\(\therefore \) \([H^+]\) changes from \(1\) to \(10^{-7}\,M\)
Accordingly \({{E}_{red.}}=\,\frac{-\,0.059}{n}\,\log \,\frac{1}{[{{H}^{+}}]}\)
\( = \,0.059\,\,\log \,{10^{ - 7}}\)
i.e., \(0.059\times (-7)\,=\,-0.41\,volts\)
$(i)\, A3^-\rightarrow A^{2-} + e; E° = 1.5 \,V$
$(ii) \,B^{+}+ e \rightarrow B; E° = 0.5 \,V$
$(iii)\, C^{2+} + e \rightarrow C^{+}; E°= 0.5\, V$
$(iv)\, D \rightarrow D^{2+}+ 2e; E° = -1.15\, V$