Q&A - Ask Doubts and Get Answers
Q

Please,please help me - Redox Reaction and Electrochemistry - JEE Main

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)
128 Views

    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.

Exams
Articles
Questions