Calculate the emf of given cell:- Ag/AgNO3// AgNO3/Ag ; E°Ag+/Ag=0.80V. [AgNO3]1= 0.005M, [AgNO3]2= 0.5M. (1) 1.18V, (2) 0.118V, (3) 2.36V, (4)0.236V

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Calculate the emf of given cell:- Ag/AgNO3// AgNO3/Ag ; E°Ag+/Ag=0.80V. [AgNO3]1= 0.005M, [AgNO3]2= 0.5M. (1) 1.18V, (2) 0.118V, (3) 2.36V, (4)0.236V

Asked by patra04011965 | 19 Jul, 2019, 08:19: AM

Expert Answer

The emf can be calculated as,

straight E subscript cell space equals space straight E subscript straight R space minus space straight E subscript straight L space space space space space space space space equals open square brackets straight E subscript Ag to the power of plus divided by Ag end subscript superscript straight o space plus space fraction numerator 0.0591 over denominator straight n end fraction space log space open square brackets Ag to the power of plus close square brackets subscript straight R close square brackets space minus space open square brackets straight E subscript Ag to the power of plus divided by Ag end subscript superscript straight o space plus space fraction numerator 0.0591 over denominator straight n end fraction space log space open square brackets Ag to the power of plus close square brackets subscript straight L close square brackets space space space space space space space space equals space fraction numerator 0.059 over denominator straight n end fraction space log open square brackets Ag to the power of plus close square brackets subscript straight R over open square brackets Ag to the power of plus close square brackets subscript straight L space space space space space space space equals space space fraction numerator 0.059 over denominator 1 end fraction space log fraction numerator 0.5 over denominator 0.005 end fraction space space space space space space space equals fraction numerator 0.059 over denominator 1 end fraction space log 100 space space space space space space space equals space 0.059 space cross times 2 space space space space space space space equals space 0.1182 space straight V space

Answered by Ramandeep | 26 Jul, 2019, 01:00: PM

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