What's the cell potential?

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

Jun 16, 2017

The new cell potential is 1.09 V.

Explanation:

The equations for the cell reactions are

$\text{Zn" → "Zn"^"2+" + 2"e"^"-}$
$\text{Cu"^"2+" + 2"e"^"-" → "2Cu}$
stackrel(————————————)("Zn + Cu"^"2+" → "Zn"^"2+" + "Cu")

During the reaction, $\left[\text{Cu"^"2+}\right]$ decreases and $\left[\text{Zn"^"2+}\right]$ increases.

If $\left[\text{Zn"^"2+}\right]$ increases by 0.20 mol/L, $\left[\text{Cu"^"2+}\right]$ decreases by 0.20 mol/L.

The cell reaction becomes

$\text{Zn + Cu"^"2+""(0.80 mol/L)" → "Zn"^"2+""(1.20 mol/L)" + "Cu}$

The equation for the cell potential is the Nernst equation:

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} {E}_{\textrm{c e l l}} = {E}_{\textrm{c e l l}}^{\circ} - \frac{R T}{n F} \ln Q \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

$Q = \left(\left[\text{Zn"^"2+"])/(["Cu"^"2+}\right]\right)$

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

${E}_{\textrm{c e l l}} = \text{1.10 V" - (8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1"))) × 298.15 color(red)(cancel(color(black)("K"))))/(2 × "96 485" color(red)(cancel(color(black)("J")))·"V"^"-1"·color(red)(cancel(color(black)("mol"^"-1")))) × 0.405 = "1.10 V - 0.0052 V" = "1.09 V}$