(i) The electrode reactions are At anode : $Mg_{(s)} \rightarrow Mg^{2+} (0.001 M) + 2e^{-}$ At cathode : $Cu^{2+} (0.0001 M) + 2e^{-} \rightarrow Cu_{(s)}$ Net reaction : $Mg_(s) + Cu^{2+} (0.001 M)$ $\rightarrow Mg^{2+} (0.0001 M) + Cu_{(s)}$ The Nernst equation for this cell at 25°C $E _{\ {cell}}=E°_{\ {cell}} -\ {0.0591}/{2} \log { [\ {Mg}^{2+} ]}/{ [\ {Cu}^{2+}\ ]}$ Where $E°_{\ { anode } }=-2.37 {V} ; E°_{\ { cathode }} =+0.34 {V}$ $∴\ E°_{\ { cell }} =E°_{\ { cathode }} -E°_{\ { anode }}$ $=(+0.34 \ {V})-(-2.37 {V})$ $=+2.71 \ {V}$ The cell emf is then given by $E_{ { cell }}=2.71- {0.0591}/{2} \log (\ {0.001}/{0.0001}\ )$ $E_{\ { cell }}=\ (2.71-\ {0.0591}/{2} \log 10\ ) \ {V}$ $= 2.71 – 0.03 = 2.68 V$ (ii) The electrode reactions are At anode : $Fe_{(s)} \rightarrow Fe^{2+} (0.001 M) + 2e^{-}$ At cathode : $2H^{+}(1M) + 2e^{-} \rightarrow H_2 (1 bar)$ Net reaction : $Fe_{(s)} + 2H^{+}(1M) \rightarrow Fe^{2+} (0.001 M) + H_2 (1 bar)$ The Nernst equation of this cell at 25°C $E_{\ {cell}}=E°_{\ {cell}} -\ {0.0591}/{2} \log \ {\ [\ {Fe}^{2+} ]\ ( {pH}_{2}\ )}/{\ [\ {H}^{+}\ ]^{2}}$ $E°_{\ { cell }} =E°_{\ { cathode }} -E°_{\ { anode }}$ $=E^{\circ}_{\ {H}^{+} | \ {H}_{2}} - E^{\circ}_{\ {Fe}^{2+} | {Fe}}$ $= 0.000 V – (–0.44 V)$ $= +0.44 V$ The cell emf is then given by $E_{c e l l}=0.44- {0.0591}/{2} \log {0.001 × 1}/{(1)^{2}}$ $=0.44-0.0296 \log (\ {1}/{1000}\ )$ $=0.44-0.0296 \log (10^{-3} )$ $=0.44+(3 × 0.0296)$ $=0.44+0.0888$ Therefore, $E_{cell} = +0.53 V$ (iii) The electrode reactions are At anode : $Sn_{(s)} \rightarrow Sn^{2+} (0.05 M) + 2e^{-}$ At cathode : $2H^{+}(0.02 M) + 2e^{-} \rightarrow H_2 (1 bar)$ Net reaction :$Sn_{(s)} + 2H^{+}(0.02 M) \rightarrow Sn^{2+} (0.05 M) + H_2 (1 bar)$ The Nernst equation of this cell at 25°C $E_{\ {cell}}=E°_{\ {cell}} -\ {0.0591}/{2} \log {\ [\ {Sn}^{2+}\ ]\ (\ {pH}_{2}\ )}/{ [\ {H}^{+}\ ]^{2}}$ $E°_{\ {cell}} =E°_{\ {H}^{+} | {H}_{2}} -E°_{\ {Sn}^{2+} | { Sn}}$ $= 0.000 V – (–0.14 V)$ $= +0.14 V$ or, $E_{\ { cell }}=E°_{\ { cell }} -0.0296 \log {0.05 × 1}/{(0.02)^{2}}$ $=E°_{\ { cell }} -0.0296 \log \ ( {0.05}/{0.0004}\ )$ $=E°_{\ {cell}} -0.0296(\log 125)$ $=E°_{ { cell }} -0.0296 × 2.0969$ $=E°_{\ { cell }} -0.06$ $E_{cell} = 0.14\ –\ 0.06 = 0.08 V$ (iv) The electrode reactions are At anode :$2Br^{-}(0.01 M) \rightarrow Br_{2(l)} + 2e^{-}$ At cathode : $2H^{+}(0.03 M) + 2e^{-} \rightarrow H_2 (1 bar)$ Net reaction : $2H^{+}(0.03 M) + 2Br^{-}(0.01 M) \rightarrow Br_{2(l)} + H_2 (1 bar)$ The Nernst equation of this cell at 25°C is $E_{\ {cell}}=E°_{ {cell}} - {0.0591}/{2} \log { [ {Br}_{2(l)} ]\ (\ {pH}_{2}\ )}/{\ [\ {H}^{+}\ ]^{2} [\ {Br}^{-} ]^{2}}$ $E_{ {cell}}^{0}=E°_{\ {H}^{+}{∣H_2}} -E°_{ {Br}_{2}{∣Br^{-}} }$ $=0-1.08$ $=-1.08-0.0296 × \log {1}/{(0.03)^{2}(0.01)^{2}}$ $=-1.08-0.0296× \log \ (\ {10^{8}}/{9}\ )$$=-1.08-0.0296\ (\log 10^{8}-\log 9\ ) {V}$$=-1.08-0.0296(8-0.9542) \ {V}$$= –1.08 – 0.0296 (7.0457)$$E_{cell }= –1.08 – 0.21 = –1.29 V$