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If you construct a Galvanic / Voltaic cell using a Ni(s) / Ni2+(aq) half cell and...

If you construct a Galvanic / Voltaic cell using a Ni(s) / Ni2+(aq) half cell and an Al(s)/ Al3+(aq)half cell,and connect the whole cell to a voltmeter by connecting the Al half cell to the positive terminal of the voltmeter, what will be the display?

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Answer #1

In a galvanic (voltaic) cell, the anode is considered negative and the cathode is considered positive. The redox reaction in such cells is spontaneous.

Standard electrode potentials ( E0) for:

Ni2+(aq) + 2e- = Ni(s) , E0 = -0.25 V     

Al3+(aq) + 3e- = Al(s) , E0 = -1.66 V ; The value is less than that for Nickel, so Nickel would get reduced and Aluminum will be oxidized:

The half cell reactions are:

At cathode: Reduction takes place:

Ni2+(aq) + 2e- = Ni(s) , E0 = -0.25 V

At anode: Oxidation takes place:

Al(s) = Al3+(aq) + 3e- , E0 = +1.66 V

Overall reaction:

3 Ni2+(aq) + 2Al(s) = 3 Ni(s) + 2 Al3+(aq) ,   

E0 cell = +1.66 V -0.25 V = 1.41 V           Note that E does not change if reaction is multiplied by any number;

When anode (Al half cell) is connected to the positive terminal of voltameter, it shows positive reading of voltage = 1.41 V

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