1. A voltaic cell prepared using aluminum and nickel has the following cell nota

Question

1. A voltaic cell prepared using aluminum and nickel has the following cell notation.

Al(s) | Al3+(aq) || Ni2+(aq) | Ni(s)

Which of the following reactions occurs at the anode?

A) Al(s) Al3+(aq) + 3e–

B) Al3+(aq) + 3e Al(s)
C) Ni(s) Ni2+(aq) + 2e–

D) Ni2+(aq) + 2e– Ni(s)

E) none of the above

2. The line notation, Al(s) | Al3+(aq) || Co2+(aq) | Co(s), indicates that?

A) Co is the reducing agent.
B) Co2+ ions are oxidized.
C) Al is oxidized.

D) Al3+ is the reducing agent.

E) Aluminum metal is the cathode.

3.

Given the voltaic cell below, find the cell potential. Zn(s), [Zn2+]=4.50M] || Cu(s), [Cu2+]=0.0120M

Cu2+(aq) + 2e– Cu(s)   E° = 0.34 V

Zn2+(aq) + 2e– Zn(s)   E° = -0.76 V

A) 1.23 V

B) 1.10 V

C) 1.18 V

D) 1.02 V

E) 1.08V

Explanation / Answer

Answer-

1) We are given a voltaic cell prepared using aluminum and nickel has the following cell notation.

Al(s) | Al3+(aq) || Ni2+(aq) | Ni(s)

We know in the voltaic cell on the left side there is anode and right side there is cathode. We also know in the voltaic cell at the anode there is oxidation and at the cathode there is reduction.

So at the anode there is oxidation of Al to Al3+

So answer is A) Al(s) Al3+(aq) + 3e–

2) In the given cell notation –

Al(s) | Al3+(aq) || Co2+(aq) | Co(s)

We know Al is get oxidized to Al3+ and Co2+ is reduced to Co

Al is the reducing agent , since it gets oxidized and CO2+ the reducing agent

So answer is C) Al is oxidized.

3) We are given,

Zn(s), [Zn2+]=4.50M] || Cu(s), [Cu2+]=0.0120M

Cu2+(aq) + 2e– Cu(s)   E° = 0.34 V

Zn2+(aq) + 2e– Zn(s)   E° = -0.76 V

We know in this cell at the anode there is Zn and Zn2+, means Zn gets oxidized and Cu2+ gets reduced, so the reduction potential of the half reaction of Zn get reversed

Zn(s) ----> Zn2+(aq) + 2e- Eo = 0.76 V

Cu2+(aq) + 2e– ---> Cu(s)   E° = 0.34 V

Zn(s) + Cu2+(aq) -----> Zn2+(aq) + Cu(s)   Eocell = 1.1 V

We know the Nernst equation-

Ecell = EoCell -0.0592/n * log Q

       = 1.1 V -0.0592/2*log 4.50 / 0.0120

    = 1.02 V

So answer is, D) 1.02 V

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