Start by looking increase the standard palliation potentials that the magnesium cation and also of the copper(II) anion, which girlfriend can discover here.

You are watching: Calculate the standard emf of a cell that uses the mg/mg2+ and cu/cu2+ half-cell reaction at 25°c.

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

Now, her goal here is to number out which facet is being reduced and which aspect is gift oxidized. To execute that, you must take a look in ~ the values you have actually for the standard reduction potentials.

When magnesium cations room being decreased to magnesium metal, the negative value of the typical reduction potential speak you the the reduction equilibrium lies to the left, meaning that magnesium cations room more challenging to reduce than hydrogen cations.

In various other words, magnesium releases electrons an ext readily than hydrogen.

When copper(II) cations space being lessened to copper metal, the positive worth of the standard reduction potential tells you the the equilibrium lies come the right, an interpretation that copper(II) cations are simpler to alleviate than hydrogen cations.

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

When you attach these two electrodes, the element that has the more negative typical reduction potential will be oxidized since it will shed electrons more readily and the facet that has actually the more positive traditional reduction potential will be reduced because it will shed electrons less readily.

In her case, girlfriend have

#E_ ("red Mg"^(2+))^

so you have the right to say the magnesium will be oxidized come magnesium cations and also copper(II) cations will be reduced to copper metal.

This means that friend have--remember, when you flip the palliation half-reaction to gain the oxidation half-reaction, you need to change the sign the the conventional 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 discover the #"EMF"# that the cell, simply include the standard potentials that the two half-reactions.

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= E_"oxi"^
+ E_"red"^

You will end up with

= +"2.372 V" + "0.337 V"#

= +"2.709 V")))#

So, your galvanic cell will consist that a magnesium electrode and a copper electrode. Due to the fact that magnesium is being oxidized and also copper is gift reduced, the magnesium electrode will certainly be the anode and the copper electrode will certainly be the cathode.