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.
You need to login to perform this action.
You will be redirected in
3 sec