A voltaic cell contains two half-cells. One half-cell contains a gold electrode

Question

A voltaic cell contains two half-cells. One half-cell contains a gold electrode immersed in a 1.00 M Au(NO_3)3 solution. The second half-cell contains a zinc electrode immersed in a 1.00 M Zn(NO_3)2 solution. Au^3(aq) + 3 e rightarrow Au(s) E_red = + 1-498 V Zn^2 + (aq) + 2 e rightarrow Zn(s) E_red = - 0.762 V Using the standard reduction potentials given above, predict the standard cell potential of the voltaic cell. Write the overall balanced equation for the voltaic cell. (Include states-of-matter under the given conditions in your answer.)

Explanation / Answer

Reaction at cathode :

Gold has high reduction potential so Au3+ reduces at cathode

Au3+(aq) + 3e- ---> Au(s)

2Au3+ (aq) + 6e- -----> 2Au(s) EoAu3+/Au = 1.498 V

Reaction at anode:

Zn has oxidize at anode

Zn(s) ---> Zn+2(aq) + 2e-

3Zn(s) ---> 3Zn+2(aq) + 6e- EoZn/Zn2+ = 0.762 V

Eocell = Eocathode + Eoanode

Eocell = 1.498 + 0.762 V

Standard cell potential, Eocell = 2.26 V

b)

Adding the both balanced half cell reactions

2Au3+(aq) + 6e- -----> 2Au(s)

3Zn(s) ---> 3Zn+2(aq) + 6e-

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2Au3+(aq) +   3Zn(s)---> 2Au(s) + 3Zn+2(aq)

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