The equation relating the standard cell potential of a voltaic cell to the stand

Question

The equation relating the standard cell potential of a voltaic cell to the standard reduction potential of the cathode and anode reactions is the following. E^compositefunction_cell = E^compositefunction_cathode = E^compositefunction_anode In this equation E^compositefunction_cell is the standard cell potential, E^compositefunction_cathode is the standard reduction potential of the cathode reaction, and E^compositefunction_anode is the standard reduction potential of the reverse of the anode reaction (the reduction corresponding to the reverse of the oxidation reaction). Use this relationship to solve the problem below. The standard reduction potential for Au^3+ (aq) is 1.50 v. The half-reaction for the reduction of Au^3+ (aq) is the following. Au^3+ (aq) + 3 e^- rightarrow AU(S) The standard reduction potential for Hg_2^2+ (aq) is 0.80 v. The half-reaction for the reduction of Hg_2^2+(aq) is the following. Hg_2^2+ (aq) + 2e^- rightarrow 2 Hg(l) Using this information, calculate E^compositefunction_cell for the voltaic cell powered by the following spontaneous redox reaction. 2 Au^3 + (aq) + 6 Hg(l) rightarrow 3 Hg_2^2 + (aq) + 2 AU(s) Identify the half-reactions occurring in the redox equation. The reduction half-reaction is the following. (Include states-of-matter under the given conditions in your answer.) The oxidation half-reaction is the following.(Include states-of-matter under the given conditions in your answer.)

Explanation / Answer

The one with higher standard reduction potential will undergo reduction

So, reduction half reaction is:

Au3+(aq) + 3e- —> Au(s)

So,

Eo cathode = 1.50 V

The oxidation half reaction is:

2Hg(l) —> Hg22+(aq) + 2e-

So,

Eo anode = 0.80 V

Eo cell = Eo cathode - Eo anode

= 1.50 V - 0.80 V

= 0.70 V

Answer: 0.70 V

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