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CBSE Class 12-science Answered

Calculate EMF of the following cell at 298 K             Ni(s) ½ Ni2+(0.01M) ½½ Ag+(0.1M) ½Ag(s)
Asked by Topperlearning User | 04 Jun, 2014, 01:23: PM
answered-by-expert Expert Answer
Given : E° cell = 1.05V
            For galvanic cell
            Ni(s) + 2Ag+(aq) → Ni2+(aq) + 2Ag(s) ,the Nernst equation is 
           
             E cell = E0 Cell- ( 2.303x RT /nF ) log10 [Ni2+] / [Ag+]2
                        = 1.05 V – (0.059)/ 2 log10 [0.01] / [0.1]2
                        = 1.05 V – 0.0 = 1.05V as log10 10 = 0
Answered by | 04 Jun, 2014, 03:23: PM

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