Electrochemical cells...please help! (1 Viewer)

[*~Unique~*]

Alright, I have conducted an experiment about electrochemical cells. The cell is:
Zn<SUP>2+</SUP>/Zn // Cu<SUP>2+</SUP>/Cu

A zinc metal electrode is placed in a solution of zinc nitrate [Zn(NO<SUB>3</SUB>)<SUB>2</SUB>], this is one half-cell. The other half-cell consists of a copper metal electrode placed in a copper nitrate [Cu(NO<SUB>3</SUB>)<SUB>2</SUB>] solution.

To the copper nitrate solution, sodium sulfide (Na<SUB>2</SUB>S) is added.

A black precipitate is formed because the copper ions, Cu<SUP>2+</SUP>, in the solution react with the sulfide ions, S<SUP>2-</SUP>, to produce insoluble copper sulfide (CuS). According to the following equation:

Cu(NO<SUB>3</SUB>)<SUB>2</SUB>(aq) + Na<SUB>2</SUB>S(aq) ® NaNO<SUB>3</SUB>(aq) + CuS(s)
This is right, no?

The voltmeter reading is ~0.1V. This reading is for the reaction between what exactly? Is it the reaction between the NaNO<SUB>3</SUB> and the zinc metal?

Can someone please tell me (or write) what reaction is causing the reading on the voltmeter.

I also need to write the reactions for the anode and cathode for this cell. Can someone please help me with that also.

Much appreciated,
Kira
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[onebytwo]

the potential diiference is produced when the zinc anode oxidises to form Zn2+, these ions then go into solution, the electrons travel along the external circuit where the ions in that half cell collect the electrons and adhere to the cathode, the site of reduction. i think you need to write these two half reactions and combine, them to form your net ionic eqtn.