Question ef8d4

Sep 20, 2015

$E = {E}^{\circ} - \frac{R T}{n F} \ln Q = 0.52 V$

Explanation:

$Z n \left(s\right) + P {b}^{2 +} \left(a q\right) \to Z {n}^{2 +} \left(a q\right) + P b \left(s\right)$

This question could be solved using Nernst Equation:
$E = {E}^{\circ} - \frac{R T}{n F} \ln Q$

where, $E$ is the cell potential at the current conditions,
${E}^{\circ}$ is the cell potential at standard conditions ${E}^{\circ} = 0.63 V$,

$n$ is the number of electrons transferred from oxidation to reduction $n = 2$,
Oxidation: $Z n \left(s\right) \to Z {n}^{2 +} \left(a q\right) + 2 {e}^{-}$
Reduction: $P {b}^{2 +} \left(a q\right) + 2 {e}^{-} \to P b \left(s\right)$

$R$ is the universal gas constant $R = 8.3145 \frac{J}{m o l . K}$

$F$ is Faraday's constant $F = 96485 \frac{C}{m o l {e}^{-}}$

$T$ is the temperature in Kelvin $T = 298 K$

and $Q$ is the reaction quotient $Q = \frac{\left[Z {n}^{2 +} \left(a q\right)\right]}{\left[P {d}^{2 +} \left(a q\right)\right]}$

Therefore, $E = {E}^{\circ} - \frac{R T}{n F} \ln \left(\frac{\left[Z {n}^{2 +} \left(a q\right)\right]}{\left[P {d}^{2 +} \left(a q\right)\right]}\right)$

$E = 0.63 - \frac{8.3145 \times 298}{2 \times 96485} \ln \left(\frac{1.0}{2.0 \times {10}^{- 4}}\right)$

$E = 0.63 - 0.11 = 0.52 V$

I recommend this video about the Nernst Equation:

Sep 20, 2015

${E}_{c e l l} = + 0.52 \text{V}$

Explanation:

The usable form of the Nernst Equation at ${25}^{\circ} \text{C}$ is:

${E}_{c e l l} = {E}^{\circ} - \frac{0.052}{n} \log \text{Q}$

$\text{Q}$ is the reaction quotient.

$n$ is the no. moles electrons transferred. which in, in this case, = 2.

${E}_{c e l l} = 0.63 - \frac{0.592}{2} \log \left(\frac{\left[Z {n}^{2 +}\right]}{\left[P {b}^{2 +}\right]}\right)$

${E}_{c e l l} = 0.63 - \frac{0.0592}{2} \log \left[\frac{1}{2 \times {10}^{- 4}}\right]$

${E}_{c e l l} = 0.63 - \left[\frac{0.0592 \times 3.699}{2}\right]$

${E}_{c e l l} = 0.63 - 0.11 = + 0.52 \text{V}$

Sep 20, 2015

Th cell potential will be equal to $+ \text{0.52 V}$.

Explanation:

Start by taking a look at the redox reaction that is used for the cell

${\text{Zn"_text((s]) + "Pb"_text((aq])^(2+) -> "Zn"_text((aq])^(2+) + "Pb}}_{\textrm{\left(s\right]}}$

Zinc metal is being oxidized to zinc cations and lead cations are being reduced to lead metal.

Before moving forward, you need to determine how many moles of electrons are being transferred in this redox reaction.

$\stackrel{\textcolor{b l u e}{0}}{{\text{Zn"_text((s])) + stackrel(color(blue)(+2))("Pb"_text((aq])^(2+)) -> stackrel(color(blue)(+2))("Zn"_text((aq])^(2+)) + stackrel(color(blue)(0))("Pb}}_{\textrm{\left(s\right]}}}$

Notice that zinc metal's oxidation number chenges from 0 on the reactants' side, to +2 on the products' side, which means that it loses two electrons.

Lead goes from a +2 oxidation state on the reactants' side, to 0 on the products' side, which of course means that it gained two electrons.

This tells you that two moles of electrons are being transferred between the species that is being oxidized and the one that is being reduced.

Now, in order to find the cell's potential, you need to use the Nernst equation - I'll use the version at ${25}^{\circ} \text{C}$, since that is the temperature at which your cell operates

${E}_{\text{cell" = E_"cell"^@ - 0.059/n * logQ_r" }}$, where

${E}_{\text{cell}}$ - the cell potential;
E_'cell"^@ - the standard cell potential
$n$ - the number of moles of electrons transferred in the reaction;
${Q}_{r}$ - the reaction quotient.

In your case, the reaction quotient will be equal to - remember that solids are not included in the expression of the reaction quotient

Q_r = (["Zn"^(2+)])/(["Pb"^(2+)]) = (1.0color(red)(cancel(color(black)("M"))))/(2.0 * 10^(-4)color(red)(cancel(color(black)("M")))) = 5.0 * 10""^3

Therefore, the cell potential will be

E_"cell" = +0.63 - 1/2 * 0.059 * log(5 * 10""^3)

E_"cell" = +0.63 - 1/2 * 0.059 * 3.699 = 0.63 - 0.109 = color(green)(+"0.52 V")#

The answer is rounded to two sig figs.

I recommend this video about the Nernst Equation: