Answer:
Explanation:
From the information given:
left half-cell = 3.00 mM M(NO)
right half-cell = 3.00 M M(NO)
Since the electrode are the same in both cells, then the concentration for the cell are also the same.
Negative electrode = Anode = lower concentration = 3.00 mM
Positive electrode = cathode = higher concentration = 3.00 M
Thus, right half cell will be postive electrode.
To determine the concentration cell:
[tex]Ecell =\Big( \dfrac{2.303\times R\times T}{nF} \Big)log\Big(\dfrac{[cathode]}{[anode]}\Big)[/tex]
SInce [Cathode] > [anode],
[tex]R = 8.314 J/K/mol, \\ \\ T = 273+20 = 293 K \\ \\ Faraday's constant (F)= 96500 C/mol[/tex]
n = 3
[tex]Ecell ={ \dfrac{2.303\times 8.314\times 293 }{(3\times 96500)}} log\dfrac{3}{0.003}[/tex]
[tex]\mathbf{E_{cell} = 0.0582 V } \\ \\ \mathbf{E_{cell} = 0.06 V}[/tex]
