For Mg^{2+} + 2e^{-} rightarrow Mg_{(s)},E^{0 | vedclass

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(D) Given: $E^{0}_{Mg^{2+}/Mg} = -2.37 \, V$ and $E^{0}_{Cu^{2+}/Cu} = +0.33 \, V$. Since $E^{0}_{Cu^{2+}/Cu} > E^{0}_{Mg^{2+}/Mg}$,$Cu$ acts as the c

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$+2.7$

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(D) Given: $E^{0}_{Mg^{2+}/Mg} = -2.37 \, V$ and $E^{0}_{Cu^{2+}/Cu} = +0.33 \, V$. Since $E^{0}_{Cu^{2+}/Cu} > E^{0}_{Mg^{2+}/Mg}$,$Cu$ acts as the cathode and $Mg$ acts as the anode. The standard cell potential is calculated as: $E^{0}_{cell} = E^{0}_{cathode} - E^{0}_{anode}$ $E^{0}_{cell} = 0.33 \, V - (-2.37 \, V)$ $E^{0}_{cell} = 0.33 \, V + 2.37 \, V = 2.7 \, V$.

For $Mg^{2+} + 2e^{-} \rightarrow Mg_{(s)}$,$E^{0} = -2.37 \, V$ and for $Cu^{2+} + 2e^{-} \rightarrow Cu_{(s)}$,$E^{0} = +0.33 \, V$. The $emf$ of the cell $(E^{0}_{Cell})$ is .......... $V$.

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