*A solid piece of Copper (Cu) was used as the electrode in each side of the cell
ID: 837927 • Letter: #
Question
*A solid piece of Copper (Cu) was used as the electrode in each side of the cell and the answers to party a and b are the solutions corresponding to the anode and the cathode.
PLEASESHOW WORK FOR PARTG IN ORDER TO GET POINTS.... THANK YOU
a. Identify the anode electrode. --> 0.1 M CuSO4
b.Identify the cathode electrode. -->1 M CuSO4
c. Write the balanced oxidation half reaction and its standard oxidation potential.
d. Write the balanced reduction half reaction and its standard oxidation potential.
e. Write the overall balanced redox reaction and its standard cell potential.
f. Write the line notation for the cell.
g. Calculate the theoretical cell voltage using the standard cell potential and the solution concentrations.
Explanation / Answer
(a) Anode: Cu(s) | Cu2+(aq, 0.1 M)
(b) Cathode: Cu2+(aq, 1 M) | Cu(s)
(c) Oxidation half reaction:
Cu(s) => Cu2+(aq, 0.1 M) + 2 e-, Eo(ox) = -0.34 V
(d) Reduction half reaction:
Cu2+(aq, 1 M) + 2 e- => Cu(s), Eo(red) = +0.34 V
(e) OVerall reaction:
Cu2+(aq, 1 M) => Cu2+(aq, 0.1 M)
Eo(cell) = Eo(ox) + Eo(red) = -0.34 + 0.34 = 0.00 V
(f) Cell notation:
Cu(s) | Cu2+(aq, 0.1 M) || Cu2+(aq, 1 M) | Cu(s)
(g) Moles of electrons transferred n = 2
Faraday constant F = 96485 C/mol
Molar gas constant R = 8.314 J/mol.K
Temperature T = 25 deg C = 298.15 K
Nernst equation:
E(cell) = Eo(cell) - RT/nF ln([Cu2+(0.1 M)]/[Cu2+(1 M)])
= 0.00 - 8.314 x 298.15/(2 x 96485) x ln(0.1/1)
= 0.0296 V (or approximately 0.030 V)
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