E^{\circ}_{cell} for the given redox reaction is 2.71 V.

Mg(s)+Cu^{2+}_{\left (0.01M \right )}\rightarrow Mg^{2+}_{\left ( 0.001M \right )}+Cu(s)

Calculate  E_{cell} for the reaction. Write the direction of flow of current when external opposite potential is applied us:

(i) less than 2.71V

(ii) greater than 2.71V

 

 

 

 
 
 
 
 

Answers (1)

According to Nernst equation

E_{cell}= E^{\circ}_{cell}- \frac{2.303RT}{nF}\log\frac{C_{2}}{C_{1}}

E_{cell}= 2.71- \frac{2.303\times 8.314\times 298}{2\times 96500}\log\frac{0.001}{0.01}

E_{cell}=2.74V

(i) When an external opposite potential is applied less than 2.71V, the direction of flow of current would remain same.

(ii) When an external opposite potenial is applied more than 2.71V, the direction of flow of current would be reversed.

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