow Fe^{2+}_{\left ( aq \right )} will be

  • Option 1)

    -0.072 \: V

  • Option 2)

    0.385 \: V

  • Option 3)

    0.770 \: V

  • Option 4)

    -0.270 \: V

 

Answers (1)

best_answer

As we have learned

M(s) is Solid -

[M] = 1

- wherein

E_{M^{n+}/M}=E_{M^{n+}/M}^{0}-\frac{RT}{nf}ln\frac{1}{[M^{n+}]}

 

 Fe^{3+}+3e^-\rightarrow Fe \; \; \; \; E_{Fe^{3+}/Fe}^{0}= -0.036 V

Fe\rightarrow Fe^{2+}+2e^- \; \; \; \; E_{Fe^{2+}/Fe}^{0}= -0.439 V

______________________________

Fe^{3+}+e^-\rightarrow Fe^{2+}

 

So, E _{Fe^{3+}/Fe2+}^{0}= 2 E_{Fe/Fe^{2+}}^{0}+ 3 E_{Fe^{3+}/Fe}^{0}

E _{Fe^{3+}/Fe2+}^{0}=0.770 V

 

 

 


Option 1)

-0.072 \: V

Option 2)

0.385 \: V

Option 3)

0.770 \: V

Option 4)

-0.270 \: V

Posted by

SudhirSol

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