What is the standard EMF of a cell that uses the #"Mg/Mg"^(2+)# and #"Cu/Cu"^(2+)# half-cell reactions at #25^@"C"# ?

Start by looking up the standard reduction potentials of the magnesium cation and of the copper(II) anion, which you can find here.

#"Mg"^(2+) + 2"e"^(-) rightleftharpoons "Mg", " "E_"red"^@ = - "2.372 V"#

#"Cu"^(2+) + 2"e"^(-) rightleftharpoons "Cu", " "E_"red"^@ = +"0.337 V"#

Now, your goal here is to figure out which element is being reduced and which element is being oxidized. To do that, you need to take a look at the values you have for the standard reduction potentials.

When magnesium cations are being reduced to magnesium metal, the negative value of the standard reduction potential tells you that the reduction equilibrium lies to the left, meaning that magnesium cations are more difficult to reduce than hydrogen cations.

In other words, magnesium releases electrons more readily than hydrogen.

When copper(II) cations are being reduced to copper metal, the positive value of the standard reduction potential tells you that the equilibrium lies to the right, meaning that copper(II) cations are easier to reduce than hydrogen cations.

In other words, copper releases electrons less readily than hydrogen.

When you connect these two electrodes, the element that has the more negative standard reduction potential will be oxidized because it will lose electrons more readily and the element that has the more positive standard reduction potential will be reduced because it will lose electrons less readily.

In your case, you have

#E_ ("red Mg"^(2+))^@ < E_ ("red Cu"^(2+))^@ implies {("Mg is being oxized to Mg"^(2+)),("Cu"^(2+)color(white)(.)"is being reduced to Cu") :}#

so you can say that magnesium will be oxidized to magnesium cations and copper(II) cations will be reduced to copper metal.

This means that you have--remember, when you flip the reduction half-reaction to get the oxidation half-reaction, you must change the sign of the standard reduction potential.

#{(color(white)(aaaaaaaa)"Mg" rightleftharpoons "Mg"^(2+) + 2"e"^(-), " "E_"oxi"^@ = - (-"2.372 V")), ("Cu"^(2+) + 2"e"^(-) rightleftharpoons "Cu", " "E_"red"^@ = +"0.337 V") :}#

To find the #"EMF"# of the cell, simply add the standard potentials of the two half-reactions.

#E_"cell"^@ = E_"oxi"^@ + E_"red"^@#

You will end up with

#E_"cell"^@ = +"2.372 V" + "0.337 V"#

#color(darkgreen)(ul(color(black)(E_"cell"^@ = +"2.709 V")))#

So, your galvanic cell will consist of a magnesium electrode and a copper electrode. Since magnesium is being oxidized and copper is being reduced, the magnesium electrode will be the anode and the copper electrode will be the cathode.