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  • Confused! kindly explain, - Redox Reaction and Electrochemistry - JEE Main-2

In a cell that utilizes the reaction

Zn_{(s)}+2H^{+}{_{\left ( aq \right )}}\rightarrow Zn^{2+}{_{\left ( aq \right )}}+H_{2\left ( g \right )}

addition of H_{2}SO_{4} to cathode compartment, will

  • Option 1)

    lower the E and shift equilibrium to the left

  • Option 2)

    lower the E and shift equilibrium to the right

  • Option 3)

    increase the E and shift the equilibrium to the right

  • Option 4)

    increase the E and shift the equilibrium to the left

 

Answers (1)

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

 

 

Zn_{(s)}+2H^{+}{_{aq}}\rightleftharpoons Zn^{2+}{_{aq}}+H_{2(g)}

E_{cell}= E^{\circ}{_{cell}}-\frac{0.059}{2}\log \frac{\left [ Zn^{2+} \right ]\times p_{H_{2}}}{\left [ H^{+} \right ]^{2}}

On adding H_{2}SO_{4} the \left [ H^{+} \right ] will increase therefore E_{cell} will also increase and the equilibrium will shift  towards the right.


Option 1)

lower the E and shift equilibrium to the left

This solution is incorrect.

Option 2)

lower the E and shift equilibrium to the right

This solution is incorrect.

Option 3)

increase the E and shift the equilibrium to the right

This solution is correct.

Option 4)

increase the E and shift the equilibrium to the left

This solution is incorrect.

Posted by

Sabhrant Ambastha

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