ZnCl_2(aq)+ Al_(s), right arrow Zn_(s)_+ AlCl_3(aq) [AI^3^+] = 0.8 M [Zn^2^+] =
ID: 951243 • Letter: Z
Question
ZnCl_2(aq)+ Al_(s), right arrow Zn_(s)_+ AlCl_3(aq) [AI^3^+] = 0.8 M [Zn^2^+] = 1.3 M NERNST EQUATION E_cell_= E ^degree_cell -0.0592V/n log Q Balance the above equation using half reactions. Explain what the n value in the Nernst equation is by showing how you solved for it. Give the equation for the reaction quotient (Q) and solve for it. Are there units for Q? Solve for the standard potential for this cell (E^degree_cell) at 295K. First give the equation. Solve the Nernst equation for this cell at 295K.Explanation / Answer
1. 3ZnCl2 + 2Al =======> 3Zn + 2AlCl3
In this reaction, 2Al =======> 2Al+3 + 3e-
3Zn+2 + 3e- =======> Zn
This indicates zinc undergoes reduction acts as cathode and aluminium undergoes oxidation acts as anode,overall n=1
2) Q = [P]/[R]= [0.8]2/[1.3]3 = 0.291
3) Ecell = Ecathode -Eanode = -0.76 -(-1.66)= 0.9V
4) apply Nernst eqn, E = 0.9-(0.0591/1) * log 0.291
= 1.216V
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