question_answer

                                        The following chemical reaction is occurring in an electrochemical cell: \[Mg(s)+2A{{g}^{+}}\,(0.0001\,M)\xrightarrow{\,}\,M{{g}^{2+}}\,(0.10M)+2Ag(s)\] Given \[E_{M{{g}^{2+}}/Mg}^{o}=-2.36\,\,V;\,\]\[E_{A{{g}^{+}}/Ag}^{o}=+0.81\,\text{V}\] For the cell, calculate/write: E° value for the electrode \[2A{{g}^{+}}/2Ag\] Standard cell potential (E°) Cell potential (E) Give the symbolic representation of the above cell Will the above cell reaction be spontaneous?            

Answer:

                                                     (i) E° value of the electrode; \[2A{{g}^{+}}/2Ag\] will remain the same i.e., 0.81 V. It does not change. (ii)           \[E_{cell}^{o}=E_{cathode}^{o}-E_{anode}^{o}=0.81-(-2.36)\,=3.17\,\text{V}\] (iii)          \[{{E}_{cell}}\,=E_{cell}^{o}-\frac{(0.0591\,V)}{2}\,\log \,\frac{[M{{g}^{2+}}(aq)]}{{{[A{{g}^{+}}(aq)]}^{2}}}\] \[=(3.17\,V)\,-\frac{(0.0591\,V)}{2}\,\log \,\frac{(0.1)}{{{(0.0001)}^{2}}}\] \[=(3.17\,V)-(0.2068\,V)\,=2.9632\,V\] (iv) Cell notation: \[Mg(s)/M{{g}^{2+}}\,(0.10\,M)\]\[||A{{g}^{+}}\,(0.0001M)/2Ag(s)\] (v) Since Mg is placed below Ag in the activity series, it is a stronger reducing agent. Therefore, the cell reaction is spontaneous.