Consider the following galvanic cell: Mg(s)|Mg^2+ (aq)| |Al^3+ (aq)|Al(s) Standa

Question

Consider the following galvanic cell: Mg(s)|Mg^2+ (aq)| |Al^3+ (aq)|Al(s) Standard Reduction Potentials: Mg^2+ +2e^1- rightarrow Mg(s) E^o -2.37V Al^3+3e^1- rightarrow Al(s) E^o-1.66V Give the equation for the spontaneous cell reaction that produces charge flow. Indicate which substance is oxidized and which is reduced in this reaction. Identify the anode and the cathode electrodes for this cell. Calculate the E. How will the potential difference (voltage) of the galvanic cell Cu(s) + 2Ag^1+ (aq) rightarrow Cu^2+ (aq) + 2Ag(s) be affected by the addition of Nacl(aq)? Explain. How does increasing the mass of the anode affect the voltage? Justify your response. Consider the reduction potentials: Na^1+ e^1+ rightarrow Na(s) E^o -2.71 V 2H_2O + 2e^1- rightarrow H_2 + 2OH^1- E^o -2.37 V The electrolysis of aqueous NaCl does not produce metallic sodium. Explain

Explanation / Answer

Given the half cell reactions are

Mg2+ + 2e-1 -------> Mg (s)        E0 = -2.37 V

Al3+ + 3e-1 --------> Al (s)          E0 = -1.66 V

(a) The more negative the reduction potential of a metal ion, the more difficult it is to reduce the metal ion. Consequently, it is easier to oxidize the metal to the corresponding metal ion.

Mg/Mg2+ system has the more negative reduction potential and therefore, Mg will be oxidized to Mg2+ easily while Al3+ will be reduced. Therefore, the spontaneous cell reaction which will produce a current is

3 Mg (s) + 2 Al3+ -------> 3 Mg2+ + 3 Al (s)

(b) Mg is oxidized to Mg2+ while Al3+ is reduced in the reaction. Since oxidation occurs at the anode, the anode will be Mg (s) or Mg (s) in conjunction with an inert Pt electrode. Reduction occurs at the cathode and hence the cathode in this case will be an Al (s) electrode or an Al (s) electrode in conjunction with a Pt electrode.

(c) For a spontaneous reaction, the E0cell must be positive.

The standard reduction potentials of the two half cells are as below (note that we will consider the oxidation potential, the reverse of the reduction potential, for Mg/Mg2+ since Mg is oxidized to Mg2+). The cell potentials are

Mg (s) --------> Mg2+ + 2e-1      E0ox = +2.37 V

Al3+ + 3e-1 ---------> Al (s)        E0red = -1.66 V

E0cell = E0red + E0ox = (-1.66 V) + (+2.37 V) = (-1.66 V) + (2.37 V) = 0.71 V (ans)

(d) The given redox reaction is

Cu (s) + 2 Ag+1 (aq) --------> Cu2+ (aq) + 2 Ag (s)

The cell potential (under non-standard conditions) is given by

E = E0 – 0.0591/2*log [Cu2+]/[Ag+1]2

E0 is the standard reduction potential of the cell.

When NaCl is added to the system, Ag+1 combines with Cl-1 to form AgCl (s) as below:

Ag+1 (aq) + Cl-1 (a) ------> AgCl (s)

The above precipitation of AgCl reduces the effective concentration of Ag+; i.e, [Ag+] decreases. The reaction quotient for the reaction, Q = [Cu+2]/[Ag+1]2 increases. Since the temperature remains constant, hence Q must stay constant. Therefore, to counter the reduced [Ag+], the denominator must decrease, i.e., the reverse reaction is favored, producing more Ag+. Therefore, the cell potential must increase and reach a value close to E0.

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