What's the cell potential?

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The E(cell) of the reaction is 1.10V.

1 Answer
Jun 16, 2017

Answer:

The new cell potential is 1.09 V.

Explanation:

The equations for the cell reactions are

#"Zn" → "Zn"^"2+" + 2"e"^"-"#
#"Cu"^"2+" + 2"e"^"-" → "2Cu"#
#stackrel(————————————)("Zn + Cu"^"2+" → "Zn"^"2+" + "Cu")#

During the reaction, #["Cu"^"2+"]# decreases and #["Zn"^"2+"]# increases.

If #["Zn"^"2+"]# increases by 0.20 mol/L, #["Cu"^"2+"]# decreases by 0.20 mol/L.

The cell reaction becomes

#"Zn + Cu"^"2+""(0.80 mol/L)" → "Zn"^"2+""(1.20 mol/L)" + "Cu"#

The equation for the cell potential is the Nernst equation:

#color(blue)(bar(ul(|color(white)(a/a)E_text(cell) = E_text(cell)^@ - (RT)/(nF)lnQcolor(white)(a/a)|)))" "#

#Q = (["Zn"^"2+"])/(["Cu"^"2+"])#

#lnQ = ln((["Zn"^"2+"])/(["Cu"^"2+"])) = ln((1.20 color(red)(cancel(color(black)("mol/L"))))/(0.80 color(red)(cancel(color(black)("mol/L"))))) = ln(1.50) = 0.405#

#E_text(cell) = "1.10 V" - (8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1"))) × 298.15 color(red)(cancel(color(black)("K"))))/(2 × "96 485" color(red)(cancel(color(black)("J")))·"V"^"-1"·color(red)(cancel(color(black)("mol"^"-1")))) × 0.405 = "1.10 V - 0.0052 V" = "1.09 V"#