Anonymous
Anonymous asked in Science & MathematicsChemistry · 2 months ago

# galvanic cell potential?

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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• Voltmeters have a positive and negative terminal and, by convention, a red wire is connected to positive terminal and a black wire is connected to the negative terminal. The sign of the reading on the voltmeter tells us the spontaneous direction of the electron flow. If the black wire is connected to the anode, then the voltmeter will show a positive value. If we reverse this, by connecting the black wire to the cathode, then the voltmeter will show a negative value. Remember: if the voltmeter is showing a positive value, then the spontaneous direction of electron flow is from the black wire to the red wire. Electrons flow from the anode (electron provider or electron source) to the cathode (electron receiver or electron sink). If the voltmeter has a positive reading, the black wire is on the anode and the red wire is on the cathode.

The above statement was abstracted from the website:

Al ⇒ Al3+ + 3e- +1.66 (anode)

Ni2+ + 2e- ==⇒ Ni -0.23 (cathode)

-----------------------------------

2Al + 3Ni2+ ⇒ 2Al3+ + 3Ni E^o = +1.43 volts

Since the Al is connected to the positive terminal of the voltmeter, the display will show -1.43 volts.

Note: your reduction potential table might have different values for the Al and Ni couples.

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