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  • Calculate the emf of the following cell at  : <img alt="Al(s)\left | Al^{3+}(0.001M) \right |\left | \right |0.1 Ni^{2+}\left | Ni(s) \

Calculate the emf of the following cell at 25^{\circ}C :

Al(s)\left | Al^{3+}(0.001M) \right |\left | \right |0.1 Ni^{2+}\left | Ni(s) \right |

Given: E^{\circ}_{\left ( Ni^{2+}/Ni \right )}= -0.25V

            E^{\circ}_{\left ( Al^{3+}/Al \right )}= -1.66V

(log 2 = 0.3010, log 3 = 0.4771)

 

 

 

 
 
 
 
 

Answers (1)

Cathode Half cell reaction:

Ni^{2+}+2e^{-}\rightarrow Ni, E^{\circ}= -0.25V

Anode Half cell reaction:

Al\rightarrow Al^{3+}+3e^{-}, E^{\circ}= +1.66V

Net reaction:

3Ni^{2+}+2Al\rightarrow 3Ni +2Al^{3+}, E^{\circ}=1.66+ (-0.25)=+1.41V

Transfer of 6e- takes place, hence n = 6

E^{\circ}=+1.41V

According to Wernest Equation, emf of a cell at 25^{\circ}C is calculated as,

E_{cell}= E^{\circ}_{cell}-\frac{0.0591}{n}\log \frac{\left [ product \right ]}{\left [ reactant \right ]}= 1.41-\frac{0.0591}{6}\log \frac{\left [ Al^{3+} \right ]}{\left [ Ni^{2+} \right ]}

          = 1.41 - 0.00985 \times log \frac{0.001}{0.1}= 1.41-0.00985 log (10^{-2})= 1.41-0.00985 \times (-2)= 1.41+0.197 = 1.43V

Posted by

Sumit Saini

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