Consider the following cell at 291 K: which has a standard cell potential of 0.0

Question

Consider the following cell at 291 K: which has a standard cell potential of 0.0400 V. What will be the potential of the cell be when [Fe^2+] changes by 0.339 M? E_cell = A concentration cell based on the following half reaction at 320 K has initial concentrations of 1.25 M Zn^2+, 0.395 M Zn^2+, and a potential of 0.01588 V at these conditions. After 9 hours, the new potential of the cell is found to be 0.004138 V. What is the concentration of Zn^2+ at the cathode at this new potential [Zn^2+]_cathode = Consider a galvanic cell based upon the following half reactions: How would the following changes alter the potential of the cell? Adding equal amounts of water to both half reactions. Adding Pb^2+ ions to the lead half reaction (assuming no volume change). Removing Pb^2+ ions from solution by precipitating them out of the lead half reaction (assume no volume change). Adding Ag^+ ions to the Silver half reaction (assume no volume change)

Explanation / Answer

From Nernst Equation

Ecell=EocellRT/nF x lnQ

Ecell=Eocell2.303 x RT/nF x logQ

Q = [C]c·[D]d / [A]a·[B]b

where A, B, C, and D are chemical species; and a, b, c, and d are coefficients in the balanced equation:

a A + b B c C + d D

Q = (0.501-0.339)/1.051 since concentration of the Fe2+ changes by 0.339M from 0.501M

Q = 0.154

RT/nF = (8.3145 J/mol·K)(291 K)/(2)(96485 C/mol)
RT/nF = 0.0125 J/C = 0.0125 V

Ecell=0.042.303 x 0.0125 x log(0.154)

Ecell = 0.04 +0.0234

Ecell = 0.0634 V

2) Ecell=Eocell2.303 x RT/nF x logQ

Eo for this cell is 0

RT/nF = (8.3145 J/mol·K)(320 K)/(2)(96485 C/mol)
RT/nF = 0.01378 J/C = 0.01378 V

Ecell = 0 - 2.303 x 0.01378 x log (0.395/1.25)

Ecell = 0.01588 V

Now the new cenn potential is

Ecell = 0.004138 V

Ecell = 0 - 2.303 x 0.01378 x log (0.395/cathode)

0.004138 V = 0 - 2.303 x 0.01378 x log (0.395/cathode)

0.004138 V = - 2.303 x 0.01378 x log (0.395/cathode)

log (0.395/cathode) = -0.004138 V/0.03173

log (0.395/cathode) = -0.1303

(0.395/cathode) = 10-0.1303

(0.395/cathode) = 0.7406

[cathode] = 0.395/0.7406

concentration at cathode is 0.533 M

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