(a) Calculate the emf of the cell Mg(s) | Mg2+ (0.1M) || Cu2+ (1 × 10-3 M) | Cu (s) Given: E° Cu2+/Cu = +0.34V, E° Mg2+/Mg = -2.37 V

First of all one should write the balanced equation taking place.
We will see that the no of electrons involved in the reaction is 2
Then, one should calculate the E⁰ of the cell, as
E⁰cell = reduction potential of electrode where reduction is taking place – Reduction potential of the electrode where oxidation is taking place
On finding the E0 of the cell, one should apply the Nernst Equation to get the answer.

Mg(s)/Mg+2 (0.1 M) //Cu+2(1×10-3M)/Cu(s)for net cell equationMg=Mg+2 + 2e- (anode)Cu+2=Cu (Cathode)...........................................Mg+Cu+2 = Mg+2 + GuE=E` cell-o.o59÷2 ln(Mg+2÷ Cu+2)E` cel=Ecathode-Eaonde=0.34-(-2.37)=2.71E = 2.71 -0.0295ln {O.1 ÷1×10^-3}= 2.71 - 0.0295×2.303 Log{o. 1÷1×l0^-3}= 2.71 - 0.0295×4.606= 2.71 -0.135877= 2.574128 V