Question

Calculate the electrode potentials for the following systems:

a) ${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\left(5.00\times {10}^{-3}\mathrm{M}\right),{\mathrm{Cr}}^{3+}\left(2.50\times {10}^{-2}\mathrm{M}\right),{\mathrm{H}}^{+}(0.100\mathrm{M})\mid \mathrm{Pt}$

b)${\mathrm{UO}}_2^{2+}(0.100\mathrm{M}),{\mathrm{U}}^{4+}(0.200\mathrm{M}),{\mathrm{H}}^{+}(0.600\mathrm{M})\mid \mathrm{Pt}$

Short answer

The $Ecell$for the given cell is$+1.20\mathrm{V}$

The $Ecell$is given by localid="1645679127118" $+0.307V$

Step-by-step solution

Concept Introduction

Here we will use the Nernst equation $E_{\text{eell}}=E^0-\frac{0.0592}{n}\log \frac{{\left[a_{\text{prodcuts}}\right]}^m}{{\left[a_{\text{reactants}}\right]}^n}$

Here,$n$is the number of moles of electrons involved in the reaction, ${\left[a_{\text{prodcuts}}\right]}^m$is represents activity concentration of products with its corresponding coefficients, m represent ${\left[a_{\text{reactants}}\right]}^n$and is represents activity concentration of reactants with its corresponding coefficients,

Applying Nernst Formula for (a)

The given cell

${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\left(4.00\times {10}^{-3}\mathrm{M}\right),{\mathrm{Cr}}^{3+}\left(2.00\times {10}^{-2}\mathrm{M}\right),{\mathrm{H}}^{+}(0.100\mathrm{M})/\mathrm{Pt}$

half reaction and standard reduction will be

${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+14{\mathrm{H}}^{+}+6{\mathrm{e}}^{-}\rightleftarrows 2{\mathrm{Cr}}^{3+}+7{\mathrm{H}}_2\mathrm{O};E^0=+1.33\mathrm{V}$

$E_{\text{eell}}=E^0-\frac{0.0592}{n}\log \frac{{\left[{\mathrm{Cr}}^{3+}\right]}^2}{\left[{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\right]{\left[{\mathrm{H}}^{+}\right]}^{14}}$

In this half-reaction the number of moles of electrons involved is 6 . Substitute the given values for concentration and the value of $E_{{\mathrm{Cr}}^2}°$to determine the value of In this half-reaction the number of moles of electrons involved is 6 . Substitute the given values for concentration and the value of $E_{{\mathrm{Cr}}^2}°$to determine the value of

$$\begin{gathered} E_{\text{cell}}=E^0-\frac{0.0592}{n}\log \frac{{\left[{\mathrm{Cr}}^{3+}\right]}^2}{\left[{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\right]{\left[{\mathrm{H}}^{+}\right]}^{14}} \\ =+1.33-\frac{0.0592}{6}\log \left[\frac{{\left(2.00\times {10}^{-2}\right)}^2}{\left(4.00\times {10}^{-3}\right)(0.100{)}^{14}}\right] \\ =+1.20\mathrm{V} \end{gathered}$$

Applying Nernst Formula (b)

The given cell

${\mathrm{UO}}_2^{2+}(0.200\mathrm{M}),{\mathrm{U}}^{4+}(0.100\mathrm{M}),{\mathrm{H}}^{+}(0.500\mathrm{M})/\mathrm{Pt}$

half reaction ad its standard reduction

${\mathrm{UO}}_2^{2+}+4{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\rightleftarrows {\mathrm{U}}^{4+}+2{\mathrm{H}}_2\mathrm{O};E^0=+0.334\mathrm{V}$

The Nernst equation $E=E^0-\frac{0.0592}{n}\log \frac{\left[{\mathrm{U}}^{4+}\right]}{\left[{\mathrm{UO}}_2^{2+}\right]{\left[{\mathrm{H}}^{+}\right]}^4}$

$$\begin{gathered} E_{\mathrm{cell}}=E^0-\frac{0.0592}{n}\log \frac{\left[{\mathrm{U}}^{4+}\right]}{\left[{\mathrm{UO}}_2^{2+}\right]{\left[{\mathrm{H}}^{+}\right]}^4} \\ =+0.334-\frac{0.0592}{6}\log \frac{\left(2.00\times {10}^{-2}\right)}{\left(4.00\times {10}^{-3}\right)(0.100{)}^{14}} \\ =+0.307\mathrm{V} \end{gathered}$$