Chemistry problem

Question

Determine the nonstandard cell potential for an electrochemical cell based on the following two half-reactions:

Oxidation (Anode): Mg(s)→Mg2+(aq, 2.75 M)+2 e−

Reduction (Cathode): Pb2+(aq, 0.200 M)+2 e−→Pb(s)

+2.27 V

+2.21 V

−2.53 V

Answer 1

Redox equation: Mg(s) + Pb²⁺(aq) ==> Mg²⁺(aq) + Pb(s)

Nernst equation:

E = Eºcell - RT/nF ln Q

@ 298K and converting to log base 10, this becomes...

E = Eºcell - 0.0592/n log Q

Find Eºcell first:

Eºcell = -0.13 -(-2.38) = 2.25 V

Next, find Q:

Q = [Mg2+] / [Pb2+] = (2.75 M) / (0.200 M) = 13.75

Plugging in to the Nernst equation:

E = 2.25 - 0.0592/2 log 13.75 = 2.25 - (0.0296)(1.14)

E = 2.25 - 0.034

E = 2.22 V

(be sure to check my math)