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